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FORCE AND NEWTON�S LAWS OF MOTION

INTRODUCTION

Force is a push or pull which tries to change or

successfully changes the state of rest or of uniform

motion of a body, i.e., force is the cause of translatorymotion.

It arises due to interaction of the bodies either due to

contact (e.g., normal reaction, friction, tension, spring

force etc.) or from a distance (e.g., gravitational or

electric force).

FUNDAMENTAL FORCES

All forces observed in nature such as muscular force,

tension, reaction, friction, weight, electric, magnetic,

nuclear, etc., can be explained in terms of only following

four basic interactions.

(a) Gravitational Force :

The force of interaction which exists between two

particles of masses m1 and m

2, due to their masses is

called gravitational force. The gravitational force acts

over long distances and does not need, any intervening

medium. Gravitational force is the weakest force of

nature.

(b) Electromagnetic Force :

Force exerted by one particle on the other because of

the electric charge on the particles is called

electromagnetic force. Following are the main

characteristics of electromagnetic force

(i) These can be attractive or repulsive.

(ii) These are long range forces.

(iii) These depend on the nature of medium between

the charged particles.

(iv) All macroscopic forces (except gravitational) which

we experience as push or pull or by contact are

electromagnetic, i.e., tension in a rope, the force

of friction, normal reaction, muscular force, and

force experienced by a deformed spring are

electromagnetic forces. These are manifestations

of the electromagnetic attractions and repulsions

between atoms/molecules.

(c) Nuclear Force :

It is the strongest force. It keeps nucleons (neutrons

and protons) together inside the nucleus inspite of

large electric repulsion between protons. Radioactivity,

fission, and fusion, etc. results because of unbalancing

of nuclear forces. It acts within the nucleus that too

upto a very small distance. It does not depends on

charge and acts equally between a proton and proton,

a neutron and neutron, and proton and neutron,

electrons does not experience this force. It acts for very

short distance order of 10�15 m.

(d) Weak Force :

It acts between any two elementary particles. Under its

action a neutron can change into a proton emitting an

electron and a particle called antineutrino. The range

of weak force is very small, in fact much smaller than

size of a proton or a neutron.

It has been found that for two protons at a distance of 1

fermi :

FN:F

EM:F

W:F

G::1:10�2:10�7:10�38

On the basis of contact forces are classified into two

categories

(i) Contact forces

(ii) Non contact or field forces

(a) Contact force :

Forces which are transmitted between bodies by short

range atomic molecular interactions are called contact

forces. When two objects come in contact they exert

contact forces on each other. e.g. Normal, Tension etc.

(b) Field force :

Force which acts on an object at a distance by the

interaction of the object with the field produced by other

object is called field force. e.g. Gravitational force,

Electro magnetic force etc.

DETAILED ANALYSIS OF CONTACT FORCE

(a) Normal force (N) :

It is the component of contact force perpendicular to

the surface. It measures how strongly the surfaces in

contact are pressed against each other. It is the

electromagnetic force.

e.g.1 A table is placed on Earth as shown in figure


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Here table presses the earth so normal force exerted

by four legs of table on earth are as shown in figure.

e.g.2 A boy pushes a block kept on a frictionlesssurface.

Here, force exerted by boy on block is electromagneticinteraction which arises due to similar chargesappearing on finger and contact surface of block, it isnormal force.

A block is kept on inclined surface. Component of itsweight presses the surface perpendicularly due towhich contact force acts between surface and block.

Normal force exerted by block on the surface of inclinedplane is shown in figure. Here normal force is acomponent of weight of the body perpendicular to theinclined surface i.e. N = mgsin

Force acts perpendicular to the surface

1. Two blocks are kept in contact on a smooth surface asshown in figure. Draw normal force exerted byA on B.

Sol. In above problem, block A does not push block B, sothere is no molecular interaction between A and B.Hence normal force exerted by A on B is zero.

Note :

� Normal is a dependent force it comes in role whenone surface presses the other.

(b) Tension :

Tension is the magnitude of pulling force exerted by astring, cable, chain, rope etc. When a string isconnected to a body and pulled out, the string said tobe under tension. It pulls the body with a force T, whosedirection is away from the body and along the length ofthe string. Usually strings are regarded to be masslessand unstretchable, known as ideal string.

Note : (i) Tension in a string is an electromagneticforce and it arises only when string is pulled. If amassless string is not pulled, tension in it is zero.(ii) String can not push a body in direct contact.

(c) Force Exerted by spring :

A spring is made of a coiled metallic wire having adefinite length. When it is neither pushed nor pulledthen its length is called natural length.At natural length the spring does not exert any force onthe objects attached to its ends.f the spring is pulledat the ends, its length becomes larger than its naturallength, it is known as stretched or extended spring.Extended spring pulls objects attached to its ends.

A B

A B

A B

Normal spring

Stretched spring

Compressed spring

Spring force on A Spring force on B

Spring force on A Spring force on B

If the spring is pushed at the ends, its length becomesless than natural length. It is known as compressedspring. A compressed spring pushes the objectsattached to its ends.

F

x

FF

x

F

F = 0 spring in naturallength does not exertsany force on its ends

F = � kx

x = compression in spring

F = � kx ;k = spring

constant or stiffness constant (unit = N/m)x = extension in spring

Fext

Fext

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Note : Spring force is also electromagnetic in nature :

(d) Friction force :

When a body is moving on a rough surface resistance

to the motion occurs because of the interaction

between the body and its surroundings. We call such

resistance as force of friction. Friction is also

considered as component of contact force which acts

parallel to the surfaces in contact.

(i) Origin of friction : The frictional force arises due to

molecular interactions between the surfaces at the

points of actual contact. When two bodies are placed

one over other, the actual area of contact is much

smaller then the total surface areas of bodies. The

molecular forces starts operating at the actual points

of contact of the surfaces. Molecular bonds are formed

at these contact points. When one body is pulled over

the other, these bonds are broken, and the material

get deformed and new bonds are formed. The local

deformation sends vibrations into the bodies. These

Vibrations ultimately dumps out and energy of

vibrations appears as heat. Hence to start or carry on

the motion, there is a need of force.

Body 1

Body 2

Actual area of contact

(ii) Statics and Kinetic Frictions :

� Experiment :

(A) Consider a block placed on a table, and a small

force F1 is acted on it. The block does not move. It

indicates that the frictional force fs starts acting in

opposite direction of applied force and its magnitude

is equal of F1(figure b). That is for the equilibrium of

the block, we have

F1 � f

s = 0 or F

1 = f

s

The force of friction when body is in state of rest over

the surface is called static friction (fs).

(B) As the applied force increases the frictional force

also increases. When the applied force is increased

up to a certain limit (F2) such that the block is on the

verge of motion. The value of frictional force at this

stage is called limiting friction flim

(figure c).

(C) Once the motion started, the smaller force is nownecessary to continue the motion (F

3) and thus

frictional force decreases. The force of frictionwhen body is in state of motion over the surface iscalled kinetic or dynamic friction f

k (figure d).

(iii) More about frictional force :

(A) About static friction

1. The limiting friction depends on the materialsof the surfaces in contact and their state ofpolish.

2. The magnitude of static friction isindependent of the apparent area of contactso long as the normal reaction remains thesame.

3. The limiting friction is directly proportional tothe magnitude of the normal reactionbetween the two surfaces i.e. f

lim=

SN. Here

s is coefficient of static friction.

We can write, s =

Nflim

(B) About kinetic friction :

1. The kinetic friction depends on the materialsof the surface in contact.

2. It is also independent of apparent area ofcontact as long as the magnitude of normalreaction remains the same.

3. Kinetic friction is almost independent of thevelocity, provided the velocity is not too largenot too small.

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4. The kinetic friction is directly proportional to

the magnitude of the normal reaction

between the surfaces.

fk =

k N. Here

k is coefficient of kinetic friction.

We can write, k =

Nfk

� There are two types of kinetic frictions:

(i) Sliding friction : The force of friction when one

body slides over the surface of the another body is

called sliding friction.

(ii) Rolling friction : When a wheel rolls without

slipping over a horizontal surface, there is no

relative motion of the point of contact of the wheel

with respect to the plane. Theoretically for a rolling

wheel the frictional force is zero. This can only

possible when bodies in contact are perfectly rigid

and contact of wheel with the surface is made

only at a point. But in practice no material body is

perfectly rigid and therefore bodies get deformed

when they pressed each other. The actual area of

their contact no longer remains a point, and thus

a small amount of friction starts acting between

the body and the surface. Here frictional force is

called rolling friction. It is clear from above

discussion that rolling friction is very much smaller

than sliding friction.

flim > fkinetic > frolling.

Note : s and

k are dimensionless quantities and

independent of shape and area of contact . It is a

property of two contact surfaces. s

will always be

greater than k .Theoretical value of can be o to but

practical value is 0 < 1.6

(a) Conservative Force :

A force is said to be conservative if the amount of work

done in moving an object against that force is

independent on the path. One important example of

conservative force is the gravitational force. It means

that amount of work done in moving a body against

gravity from location A to location B is the same

whichever path we may follow in going from A to B. This

is illustrated in figure.

A force is conservative if the total work done by the

force on an object in one complete round is zero,

i.e. when the object moves around any closed path

(returning to its initial position).

A force is conservative if there is no change in kinetic

energy in one complete round. KE = 0

This definition illuminates an important aspect of a

conservative force viz. Work done by a conservative

force is recoverable. Thus in figure, we shall have to

do mgh amount of work in taking the body from A to B.

However, when body is released from B, we recover

mgh of work.

Other examples of conservative forces are spring force,

electrostatic force etc.

(b) Non-Conservative Force :

A force is non-conservative if the work done by that force

on a particle moving between two points depends on

the path taken between the points.

The force of friction is an example of non-conservative

force. Let us illustrate this with an instructive example.

Suppose we were to displace a book between two points

on a rough horizontal surface (such as a table). If the

book is displaced in a straight line between the two

points, the work done by friction is simply FS where :

F = force of friction ;

S = distance between the points.

However, if the book is moved along any other path

between the two points (such as a semicircular path),

the work done by friction would be greater than FS.

Finally, if the book is moved through any closed path,

the work done by friction is never zero, it is always

negative. Thus the work done by a non-conservative

force is not recoverable, as it is for a conservative force.

SYSTEM

Two or more than two objects which interact with each

other form a system.

Classification of forces on the basis of boundary of

system :

(a) Internal Forces : Forces acting with in a system

among its constituents.

(b) External Forces : Forces exerted on the

constituents of a system by the outside

surroundings are called as external forces.

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FREE BODY DIAGRAM

A free body diagram consists of a diagrammaticrepresentations of single body or a subsystem ofbodies isolated from surroundings showing all theforces acting on it.

Steps for F.B.D.

Step 1 : Identify the object or system and isolate it from

other objects, clearly specify its boundary.

Step 2 : First draw non-contact external force in the

diagram, generally it is weight.

Step 3 : Draw contact forces which acts at the boundary

of the object of system. Contact forces are normal ,

friction, tension and applied force. In F.B.D, internal

forces are not drawn only external are drawn.

2. A block of mass �m� is kept on the ground as shown in

figure.

(i) Draw F.B.D. of block.

(ii) Are forces acting on block forms action- reactionpair.

(iii) If answer is no, draw action reaction pair.

Sol.(i) F.B.D. of block

(ii) �N� and mg are not action -reaction pair. Since pair

act on different bodies, and they are of same

nature.

(iii) Pair of �mg� of block acts on earth in opposite

direction.

mg earth

and pair of �N� acts on surface as shown in figure.

N

3. Two sphere A and B are placed between two verticalwalls as shown in figure. Draw the free body diagramsof both the spheres.

A

B

Sol.F.B.D. of sphere �A� :

F.B.D. of sphere �B� :

(exerted by A)

Note : Here NAB

and NBA

are the action - reaction pair

(Newton�s third law).

4. Draw F.B.D. for systems shown in figure below.

Sol.

TRANSLATORY EQUILIBRIUM

When several forces acts on a body simultaneously insuch a way that resultant force on the body is zero, i.e.,

F

= 0 with F

=

iF the body is said to be in translatory

equilibrium. Here it is worthy to note that :

(i) As if a vector is zero all its components must vanishi.e. in equilibrium as -

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F

= 0 with F

=

iF = 0

xF = 0 ; y

F = 0 ; z

F = 0

So in equilibrium forces along x axes must balanceeach other and the same is true for other directions.If a body is in translatory equilibrium it will be either atrest or in uniform motion. If it is at rest, equilibrium iscalled static, otherwise dynamic.

Static equilibrium can be divided into following threetypes :

(a) Stable equilibrium :

If on slight displacement from equilibrium position abody has a tendency to regain its original position it issaid to be in stable equilibrium. In case of stableequilibrium potential energy is minimum and so centerof gravity is lowest.

O

(b) Unstable equilibrium : If on slight displacementfrom equilibrium position a body moves in the directionof displacement, the equilibrium is said to be unstable.In this situation potential energy of body is maximumand so center of gravity is highest.

O

(c) Neutral equilibrium : If on slight displacement fromequilibrium position a body has no tendency to comeback to its original position or to move in the directionof displacement, it is said to be in neutral equilibrium.In this situation potential energy of body is constantand so center of gravity remains at constant height.

.

(a) Newtons 2nd law of motion :

The rate of change of linear momentum of a body isdirectly proportional to the applied force and the changetakes place in the direction of the applied force.

In relation

F =

ma the force

F stands for the net

external force. Any internal force in the system is not to

be included in

F .

In S.I. the absolute unit of force is newton (N) andgravitational unit of force is kilogram weight or kilogramforce (kgf.)

Note : The absolute unit of force remains the sameeverywhere, but the gravitational unit of force variesfrom place to place because it depends on the value of g.

(b) Applications of Newton�s 2nd Law

(i) When objects are in equilibrium :Steps to solve problem involving objects inequilibrium :

Step 1 : Make a sketch of the problem.

Step 2 : Isolate a single object and then draw the free-

body diagram for the object. Label all external forces

acting on it.

Step 3 : Choose a convenient coordinate system and

resolve all forces into rectangular components along x

and Y direction.

Step 4 : Apply the equations 0Fx and 0Fy .

Step 5 : Step 4 will give you two equations with several

unknown quantities. If you have only two unknown

quantities at this point, you can solve the two equations

for those unknown quantities.

Step 6 : If step 5 produces two equations with more

than two unknowns, go back to step 2 and select

another object and repeat these steps. Eventually at

step 5 you will have enough equations to solve for all

unknown quantities.

5. A �block� of mass 10 kg is

suspended with string asshown in figure. Find tension in the string.(g = 10 m/s2).

Sol.F.B.D. of blockFor equilibrium of block along Y axis

0Fy

T � 10 g = 0 T = 100 N

6. The system shown in figure is in equilibrium. Find themagnitude of tension in each string ; T

1 , T

2, T

3 and T

4.

(g = 10 m/s2).

Sol.F.B.D. of 10 kg block

For equilibrium of block along Y axis.

0Fy

T0

10g

T0 = 10 g

T0 = 100 N

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F.B.D. of point �A�

0Fy 30º

T2

T1x

T0

A

y

T2 cos 30º = T

0 = 100 N

T2 =

3

200N

0Fx

T1 = T

2 . sin 30º

= 3

200.

21

= 3

100N.

F.B.D. of point of �B�

60ºT4

T3x

B

y

30º

T2

yF = 0 T4 cos 60º = T

2 cos 30º

T4 = 200 N

and xF = 0 T3 + T

2 sin30º = T

4 sin 60º

T3 =

3

200N

7. Two blocks are kept in contact as shown in figure. Find :-

(a) forces exerted by surfaces (floor and wall) on

blocks.

(b) contact force between two blocks.

SolA : F.B.D. of 10 kg block

N1 = 10 g = 100 N .......(1)

N2 = 100 N .........(2)

F.B.D. of 20 kg block

N2 = 50 sin 30º + N

3

N3 = 100 � 25 = 75 N

& N4 = 50 cos 30º + 20 g

N4 = 243.30 N

8. Find magnitude of force exerted by string on pulley.

Sol B. F.B.D. of 10 kg block :

T = 10 g = 100 NF.B.D. of pulley :

Since string is massless, so tension in both sides

of string is same.

So magnitude of force exerted by string on pulley

= 22 100100 = 100 2 N

Note : Since pulley is in equilibrium position, so net

forces on it is zero.

(ii) Accelerating Objects :

Steps to solve problems involving objects that are in

accelerated motion :

Step 1 : Make a sketch of the problem.

Step 2 : Isolate a single object and then draw the free

- body diagram for that object. Label all external forces

acting on it. Be sure to include all the forces acting on

the chosen body, but be equally careful not to include

any force exerted by the body on some other body.

Some of the forces may be unknown , label them with

algebraic symbols.

Step 3 : Choose a convenient coordinate system, show

location of coordinate axis explicitly in the free - body

diagram, and then determine components of forces

with reference to these axis and resolve all forces into

x and y components.

Step 4 : Apply the equations xF = max & yF = ma

y.

Step 5 : Step 4 will give two equations with several

unknown quantities. If you have only two unknown

quantities at this point, you can solve the two equations

for those unknown quantities.

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PAGE # 8

Step 6 : If step 5 produces two equations with more

than two unknowns, go back to step 2 and select

another object and repeat these steps. Eventually at

step 5 you will have enough equations to solve for all

unknown quantities.

9. A force F is applied horizontally on mass m1 as shown

in figure. Find the contact force between m1 and m

2.

Sol.Considering both blocks as a system to find the

common acceleration.

Common acceleration

a = 21 mmF

.......(1)

m1 m2F a

To find the contact force between �A� and �B� we draw

F.B.D. of mass m2.

F.B.D. of mass m2

xF = max

N = m2 . a

N = 21

2

mmFm

21 mmF

acesin

10. A 5 kg block has a rope of mass 2 kg attached to its

underside and a 3 kg block is suspended from the other

end of the rope. The whole system is accelerated

upward at 2 m/s2 by an external force F0.

(a) What is F0 ?

(b) What is the net force on rope ?

(c) What is the tension at middle point of the rope ?

(g = 10 m/s2)

Sol.For calculating the value of F0.

F.B.D of whole system

(a) 2m/s2

F0

10 g = 100 N

F0 �100 = 10 × 2

F0 = 120 N ........(1)

(b) According to Newton�s second law, net force on

rope.

F = ma = 2 × 2

= 4N ............(2)

(c) For calculating tension at the middle point we draw

F.B.D. of 3 kg block with half of the rope (mass 1

kg) as shown.

T � 4 g = 4 . 2

T = 48 N

11. A block of mass 50 kg is kept on another block of mass

1 kg as shown in figure. A horizontal force of 10 N is

applied on the 1Kg block. (All surface are smooth).

Find : (g = 10 m/s2)

(a) Acceleration of blocks A and B.

(b) Force exerted by B on A.

50 kg 1 kg A

B

Sol.(a) F.B.D. of 50 kg

N2 = 50 g = 500 N

along horizontal direction, there is no force aB = 0

(b) F.B.D. of 1 kg block :

N1 N2

10 N

1g

along horizontal direction

10 = 1 aA.

aA

= 10 m/s2

along vertical direction

N1 = N

2 + 1g

= 500 + 10 = 510 N

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PAGE # 9

12. One end of string which passes through pulley and

connected to 10 kg mass at other end is pulled by 100

N force. Find out the acceleration of 10 kg mass. (g

=9.8 m/s2)

Sol.Since string is pulled by 100 N force. So tension in the

string is 100 N

F.B.D. of 10 kg block

100 � 10 g = 10 a

100 � 10 × 9.8 = 10 a

a = 0.2 m/s2.

WEIGHING MACHINE

A weighing machine does not measure the weight but

measures the force exerted by object on its upper

surface.

13. A man of mass 60 Kg is

standing on a weighing

machine placed on ground.

Calculate the reading of

machine (g = 10 m/s2). weighing machine

Sol.For calculating the reading of weighing machine, we

draw F.B.D. of man and machine separately.

F.B.D of man

N

Mg

N = Mg

F.B.D of man taking mass of man as M

weighing machine N

N1Mg

F.B.D. of weighing machine

Here force exerted by object on upper surface is N

Reading of weighing machine

N = Mg

= 60 × 10

N = 600 N.

SPRING BALANCE

It does not measure the weight. It measures the force

exerted by the object at the hook. Symbolically, it is

represented as shown in figure.

A block of mass �m� is suspended at hook. When spring

balance is in equilibrium, we draw the F.B.D. of mass

m for calculating the reading of balance.

m

spring balance

hook

F.B.D. of �m�.

mg � T = 0

T = mgMagnitude of T gives the reading of spring balance.

14. A block of mass 20 kg is suspended through two lightspring balances as shown in figure . Calculate the :

(1) reading of spring balance (1).

(2) reading of spring balance (2).

Sol.For calculating the reading, first we draw F.B.D.of 20 kg

block.

F.B.D. 20 kg

T

20 g

mg � T = 0

T = 20 g = 200 N

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PAGE # 10

Since both the balances are light so, both the scaleswill read 200 N.

15. (i) A 10 kg block is supported by a cord that runs to aspring scale, which is supported by another cordfrom the ceiling figure (a). What is the reading onthe scale ?

(ii) In figure (b) the block is supported by a cord thatruns around a pulley and to a scale. The oppositeend of the scale is attached by cord to a wall. Whatis the reading of the scale.

(iii) In figure (c) the wall has been replaced with asecond 10 kg block on the left, and the assembly isstationary. What is the reading on the scale now ?

spring balance

hook

10 kg

T

T

(a)

T

TT

(b)

10kg

T

TT

T

10kg10kg

(c)

Sol. In all the three cases the spring balance reads 10 kg.To understand this let us cut a section inside the springas shown;

As each part of the spring is at rest, so F= T. As theblock is stationary, so T= 10g = 100N.

16. Pull is easier than push

Push : Consider a block of mass m placed onrough horizontal surface. The coefficient of staticfriction between the block and surface is . Let apush force F is applied at an angle with thehorizontal.

As the block is in equilibrium along y-axis, so we have

;0Fy

or N = mg + F sin

To just move the block along x-axis, we have

F cos = N = (mg + F sin )

or F =

sin�cosmg

.......(i)

Pull : Along y-axis we have ;

;0Fy

N = mg � F sin

To just move the block along x-axis, we have

F cos = N = (mg � F sin )

or F =

sincosmg

. .......(ii)

It is clear from above discussion that pull force is

smaller than push force.

17. Discuss the direction of friction in the following cases :

(i) A man walks slowly, without change in speed.

(ii) A man is going with increasing speed.

(iii) When cycle is gaining speed.

(iv) When cycle is slowing down .

Sol. (i) Consider a man walks slowly without acceleration,

and both the legs are touching the ground as

shown in figure (a). The frictional force on rear leg

is in forward direction and on front leg will be on

backward direction of motion.

As a = 0,

Fnet

= 0 or f1 � f

2 = 0

f1 = f

2& N

1 = N

2.

N1 N2

(b)

Ground

f1 f2f1 f2

N1 N2

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PAGE # 11

(ii) When man is gaining the speed : The frictionalforce on rear leg f

1 will be greater than frictional

force on front leg f2 (fig. b).

acceleration of the man, a = m

ff 21 .

(iii) When cycle is gaining speed : In this case torque

is applied on the rear wheel of the cycle by the

chain-gear system. Because of this the slipping

tendency of the point of contact of the rear wheel is

backward and so friction acts in forward direction.

The slipping tendency of point of contact of front

wheel is forward and so friction acts in backward

direction. If f1 and f

2 are the frictional forces on rear

and front wheel, then acceleration of the cycle a =

Mf�f 21 , where M is the mass of the cycle together

with rider (fig. a).

N1 N2

f1 f2

(a)

N1 N2

f1 f2

(b)

(iv) When cycle is slowing down : When torque is not

applied (cycle stops pedaling), the slipping

tendency of points of contact of both the wheels

are forward, and so friction acts in backward

direction (fig. b). If f1 and f

2 are the frictional forces

on rear and front wheel, then retardation

a = M

ff 21

18. A block of mass 25 kg is raised by a 50 kg man in two

different ways as shown in fig.. What is the action on

the floor by the man in the two cases ? If the floor yields

to a normal force of 700 N, which mode should the

man adopt to lift the block without the floor yielding.

50g

50g

Sol. The FBD for the two cases are shown in figure.

In Ist case, let the force exerted by the man on the floor is

N1. Consider the forces inside the dotted box, we have

N1 = T + 50 g.

Block is to be raised without acceleration, so

T = 25 g.

N1 = 25 g + 50 g

= 75 g = 75 × 9.8 = 735 N

In IInd case, let the force exerted by the man on the floor

in N2 . Consider the forces inside the dotted box, we

have

N2 = 50 g � T

and T = 25 g

N2 = 50 g � 25 g

= 25 g = 25 × 9.8 = 245 N.

As the floor yields to a downward force of 700 N, so the

man should adopt mode .

19. Figure shows a weighing machine kept in a lift is

moving upwards with acceleration of 5 m/s2. A block is

kept on the weighing machine. Upper surface of block

is attached with a spring balance. Reading shown by

weighing machine and spring balance is 15 kg and 45

kg respectively.

Answer the following questions. Assume that theweighing machine can measure weight by havingnegligible deformation due to block, while the springbalance requires larger expansion. (take g = 10 m/s2)

(i) Find the mass of the object in kg and the normalforce acting on the block due to weighing machine?

(ii) Find the acceleration of the lift such that weighingmachine shows its true weight ?

Sol. (i)

T + N � Mg = Ma

45 g + 15 g = M(g + a)450 + 150 = M(10 + 5)M = 40 kgNormal force is the reaction applied by weighingmachine i.e. 15 × 10 = 150 N.

121212121212

PAGE # 12

(ii)

T + N � Mg = Ma

45 g + 40 g = 40(g + a)450 + 400 = 400 +40 a

a = 40

450=

445

m/s2

EXERCISE

Normal Force :

1. Two blocks are in contact on a frictionless table. Onehas mass m and the other 2m.A force F is applied on2m as shown in the figure. Now the same force F isapplied from the right on m. In the two casesrespectively, the ratio of force of contact between thetwo blocks will be :

(A) Same (B) 1 : 2(C) 2 : 1 (D) 1 : 3

2. Two forces of 6N and 3N are acting on the two blocks of2kg and 1kg kept on frictionless floor. What is the forceexerted on 2kg block by 1kg block ?:

2kg 1kg

6N3N

(A)1N (B) 2N(C) 4N (D) 5N

3. There are two forces on the 2.0 kg box in the overheadview of figure but only one is shown. The second forceis nearly :

y

30º

F = 20 N1

a = 12 m/s2

x

(A) �20 j�N (B) � 20 i� + 20 j� N

(C) �32 i� � 12 3 j�N (D) �21 i� � 16 j�N

4. A dish of mass 10 g is kept horizontally in air by firing

bullets of mass 5 g each at the rate of 100 per second.

If the bullets rebound with the same speed, what is the

velocity with which the bullets are fired :

(A) 0.49 m/s (B) 0.098 m/s

(C) 1.47 m/s (D) 1.96 m/s

5. A block of metal weighing 2 kg is resting on a

frictionless plank. If struck by a jet releasing water at a

rate of 1 kg/s and at a speed of 5 m/s. The initial

acceleration of the block will be :

(A) 2.5 m/s2 (B) 5.0 m/s2

(C) 10 m /s2 (D) none of the above

6. A constant force F is applied in horizontal direction as

shown. Contact force between M and m is N and

between m and M� is N� then

(A) N= N� (B) N > N�

(C) N�> N

(D) cannot be determined

� ASSERTION / REASON

7. STATEMENT-1 : Block A is moving on horizontal surface

towards right under action of force. All surface are

smooth. At the instant shown the force exerted by block

A on block B is equal to net force on block B.

STATEMENT-2 : From Newtons�s third law, the force

exerted by block A on B is equal in magnitude to force

exerted block B on A

(A) statement-1 is true, Statement 2 is true, statement-2

is correct explanation for statement-1.

(B) statement-1 is true, Statement 2 is true, statement-2

is NOT a correct explanation for statement-1.

(C) statement-1 is true, Statement 2 is false

(D) statement-1 is False, Statement 2 is True

8. A certain force applied to a body A gives it an acceleration

of 10 ms�2 . The same force applied to body B gives it

an acceleration of 15 ms�2 . If the two bodies are joined

together and same force is applied to the combination,

the acceleration will be :

(IJSO/Stage-I/2011)

(A) 6 ms�2 (B) 25 ms�2

(C) 12.5 ms�2 (D) 9 ms�2

131313131313

PAGE # 13

9. Four blocks are kept in a row on a smooth horizontaltable with their centres of mass collinear as shown inthe figure. An external force of 60 N is applied from lefton the 7 kg block to push all of them along the table.The forces exerted by them are :(IAO/Sr./Stage-I/2008)

7 kg 5 kg 2 kg 1 kg60N

P Q R S

(A) 32 N by P on Q (B) 28 N by Q on P(C) 12 N by Q on R (D) 4 N by S on R

Tension :

10. A mass M is suspended by a rope from a rigid supportat A as shown in figure. Another rope is tied at the endB, and it is pulled horizontally with a force F. If the ropeAB makes an angle with the vertical inequilibrium,then the tension in the string AB is :

(A) F sin (B) F /sin (C) F cos (D) F / cos

11. In the system shown in the figure, the acceleration ofthe 1kg mass and the tension in the string connectingbetween A and B is :

(A) g4

downward, 8g7

(B) g4

upward, g7

(C) g7

downward, 67

g (D) g2

upward, g

12. A body of mass 8 kg is hanging from another body ofmass 12 kg. The combination is being pulled by astring with an acceleration of 2.2 m s�2. The tension T

1

and T2 will be respectively :(Use g =9.8 m/s2)

(A) 200 N, 80 N (B) 220 N, 90 N(C) 240 N, 96 N (D) 260 N, 96 N

13. Two masses M1 and M

2 are attached to the ends of a

light string which passes over a massless pulleyattached to the top of a double inclined smooth plane

of angles of inclination and . If M2 > M

1 then the

acceleration of block M2 down the inclined will be :

(A) 2

1 2

M (sin )

M M

g (B)

1

1 2

M g(sin )

M M

(C)

21

12

MMsinMsinM

g (D) Zero

14. Three masses of 1 kg, 6 kg and 3 kg are connected to

each other by threads and are placed on table as

shown in figure. What is the acceleration with which

the system is moving ? Take g = 10 m s�2:

(A) Zero (B) 1 ms�2

(C) 2 m s�2 (D) 3 m s�2

15. The pulley arrangements shown in figure are identical

the mass of the rope being negligible. In case I, the

mass m is lifted by attaching a mass 2m to the other

end of the rope. In case II, the mass m is lifted by

pulling the other end of the rope with a constant

downward force F= 2 mg, where g is acceleration due

to gravity. The acceleration of mass in case I is :

(A) Zero

(B) More than that in case II

(C) Less than that in case II

(D) Equal to that in case II

141414141414

PAGE # 14

16. A 50 kg person stands on a 25 kg platform. He pullsmassless rope which is attached to the platform viathe frictionless, massless pulleys as shown in thefigure. The platform moves upwards at a steady velocityif the force with which the person pulls the rope is :

(A) 500 N (B) 250 N(C) 25 N (D) 50 N

17. Figure shows four blocks that are being pulled along a

smooth horizontal surface. The mssses of the blocks

and tension in one cord are given. The pulling force F is :

4kg 3kg 2kg 1kg60º

F30N

(A) 50 N (B) 100 N

(C) 125 N (D) 200 N

18. A10 kg monkey climbs up a massless rope that runs

over a frictionless tree limb and back down to a 15 kg

package on the ground. The magnitude of the least

acceleration the monkey must have if it is to lift the

package off the ground is :

(A) 4.9 m/s2 (B) 5.5 m/s2

(C) 9.8 m/s2 (D) none of these

19. Two blocks, each of mass M, are connected by a

massless string, which passes over a smooth

massless pulley. Forces F

act on the blocks as shown.

The tension in the string is :

(A) Mg (B) 2 Mg

(C) Mg + F (D) none of these

20. Two blocks of mass m each is connected with the

string which passes over fixed pulley, as shown in figure.

The force exerted by the string on the pulley P is :

(A) mg (B) 2 mg

(C) 2 mg (D) 4 mg

21. One end of a massless rope, which passes over a

massless and frictionless pulley P is tied to a hook C

while the other end is free. Maximum tension that rope

can bear is 360 N, with what minimum safe

acceleration (in m/s2) can a monkey of 60 kg move

down on the rope :

P

C

(A) 16 (B) 6

(C) 4 (D) 8

22. Which figure represents the correct F.B.D. of rod of

mass m as shown in figure :

(A) (B)

(C) (D) None of these

23. Two persons are holding a rope of negligible weight

tightly at its ends so that it is horizontal. A 15 kg weight

is attached to the rope at the mid point which now no

longer remains horizontal. The minimum tension

required to completely straighten the rope is :

(A) 15 kg

(B) kg2

15

(C) 5 kg

(D) Infinitely large (or not possible)

151515151515

PAGE # 15

24. In the figure, the blocks A, B and C of mass each have

acceleration a1 , a

2 and a

3 respectively . F

1 and F

2 are

external forces of magnitudes 2 mg and mg

respectively then which of the following relations is

correct :

(A) a1 = a

2 = a

3(B) a

1 > a

2 > a

3

(C) a1 = a

2 , a

2 > a

3(D) a

1 > a

2 , a

2= a

3

25. A weight is supported by two strings 1.3 and 2.0 m

long fastened to two points on a horizontal beam 2.0

m apart. The depth of this weight below the beam is :

(IAO/Jr./Stage-I/2007)

(A) 1.0 m (B) 1.23 m

(C) 0.77 m (D) 0.89 m

26. A fully loaded elevator has a mass of 6000 kg. Thetension in the cable as the elevator is accelerateddownward with an acceleration of 2ms�2 is (Take g = I0ms�2) (KVPY/2007)

(A) 7·2 × 104 N (B) 4.8 × 104 N(C) 6 × 104 N (D) 1.2 × 104 N

27. A light string goes over a frictionless pulley. At its oneend hangs a mass of 2 kg and at the other end hangsa mass of 6 kg. Both the masses are supported byhands to keep them at rest. When the masses arereleased, they being to move and the string gets taut.(Take g = 10 ms�2) The tension in the string during themotion of the masses is : (KVPY/2008)(A) 60 N (B) 30 N(C) 20 N (D) 40 N

Force Exerted by Spring :

28. In the given figure. What is the reading of the spring

balance:

(A) 10 N (B) 20 N

(C) 5 N (D) Zero

29. Two bodies of masses M1 and M

2 are connected to

each other through a light spring as shown in figure. If

we push mass M1 with force F and cause acceleration

a1 in mass M

1 what will be the acceleration in M

2 ?

(A) F/M2

(B) F/(M1 + M

2)

(C) a1

(D) (F�M1a

1)/M

2

30. A spring balance is attached to 2 kg trolley and is used

to pull the trolly along a flat surface as shown in the fig.

The reading on the spring balance remains at 10 kg

during the motion. The acceleration of the trolly is (Use

g= 9.8 m�2) :

(A) 4.9 ms�2 (B) 9.8 ms�2

(C) 49 ms�2 (D) 98 ms�2

31. A body of mass 32 kg is suspended by a spring balance

from the roof of a vertically operating lift and going

downward from rest. At the instants the lift has covered

20 m and 50 m, the spring balance showed 30 kg & 36

kg respectively. The velocity of the lift is :

(A) Decreasing at 20 m & increasing at 50 m

(B) Increasing at 20 m & decreasing at 50 m

(C) Continuously decreasing at a constant rate

throughout the journey

(D) Continuously increasing at constant rate throughout

the journey

Friction Force :

32. A ship of mass 3 × 107 kg initially at rest is pulled by a

force of 5 × 104 N through a distance of 3m. Assume

that the resistance due to water is negligible, the speed

of the ship is :

(A) 1.5 m/s (B) 60 m/s

(C) 0.1 m/s (D) 5 m/s

33. When a horse pulls a cart, the force needed to move

the horse in forward direction is the force exerted by :

(A) The cart on the horse

(B) The ground on the horse

(C) The ground on the cart

(D) The horse on the ground

161616161616

PAGE # 16

34. A 2.5 kg block is initially at rest on a horizontal surface.

A 6.0 N horizontal force and a vertical force P

are appliedto the block as shown in figure. The coefficient of staticfriction for the block and surface is 0.4. The magnitudeof friction force when P = 9N : (g = 10 m/s2)

(A) 6.0 N (B) 6.4 N(C) 9.0 N (D) zero

35. The upper half of an inclined plane with inclination isperfectly smooth while the lower half is rough. A bodystarting from rest at the top will again come to rest at thebottom, if the coefficient of friction for the lower half is :(A) 2 tan (B) tan

(C) 2 sin (D) 2 cos

36. Minimum force required to pull the lower block is (takeg = 10 m/s2) :

(A) 1 N (B) 5 N(C) 7 N (D) 10 N

37. N bullets each of mass m are fired with a velocity v m/s at the rate of n bullets per sec., upon a wall. If thebullets are completely stopped by the wall, the reactionoffered by the wall to the bullets is :(A) N m v / n (B) n m v(C) n N v / m (D) n v m / N

38. A vehicle of mass m is moving on a rough horizontalroad with momentum P. If the coefficient of frictionbetween the tyres and the road be , then the stoppingdistance is :

(A) mg2P

(B) mg2

P2

(C) gm2

P2

(D) gm2

P2

2

39. What is the maximum value of the force F such that theblock shown in the arrangement, does not move :

F60º

12 3

m = 3kg

(A) 20 N (B) 10 N(C) 12N (D) 15 N

40. A bock of mass 5 kg is held against wall by applying ahorizontal force of 100N. If the coefficient of frictionbetween the block and the wall is 0.5, the frictionalforce acting on the block is : (g =9.8 m/s2)

5kg100N

(A)100 N (B) 50 N(C) 49 N (D) 24.9 N

41. A heavy roller is being pulled along a rough road asshown in the figure. The frictional force at the point ofcontact is : (IAO/Jr./Stage-I/2007)

F

(A) parallel to F (B) opposite to F(C) perpendicular to F (D) zero

42. When a motor car of mass 1500 kg is pushed on aroad by two persons, it moves with a small uniformvelocity. On the other hand if this car is pushed on thesame road by three persons, it moves with anacceleration of 0.2 m/s2. Assume that each person isproducing the same muscular force. Then, the force offriction between the tyres of the car and the surface ofthe road is : (IAO/Jr./Stage-I/2009)

(A) 300 N (B) 600 N(C) 900 N (D) 100 N

43. A block of mass M is at rest on a plane surface inclinedat an angle to the horizontal The magnitude of forceexerted by the plane on the block is : (KVPY/2009)(A) Mg cos (B) Mg sin (C) Mg tan (D) Mg

44. A block of mass M rests on a rough horizontal table. Asteadily increasing horizontal force is applied such thatthe block starts to slide on the table without toppling.The force is continued even after sliding has started.Assume the coefficients of static and kinetic frictionbetween the table and the block to be equal. The cor-rect representation of the variation of the frictionalforces, �, exerted by the table on the block with time t is

given by : (KVPY/2010)

(A) (B)

(C) (D)

171717171717

PAGE # 17

45. A small child tries to move a large rubber toy placed on

the ground. The toy does not move but gets deformed

under her pushing force )F(

which is obliquely upward

as shown . Then (KVPY/2011)

(A) The resultant of the pushing force )F(

, weight of

the toy, normal force by the ground on the toy and the

frictional force is zero.

(B) The normal force by the ground is equal and oppo-

site to the weight of the toy.

(C) The pushing force )F(

of the child is balanced by

the equal and opposite frictional force

(D) The pushing force )F(

of the child is balanced by

the total internal force in the toy generated due to

deformation

46. On a horizontal frictional frozen lake, a girl (36 kg) and

a box (9kg) are connected to each other by means of a

rope. Initially they are 20 m apart. The girl exerts a

horizontal force on the box, pulling it towards her. How

far has the girl travelled when she meets the box ?

(KVPY/2011)

(A) 10 m

(B) Since there is no friction, the girl will not move

(C) 16 m

(D) 4m

47. Which of the following does NOT involve friction ?

(IJSO/Stage-I/2011)

(A) Writing on a paper using a pencil

(B) Turning a car to the left on a horizontal road.

(C) A car at rest parked on a sloping ground

(D) Motion of a satellite around the earth.

48. In the two cases shown below, the coefficient of kineticfriction between the block and the surface is the same,and both the blocks are moving with the same uniformspeed. Then, (IAO/Sr./Stage-I/2008)

F1

F2

(A) F1 = F

2

(B) F1 < F

2

(C) F1 > F

2

(D) F1 = 2F

2 if sin = Mg/4F

2

Weighing Machine :

49. The ratio of the weight of a man in a stationary lift and

when it is moving downward with uniform acceleration

�a� 3:2. The value of �a� is : (g = acceleration, due to

gravity)

(A) (3/2)g (B) g

(C) (2/3) g (D) g/3

50. A person standing on the floor of an elevator drops a

coin. The coin reaches the floor of the elevator in time

t1 when elevator is stationary and in time t2 if it is moving

uniformly. Then

(A) t1 = t2(B) t1 > t2(C) t1 < t2(D) t1 < t2 or t1 > t2 depending

� ASSERTION / REASON

51. STATEMENT-1 : A man standing in a lift which is moving

upward, will feel his weight to be greater than when

the lift was at rest.

STATEMENT-2 : If the acceleration of the lift is �a� upward

then the man of mass m shall feel his weight to be

equal to normal reaction (N) exerted by the lift given N

= m(g+a) (where g is acceleration due to gravity

(A) statement-1 is true, Statement 2 is true, statement-

2 is correct explanation for statement 1.

(B) statement-1 is true, Statement 2 is true, statement-

2 is NOT a correct explanation for statement-1.



52. A beaker containing water is placed on the platform of

a digital weighing machine. It reads 900 g. A wooden

block of mass 300 g is now made to float in water in

the beaker (without touching walls of the beaker). Half

the wooden block is submerged inside water. Now,

the reading of weighing machine will be :

(IAO/Jr./Stage-I/2009)

(A) 750 g (B) 900 g

(C) 1050 g (D) 1200 g

Miscellaneous :

53. An object will continue accelerating until :(A) Resultant force on it begins to decreases(B) Its velocity changes direction(C) The resultant force on it is zero(D) The resultant force is at right angles to its directionof motion

181818181818

PAGE # 18

54. In which of the following cases the net force is not zero ?(A) A kite skillfully held stationary in the sky(B) A ball freely falling from a height(C) An aeroplane rising upward at an angle of 45° with

the horizontal with a constant speed(D) A cork floating on the surface of water.

55. Figure shows the displacement of a particle going

along the X-axis as a function of time. The force acting

on the particle is zero in the region.

(A) AB (B) BC

(C) CD (D) DE

56. A 2 kg toy car can move along x axis. Graph shows force

Fx, acting on the car which begins to rest at time t = 0. The

velocity of the car at t = 10 s is :

(A) � i� m/s (B) � 1.5 i� m/s

(C) 6.5 i� m/s (D) 13 i� m/s

57. Figure shows the displacement of a particle going

along the x-axis as a function of time :

(A) The force acting on the particle is zero in the region AB

(B) The force acting on the particle is zero in the region BC

(C) The force acting on the particle is zero in the region CD

(D) The force is zero no where

58. A force of magnitude F1 acts on a particle so as to

accelerate if from rest to velocity v. The force F1 is then

replaced by another force of magnitude F2 which

decelerates it to rest.

(A) F1 must be the equal to F2

(B) F1 may be equal to F2

(C) F1 must be unequal to F2

(D) None of these

59. In a imaginary atmosphere, the air exerts a small forceF on any particle in the direction of the particle�s motion.

A particle of mass m projected upward takes a time t1in reaching the maximum height and t2 in the returnjourney to the original point. Then(A) t1 < t2(B) t1 > t2(C) t1 = t2(D) The relation between t1 and t2 depends on the massof the particle

60. A single force F of constant magnitude begins to act ona stone that is moving along x axis. The stone continuesto move along that axis. Which of the followingrepresents the stone�s position ?

(A) x = 5t � 3 (B) x = 5t2 + 8t � 3

(C) x = �5t2 + 5t � 3 (D) x = 5t3 + 4t2 � 3

61. Three forces act on a particle that moves with

unchanging velocity v = (3 i� � 4 j� ) m/s. Two of the

forces are 1F

= (3 i� + 2 j� � 4 k� ) N and 2F

= (�5 i� + 8 j�

+ 3 k� ) N. The third force is :

(A) (�2 i� + 10 j� � 7 k� ) N(B) (2 i� � 10 j� + k� ) N(C) (7 i� � 2 k� + 10 j� ) N(D) none of these

62. An 80 kg person is parachuting and experiencing a

downward acceleration of 2.5 m/s2 . The mass of the

parachute is 5.0 kg. The upward force on the open

parachute from the air is :

(A) 620 N (B) 740 N

(C) 800 N (D) 920 N

63. A block of mass m is pulled on the smooth horizontal

surface with the help of two ropes, each of mass m,

connected to the opposite faces of the block. The

forces on the ropes are F and 2F. The pulling force on

the block is :

(A) F (B) 2F(C) F/3 (D) 3F/2

191919191919

PAGE # 19

64. A body of mass 5 kg starts from the origin with an initialvelocity u

= 30 i� + 40 j� ms�1 . If a constant force

F

= �( i� + 5 j� ) N acts on the body, the time in which the

y-component of the velocity becomes zero is :

(A) 5 s (B) 20 s

(C) 40 s (D) 80 s

65. STATEMENT-1 :According to the newton�s third law of

motion, the magnitude of the action and reaction force

is an action reaction pair is same only in an inertial

frame of reference.

STATEMENT-2 : Newton �s laws of motion are

applicable in every inertial reference frame.

(A) statement-1 is true, Statement 2 is true, statement-

2 is correct explanation for statement 1.

(B) statement-1 is true, Statement 2 is true, statement-

2 is NOT a correct explanation for statement-1.



66. A body of mass 10 g moves with constant speed 2 m/s along a regular hexagon. The magnitude of changein momentum when the body crosses a corner is :

(IAO/Sr./Stage-I/2007)

(A) 0.04 kg-m/s (B) zero

(C) 0.02 kg-m / s (D) 0.4 kg-m/s

67. An object with uniform density is attached to a spring

that is known to stretch linearly with applied force as

shown below

When the spring object system is immersed in a

liquid of density 1 as shown in the figure, the spring

stretches by an amount x1 ( >

1). When the experiment

is repeated in a liquid of density 2 <

1 . the spring is

stretched by an amount x2. Neglecting any buoyant force

on the spring, the density of the object is:

(KVPY/2011)

(A) 21

2211xx

xx

(B)

12

1221xx

xx

(C) 21

1221xx

xx

(D)

21

2211xx

xx

68. A body of 0.5 kg moves along the positive x - axis under

the influence of a varying force F (in Newtons) as shown

below : (KVPY/2011)

3

3

1

F(N

)

0,0 2 4 6 8 10

x(m)

If the speed of the object at x = 4m is 3.16 ms�1 then its

speed at x = 8 m is :

(A) 3.16 ms�1 (B) 9.3 ms�1

(C) 8 ms�1 (D) 6.8 ms�1

69. A soldier with a machine gun, falling from an airplanegets detached from his parachute. He is able to resistthe downward acceleration if he shoots 40 bullets asecond at the speed of 500 m/s. If the weight of a bulletis 49 gm, what is the weight of the man with the gun ?Ignore resistance due to air and assume theacceleration due to gravity g = 9.8 m/s2 . (KVPY/2010)(A) 50 kg (B) 75 kg(C) 100 kg (D) 125 kg

20202020PAGE # 20

CARBON

INTRODUCTION

The compounds like urea, sugars, fats, oils, dyes,proteins, vitamins etc., which are isolated directly orindirectly from living organisms such as animals andplants are called organic compounds.The branch ofchemistry which deals with the study of thesecompounds is called ORGANIC CHEMISTRY.

VITAL FORCE THEORY OR

BERZELIUS HYPOTHESIS

Organic compounds cannot be synthesized in thelaboratory because they require the presence of amysterious force (called vital force) which exists onlyin living organisms.

WOHLER�S SYNTHESIS

In 1828, Friedrich Wohler synthesized urea (a wellknown organic compound) in the laboratory by heatingammonium cyanate.

(NH ) SO + 2 KCNO 2NH CNO + K SO4 2 4 4 2 4

Ammonium Potassium Ammonium Potassiumsulphate sulphatecyanatecyanate

Note :Urea is the first organic compound synthesized inthe laboratory.

MODERN DEFINITION OF ORGANIC CHEMISTRY

Organic compounds may be defined as hydrocarbonsand their derivatives and the branch of chemistrywhich deals with the study of hydrocarbons and theirderivatives is called ORGANIC CHEMISTRY.

Organic chemistry is treated as a separate branchbecause of following reasons-

(i) Organic compounds are large in number.

(ii) Organic compounds generally contain covalentbond.

(iii) Organic compounds are soluble in non polar solvents.

(iv) Organic compounds have low melting and boilingpoints.

(v) Organic compounds show isomerism .

(vi) Organic compounds exhibit homology.

VERSATILE NATURE OF CARBON

About 3 million organic compounds are known today.The main reasons for this huge number of organiccompounds are -

(i) Catenation : The property of self linking of carbonatoms through covalent bonds to form long straightor branched chains and rings of different sizes iscalled catenation.Carbon shows maximumcatenation in the periodic table due to its small size,electronic configuration and unique strength of carbon-carbon bonds.

(ii) Electronegativity and strength of bonds : Theelectronegativity of carbon (2.5) is close to a numberof other elements like H (2.1) , N(3.0) , P (2.1), Cl (3.0)and O (3.5). So carbon forms strong covalent bondswith these elements.

(iii) Tendency to form multiple bonds : Due to smallsize of carbon it has a strong tendency to form multiplebonds (double & triple bonds).

(iv) Isomerism : It is a phenomenon by the virtue ofwhich two compounds have same molecular formulabut different physical and chemical properties.

CLASSIFICATION OF ORGANIC COMPOUNDS

The organic compounds are very large in number onaccount of the self -linking property of carbon calledcatenation. These compounds have been furtherclassified as open chain and cyclic compounds.

Open chaincompounds

Closed chain compounds

Organic compounds

Alicycliccompounds

Aromaticcompounds

(a) Open Chain Compounds :

These compounds contain an open chain of carbonatoms which may be either straight chain or branchedchain in nature. Apart from that, they may also besaturated or unsaturated based upon the nature ofbonding in the carbon atoms. For example.

, ,


21212121PAGE # 21

,

n-Butane is a straight chain alkane while

2-Methylpropane is branched alkane.

(b) Closed Chain or Cyclic Compounds :

Apart from the open chains, the organic compounds

can have cyclic or ring structures. A minimum of three

atoms are needed to form a ring. These compounds

have been further classified into following types.

(i) Alicyclic compounds : Those carbocyclic

compounds which resemble to aliphatic compounds

in their properties are called alicyclic compounds .

e.g. or Cyclopropane

or Cyclobutane

or Cyclopentane

or Cyclohexane

(ii) Aromatic compounds : Organic compounds which

contain one or more fused or isolated benzene rings

are called aromatic compounds.

e.g.

Benzene Toluene Ethyl benzene

Phenol Aniline

Note :Benzene is the parent compound of majority ofaromatic organic compounds.

HYDROCARBONS

The organic compounds containing only carbon andhydrogen are called hydrocarbons. These are thesimplest organic compounds and are regarded asparent organic compounds. All other compounds areconsidered to be derived from them by thereplacement of one or more hydrogen atoms by otheratoms or groups of atoms. The major source ofhydrocarbons is petroleum.

Types of Hydrocarbons :

The hydrocarbons can be classified as :

(i) Saturated hydrocarbons :

(A) Alkanes : Alkanes are saturated hydrocarbonscontaining only carbon - carbon and carbon - hydrogensingle covalent bonds.

General formula- CnH

2n + 2(n is the number of carbon atoms)

e.g. CH4 ( Methane)

C2H

6 (Ethane)

(ii) Unsaturated hydrocarbons :

(A) Alkenes : These are unsaturated hydrocarbonswhich contain carbon - carbon double bond. Theycontain two hydrogen less than the correspondingalkanes.

General formula - CnH

2n

e.g. C2H

4(Ethene)

C3H

6(Propene)

(B) Alkynes : They are also unsaturated hydrocarbonswhich contain carbon-carbon triple bond. They containfour hydrogen atoms less than the correspondingalkanes.

General formula - CnH

2n�2

e.g. C2H

2(Ethyne)

C3H

4(Propyne)

22222222PAGE # 22

NOMENCLATURE OF ORGANIC COMPOUNDS

Nomenclature means the assignment of names toorganic compounds . There are two main systems ofnomenclature of organic compounds -(1) Trivial system(2) IUPAC system (International Union of Pure andApplied Chemistry)

(a) Basic rules of IUPAC nomenclature of

organic compounds :

For naming simple aliphatic compounds, the normalsaturated hydrocarbons have been considered asthe parent compounds and the other compounds astheir derivatives obtained by the replacement of oneor more hydrogen atoms with various functionalgroups.

(i) Each systematic name has two or three of thefollowing parts-

(A) Word root : The basic unit of a series is word rootwhich indicate linear or continuous number of carbonatoms.

(B) Primary suffix : Primary suffixes are added to theword root to show saturation or unsaturation in acarbon chain.

(C) Secondary suffix : Suffixes added after theprimary suffix to indicate the presence of a particularfunctional group in the carbon chain are known assecondary suffixes.

(ii) Names of straight chain hydrocarbons : Thename of straight chain hydrocarbon may be dividedinto two parts-

(A) Word root (B) Primary suffix

(A) Word roots for carbon chain lengths :

Chain length root length root

C Meth- C Hex-C Eth - C Hept-C Prop - C Oct-C But - C Non-C Pent- C Dec-

1 6

2 7

3 8

4 9

5 10

Word Chain Word

(B) Primary suffix :

Examples :

Eth -

E B P E M

Propene

a

Note :

The name of the compound, in general , is written in

the following sequence-

(Position of substituents )-(prefixes ) (word root)-(p -

suffix).

(iii) Names of branched chain hydrocarbon : The

carbon atoms in branched chain hydrocarbons are

present as side chain . These side chain carbon atoms

constitute the alkyl group or alkyl radicals. An alkyl group

is obtained from an alkane by removal of a hydrogen.

General formula of alkyl group = CnH

2n+1

An alkyl group is represented by R.

e.g.

(A)

H

H

H

C

Methyl

�H

(B) H

HH

CC

Ethyl

�H

H H

(C)

23232323PAGE # 23

A branched chain hydrocarbon is named using the

following general IUPAC rules :

Rule1: Longest chain rule : Select the longest possible

continuous chain of carbon atoms. If some multiple

bond is present , the chain selected must contain the

multiple bond.

(i) The number of carbon atoms in the selected chain

determines the word root .

(ii) Saturation or unsaturation determines the primary

suffix (P. suffix).

(iii) Alkyl substituents are indicated by prefixes.

e.g. CH � CH � CH � CH � CH3 2 2 3

CH3

Prefix : MethylWord root : pent-P. Suffix: - ane

e.g. CH � CH � CH � CH � CH3 2 3

CH3

Prefix : Methyl

Word root : Hept-

P. Suffix : -aneCH � CH � CH2 2 3

e.g. CH � CH � C � CH3 2 3

CH2

Prefix : MethylWord root : But-P. Suffix : �ene

e.g. CH � CH � CH � CH � CH23 2 3

CH3

Prefixes : Ethyl, MethylWord root : Pent-P. Suffix : -ane

CH � CH3

Rule 2 : Lowest number rule: The chain selected is

numbered in terms of arabic numerals and the

position of the alkyl groups are indicated by the number

of the carbon atom to which alkyl group is attached .

(i) The numbering is done in such a way that the

substituent carbon atom has the lowest possible

number.

(ii) If some multiple bond is present in the chain, the

carbon atoms involved in the multiple bond should

get lowest possible numbers.

e.g.

2�Methylbutane 3�Methylbutane

(Correct) (Wrong)

e.g.

2�Methylpentane 4�Methylpentane

(Correct) (Wrong)

e.g.

3�Methylbut�1� ene 2�Methylbut � 3 � ene

(Correct) (Wrong)

e.g.

CH3| |

CH3

11 33 44 22

3-Methylbut-1-yne (Correct)

2-Methylbut-3-yne (Wrong)

Rule 3 : Use of prefixes di, tri etc. : If the compoundcontains more than one similar alkyl groups,theirpositions are indicated separately and an appropriatenumerical prefix di, tri etc. , is attached to the nameof the substituents. The positions of the substituentsare separated by commas.

e.g. CH � CH � C � CH � CH3 2 3

CH3

CH3CH3

345 2 1

2,3 - Dimethylpentane 2,3,3 - Trimethylpentane

e.g.

2,3,5 -Trimethylhexane 2,2,4 - Trimethylpentane

Rule 4 : Alphabetical arrangement of prefixes: If thereare different alkyl substituents present in thecompound their names are written in the alphabeticalorder. However, the numerical prefixes such as di,tri etc. , are not considered for the alphabetical order.

e.g.

3-Ethyl - 2,3-dimethylpentane

Rule 5 : Naming of different alkyl substituents at theequivalent positions :

Numbering of the chain is done in such a way that thealkyl group which comes first in alphabetical ordergets the lower position.

e.g.

3-Ethyl-4-methylhexane

24242424PAGE # 24

Rule - 6 : Lowest sum ruleAccording to this rule numbering of chain is done insuch a way that the sum of positions of differentsubstituents gets lower value.

e.g.

(i)

Word root : Hex -Primary suffix : - aneSubstituent : two methyl & one ethyl groups

IUPAC name : 4-Ethyl - 2, 4 - dimethylhexane

Some other example :

(i)

Word root : Prop -P. Suffix : -aneSubstituent : two methyl groupsIUPAC name : 2, 2 - Dimethylpropane

(ii)

Word root : But -P. Suffix : - eneSubstituent : two methyl groupsIUPAC name : 2, 3 - Dimethylbut - 1 - ene

(iii)

Word root : Hex -P. Suffix : - yneSubstituent : one methyl groupIUPAC name : 4 - Methylhex - 2 - yne

FUNCTIONAL GROUP

An atom or group of atoms in an organic compoundor molecule that is responsible for the compound�scharacteristic reactions and determines its propertiesis known as functional group. An organic compoundgenerally consists of two parts -(i) Hydrocarbon radical (ii) Functional group

e.g.

Hydrocarbon radical Functional group� Functional group is the most reactive part of the

molecule.� Functional group mainly determines the chemical

properties of an organic compound.� Hydrocarbon radical mainly determines the physical

properties of the organic compound.

(a) Main Functional Groups :

(i) Hydroxyl group (� OH) : All organic compoundscontaining - OH group are known as alcohols .e.g. Methanol (CH

3OH) , Ethanol (CH

3 � CH

2 � OH) etc .

(ii) Aldehyde group (�CHO) : All organic compoundscontaining �CHO group are known as aldehydes.

e.g. Methanal (HCHO), Ethanal (CH3CHO) etc.

(iii) Ketone group (�CO�) : All organic compoundscontaining �CO� group are known as ketones.

e.g. Propanone (CH3COCH

3), Butanone

(CH3COCH

2CH

3) etc.

(iv) Carboxyl group ( � COOH) : All organic compoundscontaining carboxyl group are called carboxylic acids.e.g. CH

3COOH (Ethanoic acid)

CH3CH

2COOH(Propanoic acid)

(v) Halogen group (X = F, Cl, Br, I) : All organic

compounds containing �X (F, Cl, Br or I) group areknown as halides.e.g. Chloromethane (CH

3Cl), Bromomethane (CH

3Br)

etc .

(b) Nomenclature of Compounds Containing

Functional Group :

In case functional group (other than C = C and C C)is present, it is indicated by adding secondary suffixafter the primary suffix. The terminal �e� of the primary

suffix is removed if it is followed by a suffix beginningwith �a �, �e �, � i �, �o � , �u �. Some groups l ike

�F, � Cl, � Br and � are considered as substituentsand are indicated by the prefixes.

Some groups like � CHO, C ��

O

, � COOH, and � OH

are considered as functional groups and areindicated by suffixes.

25252525PAGE # 25

Class Functional Group

General Formula Prefix Suffix IUPAC Name

Carboxylicacid

(R = CnH2n+1)

Carboxy - oic acid Alkanoic acid

Ester Carbalkoxy alkyl (R�) - oate Alkyl alkanoate

Aldehyde � CHO R � CHO Formyl or oxo

- al Alkanal

Ketone oxo - one Alkanone

Alcohol � OH R � OH Hydroxy - ol Alkanol

Alkenes C = C CnH2n � - ene Alkene

Alkynes C C CnH2n�2 � - yne Alkyne

Halides � X

(X = F,Cl,Br,I) R � X Halo � Haloalkane

Step 4 :

The carbon atoms of the parent chain are numbered

in such a way so that the carbon atom of the functional

group gets the lowest possible number . In case the

functional group does not have the carbon atom, then

the carbon atom of the parent chain attached to the

functional group should get the lowest possible

number.

Step 5 :

The name of the compound is written as -

Prefixes - word root - primary suffix - secondary suffix

Note :

The number of carbon atoms in the parent chain

decides the word root.

Steps of naming of an organic compound

containing functional group :

Step 1:

Select the longest continuous chain of the carbon

atoms as parent chain. The selected chain must

include the carbon atoms involved in the functional

groups like � COOH, � CHO, � CN etc, or those which

carry the functional groups like � OH, � NH2,� Cl,

� NO2 etc.

The number of carbon atoms in the parent chain

decides the word root.

Step 2 :

The presence of carbon - carbon multiple bond

decides the primary suffix.

Step 3 :

The secondary suffix is decided by the functional

group.

26262626PAGE # 26

S.No. Compound Common name Derived name IUPAC Name Structure

1 CH3 � OHMethyl alcohol or Wood spirit

Carbinol Methanol

2 CH3 � CH2 � OH Ethyl alcohol Methyl carbinol Ethanol

3 CH3 � CH2 � CH2 � OH n-Propyl alcohol Ethyl carbinol 1- Propanol

4 Isopropyl alcohol Dimethyl carbinol 2 - Propanol

5 CH3 � CH2 � CH2 � CH2 � OH n-Butyl alcohol n-Propyl carbinol 1- Butanol

6 HCOOH Formic acid � Methanoic acid

7 CH3COOH Acetic acid � Ethanoic acid

8 CH3 � CH2 � COOH Propionic acid methyl acetic

acid Propanoic acid

9 CH3 � CH2 � CH2 � COOH Butyric acid ethyl acetic

acid Butanoic acid

10 CH3 � CH2 � CH2 � CH2 � COOH Valeric acid n-Propyl acetic

acid Pentanoic acid

O

H � C � C � O � H

H

H

C H3

H

Some more examples :

(i)

Word root : Hept-Primary suffix : � ane

Functional group : � OH

Secondary suffix : � ol

IUPAC Name : 2, 5-Dimethylheptan�1� ol

(ii)

Word root : Pent -Primary suffix : � ene

Secondary suffix : � oic acid

Position of double : 2nd bond

IUPAC name : Pent-2-en-1-oic acid/Pent-2-enoic acid

(iii) CH3 � CH

2 � CH

2 � NH

2

Word root : Prop -Primary suffix : - aneSecondary suffix : - amineIUPAC name : Propan - 1 - amine

(iv)Word root : Prop-Primary suffix : - aneSubstituent : nitro(prefix)IUPAC name : 1 - Nitropropane

(v)

Word root : But -Primary suffix : � ane

Prefix : � chloro

IUPAC name : 2 - Chlorobutane

27272727PAGE # 27

(vi)

Word root : But -Primary suffix : � ane

Secondary suffix : � one

Prefix : MethylIUPAC name : 3 - Methylbutan - 2- one

HOMOLOGOUS SERIES

Homologous series may be defined as a series ofsimilarly constituted compounds in which the memberspossess similar chemical characteristics and the twoconsecutive members differ in their molecular formulaby � CH

2.

(a) Characteristics of Homologous Series :

(i) All the members of a series can be represented bythe same general formula.e.g. General formula for alkane series is C

nH

2n+2 .

(ii) Any two consecutive members differ in their formulaby a common difference of � CH

2 and differ in

molecular mass by 14.

(iii) Different members in a series have a commonfunctional group.e.g. All the members of alcohol family have �OH group .

(iv) The members in any particular family have almostidentical chemical properties. Their physicalproperties such as melting point, boiling point, densityetc, show a regular gradation with the increase in themolecular mass.

(v) The members of a particular series can beprepared almost by the identical methods.

(b) Homologues :

The different members of a homologous series areknown as homologues.

e.g.(i) Homologous series of alkanesGeneral formula : C

nH

2n+2

Value of n Molecular formula IUPAC namen = 1 CH

4 Methane

n = 2 C2H

6 Ethane

n = 3 C3H

8 Propane

(ii) Homologous series of alkenesGeneral formula :C

nH

2n

Value of n Molecular IUPAC Commonformula name name

n = 2 C2H

4 Ethene Ethylene

n = 3 C3H

6 Propene Propylene

n = 4 C4H

8 But-1-ene - Butylene

(iii) Homologous series of alkynes

General formula : CnH

2n�2

Value of n Molecular IUPAC Commonformula name name

n = 2 C2H

2 Ethyne Acetylene

n = 3 C3H

4 Propyne Methyl acetylene

n = 4 C4H

6 But -1-yne Ethyl acetylene

ISOMERISM

Such compounds which have same molecularformula but different physical and chemical propertiesare known as isomers and the phenomenon isknown as isomerism.

(a) Chain Isomerism :

The isomerism in which the isomers differ from eachother due to the presence of different carbon chainskeletons is known as chain isomerism. e.g.

(i) C4H

10

,

2 - Methylpropane (Isobutane)

(ii) C5H

12

2 - Methylbutane(Isopentane)

2, 2 -Dimethylpropane(neo - pentane)

(iii) C4H

8

CH3 � CH

2 � CH = CH

2 ,

But - 1 - ene Methylpropene

(b) Position Isomerism :

In this type of isomerism, isomers differ in the structuredue to difference in the position of the multiple bondor functional group.e.g.

(i) C4H

8

CH3 � CH

2 � CH = CH

2 , CH

3 � CH = CH � CH

3

But -1 - ene But -2 - ene

(ii) C3H

8O

CH � CH � CH � OH , 3 2 2

Propan-1-ol

Propan-2-ol

CH � CH � CH3 3

OH

28282828PAGE # 28

(c) Functional Group Isomerism :

In this type of isomerism, isomers differ in the

structure due to the presence of different functional

groups.

e.g.

(i) C3H

8O

CH3 � CH

2 � O � CH

3 CH

3 � CH

2 � CH

2 � OH

Methoxy ethane Propan-1-ol

(ii) C4H

6

CH3 � CH

2 � C CH CH

2 = CH � CH = CH

2

But - 1- yne Buta - 1, 3 - diene

[or 1, 3 - Butadiene ]

ALKANES

Alkanes are aliphatic hydrocarbons having onlyC � C single covalent bonds. These are also known

as saturated hydrocarbon as they contain single bond

only. These compounds are open chain compoundswhich are also addressed as Acyclic compounds.Alkanes have the general formula C

nH

2n+2 .The carbon

atoms in alkanes are in a state of sp3 hybridization,i.e. the carbon atoms have a tetrahedral geometry.

(a) Physical Properties :

(i) Alkanes of no. of carbon atoms C1 to C

4 are gases.

Carbon atoms C5 to C

17 are liquids and C

18 & onwards

are solids.(ii) Alkanes are colourless and odourless.(iii) They are non-polar in nature, hence they dissolveonly in non-polar solvents like benzene, carbontetrachloride etc.(iv) Boiling point of alkanes increases as theirmolecular weight increases.

Note :

Alkanes are unaffected by most chemical reagents

and hence are known as paraffins (parum-little, affinisaffinity).

SOME COMMON EXAMPLES OF ALKANES

Molcular Formula Structure Trivial Name IUPAC Name

CH4 CH4 Methane Methane

C2H6 CH3�CH3 Ethane Ethane

C3H8 CH3�CH2�CH3 Propane Propane

C4H10 n-ButaneIsobutane

Butane2�Methylpropane

C5H12 n-Pentane

Isopentane

Neopentane

Pentane

2�Methylbutane

2,2�Dimethylpropane

C6H14 n-Hexane

Isohexane

�

�

Neohexane

Hexane

2�Methylpentane

3�Methylpentane

2,3�Dimethylbutane

2,2�Dimethylbutane

CH � CH � CH � CH � CH � CH3 2 2 2 2 3

CH � CH � CH � CH � CH3 2 2 3

CH3

CH � CH � CH � CH � CH3 2 2 3

CH3

CH � CH � CH � CH3 3

CH3CH3

CH3

CH � C � CH � CH3 2 3

CH3

3C H�2C H�2C H�3CH

3CH�2CH�

3CH|CH�3CH

3CH�

3CH|

|

3CH

C�3CH

3CH�2CH�2CH�3CH

33 CH�

3CH|CH�CH

29292929PAGE # 29

Note :The C � C bond distance in alkanes is 1.54 Å and the

bond energy is of the order of 80 Kcal per mole.

METHANE

It is a product of decomposition of organic matter inabsence of oxygen. It is found in coal mines (hencethe name damp fire), marshy places (hence the namemarsh gas) and the places where petroleum is found.

Note :

Methane is a major constituent of natural gas.

(a) Properties :

Methane is a colourless gas with practically no smelland is almost insoluble in water. It melts at � 183º C

and boils at �162ºC. Methane has tetrahedral

geometry in which H�atoms are situated at four

corners of the regular tetrahedron. Bond angle is109º28�. It has sp3 hybridisation.

(b) Structure :

Tetrahedral

H

H HH

C

(c) Preparation of Methane :

(i) Direct synthesis :

C + 2H2 CH4

Ni500ºC

Carbon Hydrogen Methane

(ii) Sabatier and Senderens reductive method :Methane can be prepared by passing carbonmonoxide or carbon dioxide and hydrogen over finelypowdered nickel catalyst at 300ºC.

CO + 3H2

Ni powder300ºC

CH4 + H

2O

CO2 + 4H

2 Ni powder

300ºC CH

4 + 2H

2O

(iii) Hydrolysis of aluminium carbide :Al C + 12H O4 3 2 3CH + 4Al(OH)4 3

Aluminium carbide

Water Methane Aluminium hydroxide

(iv) Reduction of methyl iodide :

CH � + 2H3 Zn�Cu Couple

H O2

CH + H4

Methyl iodide

Methane Hydrogen iodide

(v) Reduction of methanol or formaldehyde or formicacid with H

CH OH + 2H3 CH + + H O4 2Red P

Methanol Methane

HCHO + 4H CH + 2 + H O4 2Red P

Methanal Methane

(vi) Laboratory Method : Methane is prepared in thelaboratory by heating a mixture of dry sodium acetateand soda lime in a hard glass tube as shown infigure. It is a decarboxylation reaction.

Burner

Gas

Hard glass tube

Delivery tube

Gas jar

Bubbles of methane gas

Trough

WaterBeehive shelf

Iron stand

Cork

Sodium acetate and soda lime

Preparation of methane gas

Methane, so formed is collected by downwarddisplacment of water. This gas contains somehydrogen, ethylene etc. as impurities which can beremoved by passing the impure gas through alkalinepotassium permanganate solution.

(d) General Reactions :

(i) Combustion :(A) Methane burns with explosive violence in airforming carbon dioxide and water.

CH4 + 2O

2 CO

2 + 2H

2O + Heat

(B) In the presence of insufficient supply of oxygen.2CH

4 + 3O

2 2CO + 4H

2O + Heat

(ii) Halogenation :(A) In direct sunlight

CH4 + 2Cl

2 h C + 4HCl

(B) In diffused light

CH 4

Cl2CH Cl3

Methylchloride

Cl2 CH Cl2 2

Methylenedichloride

Cl2 CHCl3Chloroform

Cl2 CCl4 Carbontetrachloride

Methane

Fluorine forms similar substitution products in thepresence of nitrogen which is used as a diluentbecause of high reactivity of fluorine. Bromine vapoursreact very sluggishly while iodine vapours do not reactat all.

(iii) Nitration :

CH + HO�NO 4 2

400ºC

10 atm.CH �NO + H O3 2 2

Methane Nitric acid Nitromethane

(iv) Catalytic Air oxidation : This is a method forcommercial production of methanol.When a mixture of methane and oxygen in a ratio of 9: 1by volume is passed through a heated copper tube at200ºC and at a pressure of 100 atmospheres,

methanol is formed.CH + 1/2 O4 2 CH OH3

MethanolMethane(e) Uses :

(i) Alkanes are used directly as fuels .

(ii) Certain alkanes, such as methane, are used as asource of hydrogen.

(iii) The carbon obtained in decomposition of alkanesis in very finely divided state and is known as carbonblack. This is used in making printer�s ink, paints, boot

polish and blackening of tyres.

(iv) Alkanes are used as starting materials for anumber of other organic compounds e.g.methanol, methyl chloride, methylene dichloride etc.

30303030PAGE # 30

ALKENES

Alkenes are the simplest unsaturated aliphatichydrocarbons with one carbon - carbon double bond.Alkenes have general formula C

nH

2n. The carbon

atoms connected by the double bond are in a state ofsp2 hybridisation and this part of molecule is planar.A double bond is composed of sigma () and a pi ()bond. Alkenes are also called olefines (oil forming)becuase they form oily products with halogens.

R � CH = CH + Br 2 2 R � CH � CH2

Br Br(Oily liquid)

(a) Properties :

(i) Alkenes of C2 to C

4 are gases. Alkenes of carbon

atoms C5 to C

14 are liquids and C

14 and onwards are

solids.

(ii) Ethene is colourless gas with faint sweet smell.All other alkenes are colourless and odourless.

(iii) Alkenes are insoluble in polar solvents like water,but fairly soluble in non-polar solvents like benzene,carbon tetrachloride etc.

(iv) Boiling point of alkenes increases with increasein molecular mass.

Bond length of C = C is 1.34 Å . The energy of the

double bond is 142 Kcal mol�1, which is less thantwice the energy of a single bond i.e. 80 Kcal mole-1.

This indicates that a pi () bond is weaker than a

sigma () bond.

(b) Some common examples of alkenes -

Molecularformula

Structure Trivial Name

IUPACName

C H2 4 H C = CH2 2

C H3 6 CH � CH = CH3 2

C C4 8 CH � CH � CH = CH3 2 2

CH � C � H3

CH � C � H3

CH � C � H3

H � C � CH3

CH � C = CH3 2

CH3

C H5 10 CH = CH (CH ) CH2 2 2 3

CH CH = CHCH CH3 2 3

CH � CH � CH = CH3 2

CH3

CH � C = CH � CH3 3

CH3

CH = C � CH � CH2 2 3

CH3

Ethylene

Propylene

Ethene

Propene

-Butylene

-Butylene (cis)

-Butylene (trans)

Isobutylene

�

�

�

�

�

1-Butene

2-Butene (cis)

2-Butene (trans)

2-Methyl propene

1-Pentene

2-Pentene (cis and trans)

3-Methyl-1-butene

2-Methyl-2-butene

2-Methyl-1-butene

(b) Uses :

(i) Ethylene is mainly used in the manufacture ofethanol, ethylene oxide and higher 1-alkenes.Ethylene is used for ripening of fruits. It is also usedfor preparation of mustard gas.[Cl � CH

2 � CH

2 � S � CH

2 � CH

2 � Cl]

(ii) Polythene from ethylene, teflon from tetrafluoroethylene and polystyrene from styrene are usedas plastic materials. Acrilon or orlon obtained fromvinyl cyanide is used for making synthetic fibres.

ETHENE

Ethene occurs in natural gas, coal gas and woodgas. It is also formed during the cracking of high boilingpetroleum fractions.

(a) Properties :

Ethene is a colourless gas (B.P. = �105ºC). It is very

sparingly soluble in water but dissolves in acetone,alcohol etc. It burns with smoky flame. Ethene hastrigonal planar geometry. Bond angle is 120º. It has

sp2 hybridisation.

(b) Structure :

(c) Preparation of Ethene :

(i) By dehydration of alcohol (Lab. method) :

CH � CH � OH3 2

Conc. H SO2 4

165 � 170ºC CH = CH + H O2 2 2

EtheneEthanol

(ii) By cracking of kerosene :

CH � (CH ) � CH3 2 4 3 CH � CH � CH � CH + CH = CH3 2 2 3 2 2

Cracking

n-Hexane Butane Ethene

(iii) From alkyl halides (Dehydrohalogenation) :CH � CH + KOH2 2

H X(Alcoholic)

CH = CH + KX + H O2 2 2

Ethene(Here X = Halogen)

Ethyl halide

(d) General Reactions :

(i) Addition of halogens :

CH = CH + Cl2 2 2

CCl4CH � CH2 2

Cl Cl1,2-Dichloroethane(Ethylene dichloride)

Ethene Chlorine

CH = CH + Br2 2 2

CCl4CH � CH2 2

Br1,2-Dibromoethane (colourless)

Ethene BromineBr(red-brown colour)

31313131PAGE # 31

Note :Addition of bromine on alkenes in presence of CCl

4

is the test for unsaturation.

(ii) Addition of halogen acids (Hydrohalogenation) :

HChloroethane

CH = CH + HCl2 2 CH � CH2 2

ClEthene

(iii) Hydrogenation :

Ethane

CH = CH + H2 2 2 CH � CH3 3

Ethene

Ni or Pt

High T& P

(iv) Combustion :

C H + 3O2 4 2 2CO + 2H O2 2

Ethene

+ Heat

(v) Addition of oxygen :

(vi) Polymerisation :

nCH = CH 2 2 � (CH � CH �) 2 2 n

High T

& High P

Ethene Polyethene

ALKYNES

Alkynes are unsaturated aliphatic hydrocarbonshaving a carbon-carbon triple bond. Alkynes havegeneral formula C

nH

2n�2. Thus, they have two hydrogen

atoms less than an alkene and four hydrogen atomsless than an alkane with same number of carbonatoms. A triple bond is composed of one sigma ()and two pi () bonds. The carbon atoms connectedby a triple bond are in state of sp hybridisation.

(a) Properties :

(i) Alkynes of carbon atoms C2 to C

4 are gases. Alkynes

of carbon atoms C5 to C

12 are liquids.Alkynes of C

13 &

onwards are solids.

(ii) Alkynes are colourless and odourless, but ethynehas characteristic odour.

(iii) Boiling point and solubilities in water are relativelyhigher than those of alkanes and alkenes.

(iv) Alkynes are weakly polar in nature.

(v) Alkynes are lighter than water and soluble in non-polar solvents.

(vi) Boiling point of alkynes increases with the increasein molecular mass.

Note :The bond energy of a triple bond is 190.5 Kcal permole, which is less than thrice the energy of a single() bond.

SOME COMMON EXAMPLES OF ALKYNES :

Molecular formula Structure Derived Name IUPAC name

C2H2 Acetylene Ethyne

C3H4 Methyl acetylene Propyne

C4H6 CH3�CH2 � C CH Ethyl acetylene

Dimethyl acetylene

1�Butyne

2� Butyne

C5H8 CH3 � CH2 � CH2 � C CH n-Propyl acetylene

Ethyl methyl acetylene

Isopropyl acetylene

1�Pentyne

2-Pentyne

3-Methyl- 1-butyneCH � CH � C C H3

C H3

H � C C � H

CH � C C � H 3

ETHYNE

It is also known as acetylene. Acetylene is the first

member of alkyne series and has a linear geometry.

It has sp hybridisation.The carbon-carbon triple bond

distance and carbon-hydrogen bond distance have

been found to be 1.20 Å and 1.06 Å respectively. The

carbon-carbon hydrogen bond angle is 180º.

(a) Structure :

Linear

C C

180º

H H

32323232PAGE # 32

(b) Properties :

It is a colourless gas which is slightly soluble in water.Pure ethyne has ethereal odour. Acetylene burns withluminous flame like aromatic compounds. This is ahighly exothermic reaction.

Note :The temperature of oxyacetylene flame is about

3000ºC and is used for welding and cutting steel.

(c) Preparation :

(i) From carbon and hydrogen (Direct synthesis ) : Whenan electric arc is struck between carbon (graphite) rodsin an atmosphere of hydrogen, acetylene is formed.

2C + H2

1200ºCC H2 2

(ii) From calcium carbide (Lab. Method) :

CaC + 2H O 2 2Ca(OH) + C H2 2 2

Calciumcarbide

Calciumhydroxide

Ethyne

(iii) Dehydrohalogenation of dihaloalkanes :

(d) Chemical Properties :

(i) Addition of halogens :

(ii) Addition of Halogen acid :

HC CH + HCl

Cl

Cl

ClEthyne

H C = CH2

HClH C � CH3

Chloroethene(Vinyl chloride)

1,1-Dichloroethane(Gem dihalide)

(iii) Hydrogenation :

HC CH

Ethyne

H C = CH2 2

H2

Ni

H /Ni2 CH � CH3 3

Ethene Ethane

HC CH

Ethyne

H2

Pd/BaSO4

H C = CH2 2

Ethene

(iv) Combustion :2C

2H

2 + 5O

2 4CO

2 + 2H

2O + Heat

Ethyne

(v) Polymerisation :

3HC CHFe

H

H

H

H

H

Hor (C H )6 6

Ethyne

Benzene

TEST FOR ALKANES, ALKENES AND ALKYNES

(a) Alkanes :

(i) Bromine water test: It does not decolourise thebromine water.

(ii) Baeyer�s test: It does not, react with Baeyer�sreagent (alkaline solution of KMnO

4).

(iii) Silver nitrate Test: No reaction

(b) Alkenes:

(i) Bromine water test: It decolourises the orangecolour of Bromine water.

C = C C � C + Br 2

H H H Br Br H

H H H H

EtheneBromine water(red-brown colour)

1,2-Dibromoethane(Colourless)

CCl4

(ii) Baeyer�s test: It decolourises the purple colour ofBaeyer�s reagent.

Ethylene glycol

C = C H O + [O] 2

H H

H H H � C � C � H

OH OH

H H

Ethene

(iii) Silver nitrate Test: No reaction

(c) Alkynes :

(i) Bromine water test : It decolourises the Br2 water.

H � C C � H + Br 2 Br � C � C � Br C = C Br

Br Br

Br 2

Br

H H H H

Ethyne Bromine water 1,2-Dibromoethene 1,1,2,2-Tetrabromoethane

(Colourless)

(ii) Baeyer�s test : It also decolourises the purplecolour of alkaline KMnO

4 .

(iii) Silver nitrate Test : It gives white precipitate

Tollen's reagent

White ppt.

+ 2NH NO + 2NH 4 3 3H � C C � H + 2 [Ag (NH ) ]NO 3 2 3Ag � C C � Ag

Ethyne

EXERCISE

1. The IUPAC name of the compound having the formula

(CH3)

3 CCH = CH

2 is -

(A) 3,3,3-Trimethyl -1-propane

(B) 1,1,1-Trimethyl-1-butene

(C) 3,3-Dimethyl-1-butene

(D) 1,1�Dimethyl- 3 -butene

2. Which of the following is not an example of aromaticcompound ?

(A) Benzene (B) Naphthalene

(C) Cyclobutane (D) Phenol

33333333PAGE # 33

3. The IUPAC name of the following compound is -CH

2 = CH � CH (CH

3)

2

(A) 1,1-Dimethyl -2-propene(B) 3-Methyl-1-butene(C) 2-Vinylpropane(D) 1-Isopropylethylene

4. Which of the following is an alkyne ?(A) C

4H

8(B) C

5H

8

(C) C7H

19(D) None of these

5. The IUPAC name of the following compound is -

CH � CH � C = CH3 2

C H2 5

CH3

(A) 3-Ethyl-2-methylbut-3-ene(B) 2-Ethyl-3- methylbut -1-ene(C) 2-Methyl-3-ethylbut-3-ene(D) 3-Methyl-2-ethylbut-1-ene

6. The first organic compound synthesized in thelaboratory was -(A) urea (B) glucose(C) alcohol (D) None of these

7. Propane is an -(A) acyclic compound(B) open chain compound(C) alipthatic compound(D) All of these

8. The scientist who gave vital force theory was -(A) Berzelius (B) Avogadro(C) Wohler (D) Lavoisier

9. Which one of the following is not an organic compound ?(A) Hexane (B) Urea(C) Ammonia (D)Ethyl alcohol

10. Vast number of carbon compounds is due to the factthat carbon has -(A) variable valency(B) property of catenation(C) great chemical affinity(D) None of these

11. The IUPAC name for CH � C � H3

||O

is -

(A) Acetal (B) Methanal(C) Ethanal (D) Acetaldehyde

12. The IUPAC name of compound

CH � C � CH � CH 3 2 3

H

COOH

is -

(A) Butan -3- oic acid(B) Butan -2- oic acid(C) 3-Methylbutanoic acid(D) 2-Methylbutanoic acid

13. The functional group, present in CH3COOC

2H

5 is -

(A) ketonic (B) aldehydic(C) ester (D) carboxylic

14. How many chain isomers (non- cyclic aliphatic) couldbe obtained from the alkane C

6H

14 ?

(A) 6 (B) 5(C) 4 (D) 3

15. The isomerism exhibited by n-propyl alcohol andisopropyl alcohol is -(A) chain isomerism(B) position isomerism(C) functional isomerism(D) None of these

16. Any two consecutive members in a homologousseries differ in molecular mass by-(A) 8 (B) 14(C) 24 (D) 12

17. The IUPAC name of-CH

3 � C(CH

3) (OH) CH

2 � CH(CH

3) CH

3 is -

(A) 2,4-Dimethylpentan -2-ol(B) 2,4- Dimethylpentan-4-ol(C) 2,2-Dimethylbutane(D) Isopentanol

18. The IUPAC name of (CH3)

2 CHCH

2 CH

2 Br is-

(A) 1-Bromopentane(B) 2-Methyl-4-bromopentane(C) 1-Bromo -3- methylbutane(D) 2-Methyl-3-bromopentane

19. Which of the following does not belong to homologousseries of alkanes ?(A) C

2H

6(B) C

3H

4

(C) C4H

10(D) C

5H

12

20. Isomers have -(A) same molecular formula & same structure.(B) different molecular formula & different structure.(C) same molecular formula & different structure.(D) different molecular formula & same structure.

21. Which of the following properties is not true regardingorganic compounds ?(A) They are generally covalent compounds.(B) They have high melting and boiling points.(C) They are generally insoluble in water.(D) They generally show isomerism.

22. The nature of linkage in organic compounds isgenerally -(A) ionic bond(B) covalent bond(C) co-ordinate covalent(D) metallic bond

23. Which of the following statements is incorrect ?The members of the homologous series of alkanes(A) are all straight chain compounds.(B) have the general formula C

nH

2n+2.

(C) have similar chemical properties .(D) show a regular gradation of physical properties.

34343434PAGE # 34

24. Which of the following pairs is an example of chainisomer ?

(A) CH3 � CH

2 � OH & CH

3OCH

3

(B) CH3 � CH

2 � CHO & CH

3 � CO-CH

3

(C)CH3�CH

2�CH

2�CH

2�CH

3 &

(D) All of the above

25. Which of the following is a functional isomer ofCH

3 � COOH ?

(A) CH3 � CH

2 � OH (B)

(C) (D) All of these

26. Which of the following forms a homologous series ?(A) Ethane, Ethylene, Acetylene(B) Ethane, Propane, Butanol(C) Methanal, Ethanol, Propanoic acid(D) Butane, 2-Methylbutane, 2,3-Dimethylbutane

27. Homologous have the same -(A) empirical formulae(B) molecular formulae(C) chemical properties(D) physical properties

28. Write down the general formula of homologous serieswhose third homologue is C

4H

6 ?

(A) CnH

2n � 2(B) C

n H

2n + 2

(C) Cn + 1

H2n � 2

(D) Cn H

2n

29. The general formula of saturated hydrocarbons is -(A) C

nH

2n(B) C

nH

2n+2

(C) CnH

2n�2(D) C

nH

2n+1

30. The third member of methyl ketone homologousseries is -(A) Acetone (B) 2�Butanone

(C) 2�Pentanone (D) 3�Pentanone.

31. The values of bond energies of single, double andtriple bonds are in the order -(A) C � C > C = C > C C(B) C = C > C � C > C C(C) C C > C = C > C � C

(D)C = C > C C > C � C

32. Ethyne on passing through a red hot iron tube gives -

(A) mesitylene (B) benzene

(C) butenyne (D) None

33. The general molecular formula of alkynes is -

(A) CnH

2n(B) C

nH

2n�4

(C) CnH

2n�2(D) C

nH

2n+2

34. Which of the following is not a pair of homologues ?

(A) Ethylacetylene and Dimethylacetylene

(B) Methylacetylene and Ethylacetylene

(C) 1-Butyne and 2-Pentyne

(D) 1-Pentyne and 3-Hexyne

35. Ethyne is obtained by dehydrobromination of -

(A) CHBr = CHBr (B) CH3CHBr

2

(C) CH3�CH

2Br (D) None of these

36. The temperature of oxy-acetylene flame is

(A) 200ºC (B) 1600ºC

(C) 2000ºC (D) > 2500º

37. Ethyne is isoelectronic with-

(A) chlorine (B) oxygen

(C) nitrogen gas (D) CO2

38. C C bond length is -

(A) 1.54 Å (B) 1.20 Å

(C) 1.34 Å (D) 1.39 Å

39. Which of the following gives silver nitrate test ?

(A) Methane (B) Ethene

(C) Ethyne (D) All

40. Which of the following does not decolourise bromine

water ?

(A) Alkanes only (B) Alkenes only

(C) Alkynes only (D) (B) and (C) both

41. Unsaturated fatty acids contain -

[IJSO-State-I/2011]

(A) atleast one double bond

(B) two double bonds

(C) more than two double bonds

(D) no double bond

35PAGE # 35

Pre-requisite : Before going through this chapter, youshould be thorough with the basic concepts of thesame chapter explained in IX NCERT.

LOGARITHM

If �a� is a positive real number, other than 1 and x is arational number such that ax = N, then x is thelogarithm of N to the base a.

If ax = N then loga N = x. [ Remember N will be +ve]

Systems of Logarithm :

There are two systems of logarithm which are generallyused.

(i) Common logarithm : In this system base is alwaystaken as 10.

(ii) Natural logarithm : In this system the base of thelogarithm is taken as �e�. Where �e� is an irrationalnumber lying between 2 and 3. (The approximate valueof e upto two decimal places is e = 2.73)

Some Useful Results :

(i) If a > 1 then(a) log

a x < 0 [for all x satisfying 0 < x < 1]

(b) loga x = 0 for x = 1

(c) loga x > 0 for x > 1

(d) x > y loga x > log

a y i.e. log

ax is an increasing

function.

Graph of y = loga x, a > 1

x'

y'

x0 (1,0)

y = log x, a > 1a

y

(ii) If 0 < a < 1, then(a) log

a x < 0 for all x > 1

(b) loga x = 0 for x = 1

(c) logax > 0 for all x satisfying 0 < x < 1

(d) x > y logax < log

a y i.e. log

a x is a decreasing

function.

Graph of y = loga x, 0 < a < 1.

x'

y'

x0(1,0)

y = log x, 0 < a < 1. a

y

NUMBER SYSTEM

Fundamental Laws of Logarithm :

Logarithm to any base a (where a > 0 and 1a ).

(i) loga a = 1

(ii) loga 0 = not defined

[As an = 0 is not possible, where n is any number]

(iii) loga (�ve no.) = not defined.

[As in loga N, N will always be (+ ve)]

(iv) loga (mn) = log

a m + log

an

[Where m and n are +ve numbers]

(v) loga

nm

= logam � log

an

(vi) loga(m)n = n log

am

(vii)ablog

mblogmalog

(viii) logam . log

ma = 1

(ix) If �a� is a positive real number and �n� is a positive

rational number, then

na nloga

(x) If �a� is a positive real number and �n� is a positive

rational number, then

qalog np nlogqp

a

(xi) plogqlog aa qp

(xii) logax = log

ay x = y

Ex.1 If log3a = 4, find value of a.

Sol. log3a = 4

a = 34

a = 81.

Ex.2 Find the value of 43

log3227

log�89

log

Sol. Given :

43

log3227

log�89

log 43

log3227

89

log

43

2732

89

log

= log1 = 0. [ loga1 = 0]


36PAGE # 36

Ex.3 If 2log4x = 1 + log

4(x � 1), find the value of x.

Sol. Given 2log4x = 1 + log

4(x � 1)

log4x2 � log

4(x � 1) = 1

log4 1�x

x2

= 1 41 = 1�x

x2

x2 = 4x � 4 x2 � 4x + 4 = 0

(x � 2)2 = 0 x = 2.

Ex.4 Evaluate : 5log�2 33 .

Sol. Given 5log�2 33 = 5log�2 33.3 [am + n = am.an]

= 9.1�

3 5log3

= 9 × 5�1

= 59

.

Ex.5 If A = log27

625 + 13log117 and B = log9125 + 7log1113 ,

then find the relation between A and B.

Sol. A = log27

625 + 13log117 = 4

35log 3 +

13log117

or, A = 34

log35 + 13log117 ....(i)

and,B = log9125 + 7log1113

or, B = 33

5log 2 + 13log117

or, B = 23

log35 + 13log117 ...(ii)

By (i) and (ii) we have,

A � 34

log35 = B �

23

log35

3

4log

35 <

2

3log

35

A < B.

Ex.6 Find the value of log25

125 � log84

Sol. Given, log25

125 � log84

= 22

35

2log�5log 32

= 23

log55 �

32

log2 2

= 23�

32

[ 1aloga ]

= 65

.

FACTORS AND MULTIPLES

Factors : �a� is a factor of �b� if there exists a relation

such that a × n = b, where �n� is any natural number.

1 is a factor of all numbers as 1 × b = b.

Factor of a number cannot be greater than the number

(in fact the largest factor will be the number itself).

Thus factors of any number will lie between 1 and the

number itself (both inclusive) and they are limited.

Multiples : �a� is a multiple of �b� if there exists a relation

of the type b × n = a. Thus the multiples of 6 are

6 × 1 = 6, 6 × 2 = 12, 6 × 3 = 18, 6 × 4 = 24, and so on.

The smallest multiple will be the number itself and the

number of multiples would be infinite.

NOTE :

To understand what multiples are, let�s just take an

example of multiples of 3. The multiples are 3, 6, 9, 12,

.... so on. We find that every successive multiples

appears as the third number after the previous.

So if one wishes to find the number of multiples of 6

less than 255, we could arrive at the number through

6255

= 42 (and the remainder 3). The remainder is of

no consequence to us. So in all there are 42 multiples.

If one wishes to find the multiples of 36, find 36

255 = 7

(and the remainder is 3).

Hence, there are 7 multiples of 36.

Ex.7 How many numbers from 200 to 600 are divisible by

4, 5, 6 ?

Sol. Every such number must be divisible by L.C.M. of

(4, 5, 6) = 60.

60600

�

60200

= 10 � 3 = 7.

Such numbers are 240, 300, 360, 420, 480, 540 and

600.

Clearly, there are 7 such numbers.

Factorisation : It is the process of splitting any number

into a form where it is expressed only in terms of the

most basic prime factors.

For example, 36 = 22 × 32. It is expressed in the

factorised form in terms of its basic prime factors.

Number of factors : For any composite number C,

which can be expressed as C = ap × bq × cr ×....., where

a, b, c ..... are all prime factors and p, q, r are positive

integers, the number of factors is equal to

(p + 1) × (q + 1) × (r + 1).... e.g. 36 = 22 × 32. So the

factors of 36 = (2 +1) × (2 + 1) = 3 × 3 = 9.

37PAGE # 37

Ex.8 If N = 123 × 34 ×52, find the total number of even factorsof N.

Sol. The factorised form of N is(22 × 31)3 × 34 × 52 26 × 37 × 52.Hence, the total number of factors of N is(6 + 1) (7 + 1) (2 + 1) = 7 × 8 × 3 = 168.

Some of these are odd multiples and some are even.The odd multiples are formed only with the combinationof 3s and 5s.So, the total number of odd factors is(7 + 1) (2 + 1) = 24.Therefore, the number of even factors is168 � 24 = 144.

Ex.9 A number N when factorised can be writtenN = a4 × b3 × c7. Find the number of perfect squareswhich are factors of N (The three prime numbersa, b, c > 2).

Sol. In order that the perfect square divides N, the powersof �a� can be 0, 2 or 4, i.e. 3.

Powers of �b� can be 0, 2, i.e. 2. Power of �c� can be 0, 2,

4 or 6, i.e. 4.Hence, a combination of these powers given 3 × 2 × 4

i.e. 24 numbers.So, there are 24 perfect squares that divides N.

Ex.10 Directions : (i to iv) Answer the questions based onthe given information.There are one thousand lockers and one thousandstudents in a school. The principal asks the first studentto go to each locker and open it. Then he asks thesecond student go to every second locker and close it.The third student goes to every third locker, and if it isclosed, he opens it, and it is open, he closes it. Thefourth student does it to every fourth locker and so on.The process is completed with all the thousandstudents.(i) How many lockers are closed at the end of theprocess ?(ii) How many students can go to only one locker ?(iii) How many lockers are open after 970 studentshave done their job ?(iv) How many student go to locker no. 840 ?

Sol. (i to iv) : Whether the locker is open or not depends onthe number of times it is accessed. If it is accessed oddnumber of times, then it is open while if it is accessedeven number of times then it is closed.How many times a locker will be accessed dependson the locker no. If it contains odd number of factors,then it will be open and if it contains even number offactors. Then it will be closed. We know that a perfectsquare contains odd number of factors while anon-perfect square contains even number of factors.Thus the lockers with perfect square number will beopen and the number of these perfect squares from 1to 1000 determines the no. of open lockers.

(i) No. of closed lockers = No. of non-perfect squarenumbers from 1 to 1000 = 1000 � 31 = 969.

(ii) Upto 500 students they can go to two or more than twolockers, while the rest 500 can go to only one locker.(iii) The 31 perfect squares ( the last being 312 = 961)will be open while the lockers from 971 to 1000 is yetto be accessed last time so they all are open. The totalbeing = 31 + 30 = 61(iv) The no. of students that have gone to locker no.840 is same as the no. of factors of 840.

840 = 23 × 3 × 5 × 7.

So, the no. of factors = (3 + 1) (1 + 1) (1 + 1) (1 + 1) = 32.

HCF AND LCM

LCM (least Common Multiple) : The LCM of givennumbers, as the name suggests is the smallestpositive number which is a multiple of each of the givennumbers

HCF (Highest Common factor) : The HCF of givennumbers, as the name suggests is the largest factorof the given set of numbers.

Consider the numbers 12, 20 and 30. The factors andthe multiples are

FactorsGiven

numbersMultiples

1, 2, 3, 4, 6, 12 12 12, 24, 36, 48, 60, 72, 84, 96, 108, 120....

1, 2, 4, 5, 10, 20 20 20, 40, 60, 80, 100, 120.....

1, 2, 3, 5, 6, 10, 15, 30 30 30, 60, 90, 120....

The common factors are 1 and 2 and the commonmultiples are 60, 120...Thus the highest common factor is 2 and the leastcommon multiple meaning of HCF it is the largestnumber that divides all the given numbers.Also since a number divides its multiple, the meaningof LCM is that it is the smallest number which can bedivided by the given numbers.

HCF will be lesser than or equal to the least of thenumbers and LCM will be greater than or equal to thegreatest of the numbers.

Ex.11 Find a number greater than 3 which when divided by4, 5, and 6 always leaves the same remainder 3.

Sol. The smallest number which, when divided by 4, 5 and6, leaves the remainder 3 in each case isLCM (4, 5 and 6) + 3 = 60 + 3 = 63.

Ex.12 In a school 437 boys and 342 girls have been dividedinto classes, so that each class has the same numberof students and no class has boys and girls mixed.What is the least number of classes needed?

Sol. We should have the maximum number of students ina class. So we have to find HCF (437, 342) = 19.HCF is also the factor of difference of the number.

Number of classes = 19437

+ 19342

` = 23 + 18 = 41 classes.

38PAGE # 38

For any two numbers x and y,

x × y = HCF (x, y) × LCM (x, y).

HCF and LCM of fractions :

LCM of fractions = atorsmindenoofHCFnumeratorsofLCM

HCF of fractions = atorsmindenoofLCMnumeratorsofHCF

Make sure the fractions are in the most reducible form.

Ex.13 Find the H.C.F. and L.C.M. of 98

, 8116

, 32

and 2710

.

Sol. H.C.F. of given fractions = )27,81,9,3(of.M.C.L)10,16,8,2(of.F.C.H

=812

,

L.C.M. of given fractions = )27,81,9,3(of.F.C.H)10,16,8,2(of.M.C.L

=3

80.

Ex.14 Find the least number which when divided by 6, 7, 8, 9

and 10 leaves remainder 1.

Sol. As the remainder is same

Required number = LCM of divisors + Remainder

= LCM (6, 7, 8, 9, 10) +1

= 2520 + 1

= 2521.

Ex.15 Six bells start tolling together and they toll at intervals

of 2, 4, 6, 8, 10, 12 sec. respectively, find

(i) after how much time will all six of them toll together ?

(ii) how many times will they toll together in 30 min ?

Sol. The time after which all six bells will toll together must

be multiple of 2, 4, 6, 8, 10, 12.

Therefore, required time = LCM of time intervals.

= LCM (2, 4, 6, 8, 10, 12) = 120 sec.

Therefore after 120 s all six bells will toll together.

After each 120 s, i.e. 2 min, all bell are tolling together.

Therefore in 30 min they will toll together

1

230

= 16 times

1 is added as all the bells are tolling together at the

start also, i.e. 0th second.

Ex.16 LCM of two distinct natural numbers is 211. What is

their HCF ?

Sol. 211 is a prime number. So there is only one pair of

distinct numbers possible whose LCM is 211,

i.e. 1 and 211. HCF of 1 and 211 is 1.

Ex.17 An orchard has 48 apple trees, 60 mango trees and

96 banana trees. These have to be arranged in rows

such that each row has the same number of trees and

all are of the same type. Find the minimum number of

such rows that can be formed.

Sol. Total number of trees are 204 and each of the trees

are exactly divisible by 12. HCF of (48, 60, 96).

12204

= 17 such rows are possible.

DIVISIBLITY

Division Algorithm : General representation of result

is,

DivisormainderRe

QuotientDivisor

Dividend

Dividend = (Divisor × Quotient ) + Remainder

Ex.18 On dividing 15968 by a certain number, the quotient

is 89 and the remainder is 37. Find the divisor.

Sol. Divisor = 89

3715968Quotient

mainderReDividend

= 179.

NOTE :

(i) (xn � an) is divisible by (x � a) for all the values of n.

(ii) (xn � an) is divisible by (x + a) and (x � a) for all the

even values of n.

(iii) (xn + an) is divisible by (x + a) for all the odd values of n.

Test of Divisibility :

No. Divisiblity Test

2 Unit digit s hould be 0 or even

3 The s um of digits of no. s hould be divis ible by 3

4 The no form ed by las t 2 digits of given no. s hould be divis ible by 4.

5 Unit digit s hould be 0 or 5.

6 No s hould be divis ible by 2 & 3 both

8 The num ber form ed by las t 3 digits of given no. s hould be divis ible by 8.

9 Sum of digits of given no. s hould be divis ible by 9

11The difference between s um s of the digits at even & at odd placesshould be zero or m ultiple of 11.

25 Las t 2 digits of the num ber s hould be 00, 25, 50 or 75.

Rule for 7 : Double the last digit of given number and

subtract from remaining number the result should be

zero or divisible by 7.

Ex.19 Check whether 413 is divisible by 7 or not.

Sol. Last digit = 3, remaining number = 41, 41 � (3 x 2) = 35

(divisible by 7). i.e. 413 is divisible by 7.

This rule can also be used for number having more

than 3 digits.


Sol. Last digit = 5, remaining number 654, 654 � (5 x 2)

= 644; 64 � (4 x 2) = 56 divisible by 7. i.e. 6545 is

divisible by 7.

Rule for 13 : Four times the last digit and add to

remaining number the result should be divisible by

13.

Ex.21 Check whether 234 is divisible by 13 or not .

Sol. 234, (4 x 4) + 23 = 39 (divisible by 13), i.e. 234 is divisible

by 13.

39PAGE # 39

Rule for 17 : Five times the last digit of the number and

subtract from previous number the result obtained

should be either 0 or divisible by 17.


Sol. 357, (7 x 5) � 35 = 0, i.e. 357 is divisible by 17.

Rule for 19 : Double the last digit of given number and

add to remaining number The result obtained should

be divisible by 19.


Sol. 589, (9 x 2) + 58 = 76 (divisible by 19), i.e. the number

is divisible by 19.

Ex.24 Find the smallest number of six digits which is exactly

divisible by 111.

Sol. Smallest number of 6 digits is 100000.

On dividing 100000 by 111, we get 100 as remainder.

Number to be added = (111 � 100) = 11.

Hence, required number = 100011.

Ex.25 Find the largest four digit number which when

reduced by 54, is perfectly divisible by all even natural

numbers less than 20.

Sol. Even natural numbers less than 20 are 2, 4, 6, 8, 10,

12, 14, 16, 18.

Their LCM = 2 × LCM of first 9 natural numbers

= 2 × 2520 = 5040.

This happens to be the largest four-digit number

divisible by all even natural numbers less than 20. 54

was subtracted from our required number to get this

number.

Hence, (required number � 54) = 5040

Required number = 5094.

Ex.26 Ajay multiplied 484 by a certain number to get the

result 3823a. Find the value of �a�.

Sol. 3823a is divisible by 484, and 484 is a factor of 3823a.

4 is a factor of 484 and 11 is also a factor of 484.

Hence, 3823a is divisible by both 4 and 11.

To be divisible by 4, the last two digits have to be

divisible by 4.

�a� can take two values 2 and 6.

38232 is not divisible by 11, but 38236 is divisible by

11.

Hence, 6 is the correct choice.

Ex.27 Which digits should come in place of and $ if the

number 62684$ is divisible by both 8 and 5 ?

Sol. Since the given number is divisible by 5, so 0 or 5 must

come in place of $. But, a number ending with 5 in

never divisible by 8. So, 0 will replace $.

Now, the number formed by the last three digits is 40,

which becomes divisible by 8, if is replaced by 4 or 8.

Hence, digits in place of and $ are (4 or 8 or 0) and 0

respectively.

REMAINDERS

The method of finding the remainder without actually

performing the process of division is termed as

remainder theorem.

Remainder should always be positive. For example if

we divide �22 by 7, generally we get �3 as quotient

and �1 as remainder. But this is wrong because

remainder is never be negative hence the quotient

should be �4 and remainder is +6. We can also get

remainder 6 by adding �1 to divisor 7 ( 7�1 = 6).

Ex.28 Two numbers, x and y, are such that when divided by

6, they leave remainders 4 and 5 respectively. Find the

remainder when (x2 + y2) is divided by 6.

Sol. Suppose x = 6k1 + 4 and y = 6k

2 + 5

x2 + y2 = (6k1 + 4)2 + (6k

2 + 5)2

= 36k1

2 + 48k1 + 16 + 36k

22 + 60k

2 + 25

= 36k1

2 + 48k1 + 36k

22 + 60k

2 + 41

Obviously when this is divided by 6, the remainder will

be 5.

Ex.29 A number when divided by 259 leaves a remainder

139. What will be the remainder when the same

number is divided by 37 ?

Sol. Let the number be P.

So, P � 139 is divisible by 259.

Let Q be the quotient then, 259

139P = Q

P = 259Q + 139

37P

= 37

139Q259

259 is divisible by 37,

When 139 divided by 37, leaves a remainder of 28.

Ex.30 A number being successively divided by 3, 5 and 8

leaves remainders 1, 4 and 7 respectively. Find the

respective remainders if the order of divisors be

reversed.

Sol.

714z81y5

x3

z = (8 × 1 + 7) = 15 ; y = (5z + 4) = (5 × 15 + 4) = 79 ;

x = (3y + 1) = (3 × 79 + 1) = 238.

Now,

214536295

2388

Respective remainders are 6, 4, 2.

40PAGE # 40

Ex.31 A number was divided successively in order by 4, 5and 6. The remainders were respectively 2, 3 and 4.Then find out the number.

Sol.

413z62y5

x4

z = (6 × 1 + 4) = 10

y = (5 × z + 3) = (5 × 10 + 3) = 53

x = (4 × y + 2) = (4 × 53 + 2) = 214

Hence, the required number is 214.

Ex.32 In dividing a number by 585, a student employed themethod of short division. He divided the numbersuccessively by 5, 9 and 13 (factors of 585) and got theremainders 4, 8 and 12. If he had divided number by585, then find out the remainder.

Sol.

1218z134y9

x5

Now, 1169 when divided by 585 gives remainder= 584.

To find the remainder of big number

NOTE :

(i) Binomial Expansion :

(a + b)n = an + 1!

nan�1b +

2!

1)n(n an � 2b2 + .... + bn, or

(a � b)n = an � 1!

nan�1b +

2!

1)n(n an� 2b2 � ......+ (� 1)nbn.

Hence, first term is purely of a i.e an and last digit ispurely of b, i.e. bn.

(ii) Total number of terms in the expansion of (a + b)n is(n + 1).

Ex.33 What is the remainder when 738 is divided by 48.

Sol.48738

=

487

192

=

4849 19

=

48148 19

so by using

binomial expansion, we can say that 18 terms arecompletely divisible by 48 but the last term which is

481 19

is not divisible. So, 191 = 1 is the remainder..

Ex.34 What is the remainder if 725 is divided by 4?Sol. 725 can be written (8�1)25. There are 26 terms in all and

first 25 terms are divisible by 8, hence also by 4. Thelast term is (�1)25. Hence, (8 �1)25 can be written8X � 1 or 4Y �1 ( where Y = 2X). So, 4Y � 1 divided by

4 leaves the remainder 3.

Ex.35 What is the remainder if 345 is divided by 8 ?Sol. 345 can be written as 922 × 3. 9 can be written as (8 + 1).

Hence, any power of 9 can be written as 8N + 1. Inother words, any power of 9 is 1 more than a multipleof 8. Hence, (8N + 1) × 3 leaves remainder 3 when

divided by 8.

Ex.36 What is the remainder when 161514 is divided by 5 ?

Sol.161514 = (15 �1)odd = 15n + (�1)odd, i.e. a (multiple of 5)

�1. Thus when divided by 5 the remainder will be (�1),

i.e. 4.

Ex.37 What is the remainder when 357 + 27 is divided by28?

Sol. 357 = (33)19

357 + 27 = (27)19 + 27= (28 � 1)19 + 27= 28M + (�1)19 + 27 [Expand by binomial theorem]= 28M � 1 + 27

= 28M + 26When 28M + 26 divided by 28, the remainder is 26.Hence, the required remainder is 26.

Ex.38 What is the remainder when 82361 + 83361 + 84361

+ 85361 + 86361 is divided by 7?Sol. 82361 + 83361 + 84361 + 85361 + 86361 = [(84 � 2)361

+ (84 � 1)361 + 84361 + (84 + 1)361 + (84 + 2)361]Since, 84 is a multiple of 7, then the remainder will bewhen, (� 2)361 + (�1)361 + 1361 + 2361 is divided by 7 is(� 2)361 + (�1)361 + 1361 + 2361 = 0. So the remainder iszero.

CYCLICITY

We are having 10 digits in our number systems andsome of them shows special characteristics like they,repeat their unit digit after a cycle, for example 1 repeatits unit digit after every consecutive power. So, itscyclicity is 1 on the other hand digit 2 repeat its unitdigit after every four power, hence the cyclicity of 2 isfour. The cyclicity of digits are as follows :

Digit Cyclicity

0, 1, 5 and 6 1

4 and 9 2

2, 3, 7 and 8 4

So, if we want to find the last digit of 245, divide 45 by 4.The remainder is 1 so the last digit of 245 would besame as the last digit of 21 which is 2.

To Find the Unit Dig i t in Expone nt ia l

Expre s s ions :

(i) When there is any digit of cyclicity 4 in unit�s place.

Since, when there is 2 in unit�s place then in 21 unitdigit is 2, in 22 unit digit is 4, in 23 unit digit is 8, in 24 unitdigit is 6, after that the unit�s digit repeats. e.g. unit digit

(12)12 is equal to the unit digit of, 24 i.e.6.

Ex.39 In (32)33 unit digit is equal to the unit digit of 21 i.e. 2.




41PAGE # 41

(ii) When there is any digit of cyclicity 2 in unit�s place.

Since, when there is 4 in unit�s place then in 41 unit

digit is 4, in 42 unit digit is 6 and so on.

Ex.43 In (34)33 unit digit is 4.



(iii) When there is any digit of cyclicity 1 in unit�splace.Since, when there is 5 in unit�s place then in 51 unit

digit is 5, in 52 unit digit is 5 and so on.



Ex.48 Find the last digit of

(i) 357 (ii) 1359

Sol. (i) The cyclicity of 3 is 4. Hence, 4

57gives the remainder

1. So, the last digit of 357 is same as the last digit of 31,

i.e. 3.

(ii) The number of digits in the base will not make a

difference to the last digit. It is last digit of the

base which decides the last digit of the number itself.

For 1359, we find 4

59 which gives a remainder 3. So

the last digit of 1359 is same as the last digit of 33, i.e. 7.

Ex.49 Find unit�s digit in y = 717 + 734

Sol. 717 + 734 = 71 + 72 = 56, Hence the unit digit is 6

Ex.50 What will be the last digit of 766475)73(

Sol. Let 766475)73( = (73)x where x =

766475 = (75)even power

Cyclicity of 3 is 4

To find the last digit we have to find the remainder

when x is divided by 4.

x = (75)even power = (76 � 1)even power , where n is divided by

4 so remainder will be 1.

Therefore, the last digit of 766475)73( will be 31 = 3.

Ex.51 What will be the unit digit of 556375)87( .

Sol. Let 556375)87( = (87)x where x =

556375 = (75)odd

Cyclicity of 7 is 4.

To find the last digit we have to find the remainder

when x is divided by 4.

x = (75)odd power = (76 � 1)odd power

where x is divided by 4 so remainder will be �1 or 3, but

remainder should be always positive.

Therefore, the last digit of 556375)87( will be 73 = 343.

Hence, the last digit is of 556375)87( is 3.

HIGHEST POWER DIVIDING A FACTORIAL

Factorial n : Product of n consecutive natural numbers

is known as �factorial n� it is denoted by �n!�.

So, n! = n(n � 1)(n � 2)...321. e.g. 5! = 5 × 4 × 3 × 2 × 1 = 120.

The value of factorial zero is equal to the value of

factorial one. Hence 0! = 1 = 1!

The approach to finding the highest power of x dividing

y! is

32 x

y

x

yxy

......., where [ ] represents just

the integral part of the answer and ignoring the fractional

part.

Ex.52 What is the highest power of 2 that divides 20!

completely?

Sol. 20! = 1 × 2 × 3 × 4 ×....× 18 × 19 × 20 = 1 × (21) × 3 × (22)

× 5 × (21 × 31) × 7 × (23) × ..... so on. In order to find the

highest power of 2 that divides the above product, we

need to find the sum of the powers of all 2 in this

expansion. All numbers that are divisible by 21 will

contribute 1 to the exponent of 2 in the product

12

20 = 10. Hence, 10 numbers contribute 21 to the

product. Similarly, all numbers that are divisible by

22 will contribute an extra 1 to the exponent of 2 in the

product, i.e 22

20 = 5. Hence, 5 numbers contribute an

extra 1 to exponents. Similarly, there are 2 numbers

that are divisible by 23 and 1 number that is divisible

by 24. Hence, the total 1s contributed to the exponent

of 2 in 20! is the sum of ( 10 + 5 +2 +1) = 18. Hence,

group of all 2s in 20! gives 218 x (N), where N is not

divisible by 2.

If 20! is divided by 2x then maximum value of x is 18.

Ex.53 What is the highest power of 5 that divides of

x = 100! = 100 × 99 × 98 × ...... × 3 × 2 × 1.

Sol. Calculating contributions of the different powers of 5,

we have 15

100= 20, 25

100 = 4.

Hence, the total contributions to the power of 5 is 24, or

the number 100! is divisible by 524.

Ex.54 How many zeros at the end of first 100 multiples

of 10.

Sol. First 100 multiple of 10 are = 10 × 20 × 30 × ......× 1000

= 10100 (1 × 2 × 3 × .......× 100)

= 10100 × 1024 × N

= 10124 × N

Where N is not divisible by 10

So, there are 124 zero at the end of first 100 multiple of

10.

42PAGE # 42

Ex.55 What is the highest power of 6 that divides 9!

Sol. By the normal method. 69

= 1 and 26

9 = 0. Thus

answers we get is 1 which is wrong. True there is just

one multiple of 6 from 1 to 9 but the product 2 × 3 = 6

and also 4 × 9 = 36, can further be divided by 6. Thus,

when the divisor is a composite number find the

highest power of its prime factors and then proceed. In

this case, 9! can be divided by 27 and 34 and thus by 64

(In this case we need not have checked power of 2 as

it would definitely be greater than that of 3).

Ex.56 What is the largest power of 12 that would divide 49! ?

Sol. To check the highest power of 12 in 49!, we need to

check the highest powers of 4 and 3 in it.

Highest power of 3 in 49! = 22



46 = 23

Highest power of 12 will be 22. (Since the common

power between 3 and 4 is 22).

Ex.57 How many zeros will be there at the end of !36!36 ?

Sol. Highest power of 5 in 36! is 8.

So, there will be 8 zeros at the end of 36!.

So, at the end of !36!36 , there will be 8 × 36! zeros.

BASE SYSTEM

The number system that we work in is called the

�decimal system�. This is because there are 10 digits

in the system 0-9. There can be alternative system that

can be used for arithmetic operations. Some of the

most commonly used systems are : binary, octal and

hexadecimal.

These systems find applications in computing.

Binary system has 2 digits : 0, 1.

Octal system has 8 digits : 0, 1,..., 7.

Hexadecimal system has 16 digits : 0, 1, 2,..., 9, A , B,

C, D, E, F.

After 9, we use the letters to indicate digits. For instance,

A has a value 10, B has a value 11, C has a value 12,...

so on in all base systems.

The counting sequences in each of the systems would

be different though they follow the same principle.

Conversion : Conversion of numbers from (i) decimal

system to other base system. (ii) other base system to

decimal system.

(i) Conversion from base 10 to any other base :

Ex.58 Convert (122)10

to base 8 system.

Sol.

10

718

2158

1228

The number in decimal is consecutively divided by the

number of the base to which we are converting the

decimal number. Then list down all the remainders in

the reverse sequence to get the number in that base.

So, here (122)10

= (172)8.

Ex.59 Convert (169)10

in base 7.

Sol. 03241697

77

133

Remainder

(169)10 =(331)7

Ex.60 Convert (0.3125)10

to binary equivalent.

Sol. Integer

2 0.3125 = 0.625 0

2 0.625 = 1.25 1

2 0.25 = 0.50 0

2 0.50 = 1.00 1

Thus

(0.3125)10

= (0.1010)2

Ex.61 Convert (1987.725)10

(........)8

Sol. First convert non-decimal part into base 8.

30

738

0318

32488

19878

(1987)10

= (3703)8

Now we have to convert (0.725)10 (........)

8

Multiply

0.725 × 8 = [5.8] ...5

0.8 × 8 = [6.4]

0.4 × 8 = [3.2]

0.2 × 8 = [1.6]

0.6 × 8 = [4.8]

...6

...3

...1

...4

Keep on accomplishing integral parts after

multiplication with decimal part till decimal part is zero.

(0.725)10

= (0.56314...)8

(1987.725)10

= (3703.56314...)8

43PAGE # 43

(ii) Conversion from any other base to decimalsystem :

Ex.62 Convert (231)8 into decimal system.

Sol. (231)8 , the value of the position of each of the numbers

( as in decimal system) is :1 = 80 × 1

3 = 81 × 3

2 = 82 × 2

Hence, (231)8 = (80 × 1 + 81 × 3 + 82 × 2)

10

(231)8

= (1 + 24 + 128)10

(231)8

= (153)10

Ex.63 Convert (0.03125)10

to base 16.Sol. 16 0.03125 = 0.5 0

16 0.5 = 8.0 8So (0.03125)

10 = (0.08)

16

Ex.64 Convert (761.56)8 (......)

16

Sol. In such conversion which are standard formconversions, it is easier to(761.56)

8 (.....)

2 (.....)

16

Converting every digit in base 8 to base 2,(111110001.101110)

2 (1F1.B8)

16

Ex.65 Convert (3C8.08)16

to decimalSol. (3C8.08)

16 = 3 162 + C 161 + 8 16 + 0 16�1 + 8 16�2

= 768 + 192 + 8 + 0 + 0.03125 = (968.03125)

10

So, (3C8.08)16

= (968.03125)10

ALPHA NUMERICS NUMBERS

Ex.66 If a � b = 2, and

0cc

bb

aa

then find the value of a, b and c.

Sol. These problems involve basic number(i) aa + bb = 11(a + b) (ii) aa, bb are two-digit numbers.Hence, their sum cannot exceed 198. So, c must be 1.(iii) Hence, cc0 = 110. This implies a + b = 10 or a = 6and b = 4.Such problems are part of a category of problems calledalpha numerics.

Ex.67 If _____9aa

ca

b3a

then find a, b and c if each of them is

distinctly different digit.Sol. (i) since the first digit of (a 3 b) is written as it is after

subtracting ac carry over from a to 3.(ii) there must be a carry over from 3 to b, because if nocarry over is there, it means 3 � a = a.

2a = 3 a = 23

which is not possible because a is a digit. For a carryover 1, 2 � a = a

a = 1(iii) it means b and c are consecutive digit (2, 3),(3, 4),.... (8, 9)

Ex.68 The sum of first n natural numbers is a three-digit

number, all of whose digits are the same. What is the

value of n?

Sol. In 5 seconds, you can solve the equation

2)1n(n

= aaa (111, 222, etc) . How do you proceed

next ? If you think it's hit-and-trial from this point, you

are wrong. Here goes the simple logic. It might strike

you instantly if you have been working with numbers:

2)1n(n

= aaa = a × 111 = a × 3 × 37

n(n + 1) = 6a x 37

Look at the L.H.S. of the equation, n(n + 1) is a product

of two consecutive natural numbers. Therefore, R.H.S.

should also be a product of two consecutive natural

numbers. One of the numbers is 37. Therefore, what

could the other number 6a, consecutive to 37 be? It can

only be 36, giving a = 6 and n = 36. Therefore, 36

numbers have been summed up and their sum is equal

to 666.

Ex.69 If ABC x CBA = 65125, where A, B and C are single

digits, then A + B + C = ?

Sol. As the unit digit of the product is 5, therefore, the unit

digit of one of the numbers is 5 and the unit digit of the

other number is odd. Therefore, AB5 x 5BA = 65125,

where A = 1, 3, 5, 7 or 9.

As the product of two three-digit numbers is a five-digit

number, and not a six-digit number, A can only be equal

to 1. IB5 x 5B1 = 65125.

The digit sums of both numbers, 1B5 and 5B1 will be

same. Therefore, the product would give digit sum of a

perfect square. The digit sum on the R.H.S. is 1.

Therefore, the digit sum of each number can be 1or 8.

Correspondingly B will be 4 or 2 (as digit sum cannot

be equal to 1).

Keeping B = 2, we can see that 125 x 521 = 65125.

Ex.70 Find the four-digit number ABCD such that

ABCD x 4 = DCBA.

Sol. Any number multiplied by 4 will give us an even number.

Hence, the digit D when multiplied by 4 will give us an

even number.

Since A is the unit digit of the product it is even. Hence,

A = 2, 4, 6 or 8 (It cannot be 0). A is also the first digit of

the multiplicand and if A = 4, 6 or 8 the product

ABCD x 4 will become a 5 digit number. Hence A = 2.

Writing the value of A we get 2BCD x 4 = DCB2.

44PAGE # 44

Now for the value of D looking at the first and last digits

of the multiplicand, we can see that 4 x D gives the unit

digit of 2 and 4 x 2 gives the first digit of D. Yes, you got

it right D = 8. Writing the multiplication again with the

value of D we get 2BC8 x 4 = 8CB2.

Now for the value of B. A number is divisible by 4 if the

number formed by the last two digits is divisible by 4.

Since the number 8CB2 is a multiple of 4, the number

B2 should be divisible by 4. Or, the number B2 = 12,

32, 52, 72 or 92. Hence the original number ABCD is

21C8, 23C8, 25C8, 27C8 or 29C8. But the last 4

numbers when multiplied by 4 will not give you the first

digit of 8 in the product. Therefore B = 1 and the original

number is 21C8. We write the multiplication again

21C8 x 4 = 8C12.

Now for the value of C notice that when you multiply 8,

the unit digit of 21C8, by 4 you write 2 in the unit digit of

the product and carry 3. The tenth digit of the product is

1. Therefore, 4 x C + 3 (carry over) gives a unit digit of 1.

Hence, C is 2 or 7. You can easily check by the hundreds

digit in the product (which is C again) that C = 7.

Therefore, our answer is 2178 x 4 = 8712.

1. Convert 0.225 in to form p/q.

(A) 103

(B) 409

(C) 509

(D) 4009

2. When (55)10

is represented in base 25 then the

expression is :

(A) (25)25

(B) (35)25

(C) (55)25

(D) none of these

3. There are four prime numbers written in ascending

order. The product of the first three is 385 and that of

the last three is 1001. The last number is :

(A) 11 (B) 13

(C) 17 (D) 19

4. If logxy = 100 and log

2x = 10, then the value of y is :

(A) 21000 (B) 2100

(C) 22000 (D) 210000

5. Find the value of �x� if 2logx 7 + log

7x 7 + 3log

49x 7 = 0

(A) x = 34

(B) x = 7�1/2

(C) x = 7�4/3 (D) Either (B) or (C)

6. How many numbers between 200 and 600 are divisible

by 14?

(A) 28 (B) 29

(C) 27 (D) None of these

7. The number of prime factors of (3 × 5)12 (2 × 7)10 (10)25

is :

(A) 47 (B) 60

(C) 72 (D) 94

8. How many three-digit numbers would you find, which

when divided by 3, 4, 5, 6, 7 leave the remainders 1, 2,

3, 4, and 5 respectively ?

(A) 4 (B) 3

(C) 2 (D) 1

9. Six strings of violin start vibrating simultaneously and

they vibrate at 3, 4, 5, 6,10 and 12 times in a minute,

find :

i. After how much time will all six of them vibrate

together ?

ii. How many times will they vibrate together in 30 min ?

(A) 60 min, 31 times (B) 60 sec, 31 times

(C) 120 sec, 15 times (D) None of these

10. The HCF of 2 numbers is 11 and their LCM is 693.

If their sum is 176, find the numbers.

(A) 99,77 (B) 110, 66

(C) 88,77 (D) 121, 44

11. If P is a prime number, then the LCM of P and (P + 1) is

(A) P(P +1) (B) (P + 2)P

(C) (P + 1)(P � 1) (D) None of these

12. Find out (A + B + C + D) such that AB x CB = DDD, where

AB and CB are two-digit numbers and DDD is a three-

digit number.

(A) 21 (B) 19

(C) 17 (D) 18

13. Three pieces of cakes of weights 21

4 Ibs, 43

6 Ibs and

51

7 Ibs respectively are to be divided into parts of equal

weights. Further, each must be as heavy as possible.

If one such part is served to each guest, then what is

the maximum number of guests that could be

entertained ?

(A) 54 (B) 72

(C) 20 (D) 41

14. How many natural numbers between 200 and 400 are

there which are divisible by

i. Both 4 and 5?

ii. 4 or 5 or 8 or 10 ?

(A) 9, 79 (B) 10, 80

(C) 10, 81 (D) None of these

45PAGE # 45

15. 461 + 462 + 463 + 464 is divisible by :

(A) 3 (B) 10

(C) 11 (D) 13

16. If x is a whole number, then x2 (x2 � 1) is always divisible

by :

(A) 12 (B) 24

(C) 12 � x (D) Multiple of 12

17. If 653 xy is exactly divisible by 80, then the find the value

of (x + y).

(A) 2 (B) 3

(C) 4 (D) 6

18. Find the unit digit of (795 � 358).

(A) 6 (B) 4


19. When a number P is divided by 4 it leaves remainder

3. If the twice of the number P is divided by the same

divisor 4 than what will be the remainder ?

(A) 0 (B) 1

(C) 2 (D) 6

20. If (232 +1) is divisible by a certain number then which of

the following is also divisible by that number.

(A) (216 � 1) (B) 216 + 1

(C) 296 + 1 (D) None of these

21. If the number 357y25x is divisible by both 3 and 5, then

find the missing digit in the unit�s place and the

thousand place respectively are :

(A) 0, 6 (B) 5, 6

(C) 5, 4 (D) None of these

22. A number when divided by 342 gives a remainder 47.

When the same number is divided by 19, what would

be the remainder ?

(A) 3 (B) 5


23. What is the remainder when 9875347 × 7435789

× 5789743 is divided by 4 ?

(A) 1 (B) 2


24. What is remainder when 784 is divided by 2402?

(A) 1 (B) 6


25. P is a prime number greater than 5. What is the

remainder when P is divided by 6?

(A) 5 (B) 1

(C) 1 or 5 (D) None of these

26. What is the remainder when 3040 is divided by 17?

(A) 1 (B) 16

(C) 13 (D) 4

27. What is the remainder when 650 is divided by 215?(A) 1 (B) 36(C) 5 (D) 214

28. What is the remainder when 7413 � 4113 + 7513 � 4213 isdivided by 66?(A) 2 (B) 64(C) 1 (D) 0

29. A number when divided successively by 4 and 5 leavesremainders 1 and 4 respectively. When it issuccessively divided by 5 and 4, then the respectiveremainders will be :(A) 1, 2 (B) 2, 3(C) 3, 2 (D) 4, 1

30. When Sholey screened on the TV there was acommercial break of 5 min after every 15 min of themovie. If from the start of the movie to the end of themovie there was in all 60 min of commercials that wasscreened what is the duration the movie ?(A) 180 min (B) 195 min(C) 169 min (D) 165 min

Directions : (31 to 35) Read the following informationcarefully and answer the questions given below.In a big hostel, there are 1,000 rooms. In that hostelonly even numbers are used for room numbers, i.e.the room numbers are 2, 4, 6, ...., 1998, 2000. All therooms have one resident each. One fine morning, thewarden calls all the residents and tells them to goback to their rooms as well as multiples of their roomnumbers. When a guy visits a room and finds the dooropen, he closes it, and if the door is closed, he opensit, All 1,000 guys do this operation. All the doors wereopen initially.

31. The last room that is closed is room number ?(A) 1936 (B) 2000(C) 1922 (D) None of these

32. The 38th room that is open is room number :(A) 80 (B) 88(C) 76 (D) None of these

33. If only 500 guys, i.e. residents of room number 2 to1000 do the task, then the last room that is closed isroom number(A) 2000 (B) 1936(C) 1849 (D) None of these

34. In the case of the previous question, how many roomswill be closed in all ?(A) 513 (B) 31(C) 13 (D) 315

35. If you are a lazy person, you would like to stay in a roomwhose number is :(A) more than 500 (B) more than 1000(C) 500 (D) 2000

46PAGE # 46

36. A 4-digit number is formed by repeating a 2-digit

number such as 2525, 3232 etc. Any number of this

from is exactly divisible by :

(A) 7 (B) 11

(C) 13

(D) Smallest 3-digit prime number

37. How many numbers between 400 and 600 begin with

or end with a digit of 5 ?

(A) 40 (B) 100

(C) 110 (D) 120

38. If (12 + 22 + 32 + .....+ 102) = 385, then the value of

(22 + 42 + 62 +...... + 202).

(A) 770 (B) 1155

(C) 1540 (D) (385 × 385)

39. Find the total number of prime factors in the expression

(4)11 × (7)5 × (11)2.

(A) 37 (B) 33

(C) 26 (D) 29

40. The largest number which exactly divides the product

of any four consecutive natural numbers is :

(A) 6 (B) 12

(C) 24 (D) 120

41. The largest natural number by which the product of

three consecutive even natural numbers is always

divisible, is :

(A) 6 (B) 24

(C) 48 (D) 96

42. A 3-digit number 4a3 is added to another 3-digit

number 984 to give the four-digit number 13b7, which

is divisible by 11. Then ,(a + b) is :

(A) 10 (B) 11

(C) 12 (D) 15

43. Anita had to do a multiplication. Instead of taking 35 as

one of the multipliers, she took 53. As a result, the

product went up by 540. What is the new product?

(A) 1050 (B) 540

(C) 1440 (D) 1590

44. Three friends, returning from a movie, stopped to eat

at a restaurant. After dinner, they paid their bill and

noticed a bowl of mints at the front counter. Sita took

1/3 of the mints, but returned four because she had a

monetary pang of guilt. Fatima then took 1/4 of what

was left but returned three for similar reasons. Eswari

then took half of the remainder but threw two back into

the bowl. The bowl had only 17 mints left when the raid

was over. How many mints were originally in the bowl?

(A) 38 (B) 31

(C) 41 (D) 48

45. In a number system, the product of 44 and 11 is 3414.

The number 3111 of this system, when converted to

the decimal number system, becomes :

(A) 406 (B) 1086

(C) 213 (D) 691

46. A set of consecutive positive integers beginning with 1

is written on the blackboard. A student came and erased

one number. The average of the remaining numbers

is 177

35 . What was the number erased?

(A) 7 (B) 8


47. Let D be a recurring decimal of the form D = 0. a1 a

2 a

1

a2 a

1 a

2 ....., where digits a

1 and a

2 lie between 0 and 9.

Further, at most one of them is zero. Which of the

following numbers necessarily produces an integer,

when multiplied by D?

(A) 18 (B) 108

(C) 198 (D) 288

48. What is the value of the following expression

)120(

1.....

)16(

1

)14(

1

)12(

12222 ?

(A) 199

(B) 1910

(C) 2110

(D) 2111

49. Let N = 1421 × 1423 × 1425. What is the remainder

when N is divided by 12?

(A) 0 (B) 9

(C) 3 (D) 6

50. Let N = 553 + 173 � 723, then N is divisible by :

(A) both 7 and 13 (B) both 3 and 13

(C) both 17 and 7 (D) both 3 and 17

51. Convert the number 1982 from base 10 to base 12.

The results is :

(A) 1182 (B) 1912

(C) 1192 (D) 1292

52. If n2 = 12345678987654321, find the value of n ?

(A) 12344321 (B) 1235789

(C) 11111111 (D) 111111111

53

81

171

161

151

141

131

121

1

is equal to :

(A) 9 (B) 8

(C) 4.5 (D) None of these

54. The LCM of two numbers is 567 and their HCF is 9. If

the difference between the two numbers is 18, find the

two numbers :

(A) 36 and 18 (B) 78 and 60

(C) 63 and 81 (D) 52 and 34

47PAGE # 47

55. If a, a + 2, and a + 4 are prime numbers, then thenumber of possible solution for a is :(A) three (B) two(C) one (D) more than three

56. Find the square root of 7 � 4 3 .

(A) 3�2 (B) 3�5

(C) 5�2 (D) None of these

57. How many even integers n, where 100 n 200, aredivisible neither by seven nor by nine ?(A) 40 (B) 37(C) 39 (D) 38

58. The number of positive n in the range 12 n 40 suchthat the product (n �1) (n � 2).... 3.2.1 is not divisible by

n is :(A) 5 (B) 7(C) 13 (D) 14

59. A rich merchant had collected many gold coins. He didnot want any body to know about him. One day, hiswife asked, �How many gold coins do we have?� After

pausing a moment he replied, �Well ! if divide the coins

into two unequal numbers, then 48 times the differencebetween the two numbers equals the differencebetween the square of the two numbers. � The wife

looked puzzled. Can you help the merchant�s wife by

finding out how many gold coins the merchant has ?(A) 96 (B) 53(C) 43 (D) 48

60. 76n � 66n, where n is an integer > 0, is divisible by :(A) 13 (B) 127(C) 559 (D) All of these

61. The value of 2251541082510 is :

(A) 4 (B) 6(C) 8 (D) 10

62. log10

21

1 + log10

31

1 + log10

41

1 + ... +

log10

19991

1 . When simplified has the value equal

to :(A) 1 (B) 3(C) 10 (D) 100

63. Arrange the following rational number in ascending

order 21

,97

,54

,73

.

(A) 21

,93

,57

,54

(B) 54

,97

,21

,73

(C) 73

,21

,97

,54

(D) 54

,97

,73

,21

64. Which of the following surds is greatest in magnitude

3126 4,25,2,17 .

(A) 6 17 (B) 12 25

(C) 3 4 (D) 2

65. If 5.2Nlog 10 then, find out total number of digits in N.

(A) 3 (B) 4(C) 5(D) cannot be determine

66. If log x = n then 2n is equal to :(A) log (x2) (B) (logx)2

(C) log (x+2) (D) log 2x

67. Given log2 = 0.3010, then log 16 is :(A) 2.4080 (B) 1.2040(C) 0.2408 (D) 1.9030

68. The value of [log10

(5 log10

100)]2 is :(A) 0 (B) 1(C) 2 (D) 10

69. If log10

[log10

(log10

x)] = 0.(A) x = 103 (B) x = 1010

(C) x = 155 (D) None

70. If n = 67 then find the unit digit of [3n + 2n ].(A) 1 (B) 10(C) 5 (D) None

71. What is the decimal equivalent of the 25 digits of

hexadecimal number (100.....001)16

?

(A) 223 + 1 (B) 224 + 1

(C) 292 + 1 (D) 296 + 1

72. If the decimal number 2111 is written in the octal system,

then what is its unit place digit ?

(A) 0 (B) 1

(C) 2 (D) 3

73. If 33 log3Mlog31

N = 1+ log0.008

5, then :

(A) N9

M9 (B)

M9

N9

(C) N3

M3 (D) M3

N9

74. The value of x, when log3(log

2 x) + 2 log

9(log

7 8) = 2, is :

(A) 243 (B) 27

(C) 343 (D) 64

75. Find x if log10

1250 + log10

80 = x.

(A) 5 (B) 4

(C) 8 (D) 7

76. P, Q and R are three natural numbers such that P andQ are primes and Q divides PR. Then out of thefollowing the correct statement is : [IJSO-2008](A) Q divides R (B) P divides R(C) P divides QR (D) P divides PQ

48PAGE # 48

77. It is required to decide if 1107 is a prime number ornot. The number of trials of division necessary is :

[IJSO-2008](A) 10 (B) 11(C) 12 (D) 235

78. The number of integers between 8� and 32 is :

[NSTSE-2009](A) 5 (B) 6(C) 7 (D) 8

79. When expanded, the number of zeros in 100010 is : [NSTSE-2009]

(A) 13 (B) 30(C) 4 (D) 10

80. If a2 + 2b = 7, b2 + 4c = � 7 and c2 + 6a = � 14, then the

value of (a2 + b2 + c2) is : [IJSO-2009](A) 14 (B) 25(C) 36 (D) 47

81. Let N = 28, the sum of All distinct factors of N is : [IJSO-2009]

(A) 27 (B) 28(C) 55 (D) 56

82. The units digit of (1 + 9 + 92 + 93 + --------- + 92009) is :

[IJSO-2009](A) 0 (B) 1

(C) 9 (D) 3

83. The biggest among the following is : [IJSO-2009](A) 21/2 (B) 31/3

(C) 61/6 (D) 81/8

84. If a, b 1, ab > 0, a b and logba = log

ab, then ab = ?

[IAO- 2009](A) 1/2 (B) 1

(C) 2 (D) 10

85. If 2009 = pa.qb, where "p" and "q" are prime numbers,

then find the value of p + q. [NSTSE 2009](A) 3 (B) 48

(C) 51 (D) 2009

86. If HCF (p, q) = 12 and p × q = 1800 n then LCM (p, q) is :

[NSTSE 2010](A) 3600 (B) 900

(C) 150 (D) 90

87. If yx

+ xy

= 3

10 and x + y = 10, then the value of xy will

be : [NSTSE 2010](A) 16 (B) 9

(C) 2 (D) 10

88. The value of log10

(3/2) + log10

(4/3) + ......... up to 99

terms. [IAO 2008](A) 0 (B) 2

(C) 2.5 (D) None of the above

89. In the familiar decimal number system the base is 10.

In another number system using base 4, the counting

proceeds as 1, 2, 3, 10, 11, 12, 13, 20, 21 ....

The twentieth number in this system will be :

[IJSO-2010](A) 40 (B) 320

(C) 210 (D) 110

90. If the eight digit number 2575d568 is divisible by 54

and 87, the value of the digit �d� is : [IJSO-2011](A) 4 (B) 7

(C) 0 (D) 8

91. If x < 0 and log7 (x2 � 5x � 65) = 0, then x is :

[IJSO-2011] (A) �13 (B) �11

(C) � 6 (D) � 5

49PAGE # 49

ANGLE

An angle is the amount of rotation of a revolving line

with respect to a fixed line. If the rotation is in

anticlock-wise sense, then the angle measured is

positive and if the rotation is in clock-wise sense,

then the angle measured is negative.

QUADRANTS

Let X�OX and YOY� be two lines at right angles in a

plane. These lines divide the plane into four equal

parts are known as quadrants. The lines X�OX and

YOY� are known as X-axis and Y-axis respectively.

These two lines taken together are known as the co-

ordinate axes. The regions XOY, YOX�, X�OY� and Y�OX

are known as first, second, third and fourth quadrants

respectively.

(a) Systems of measurement of angles :

(i) Sexagesimal system (ii) Centesimal system

(iii) Circular system

(i) Sexagesimal system : In this system a right angle

is divided into 90 equal parts called degrees. Each

degree is divided into 60 equal parts called minutes

and each minute is divided into 60 equal parts called

seconds.

Thus, 1 right angle = 90 degrees ( 90º)

1º = 60 minutes (60�)

1� = 60 seconds (60�)

TRIGONOMETRY

(ii) Centesimal system : In this system a right angle is

divided into 100 equal parts, called grades. Each gradeis sub divided into 100 minutes, and each minute into

100 seconds.

Thus, 1 right angle = 100 grades (100g) 1 grade = 100 minutes (100�) 1 minute = 100 seconds (100�)

(iii) Circular system : In this system the unit of

measurement is radian. One radian, written as 1c, is

the measure of an angle subtended at the centre of a

circle by an arc of length equal to the radius of the

circle.

The number of radians is an angle subtended by an

arc of a circle at the centre is equal to radius

arc of length.

= rs

Where, = angle in radian, s = arc length and r = radius.

(b) Relation Between Three System of

Measurement of Angles :

2R

100

G

90

D

Where, D = number of degrees,

G = number of grades,

and R = number of radians.

NOTE :

(i) The angle between two consecutive digits in a clock

is 30º = (/6 radians).

(ii) The hour hand rotates through an angle of 30º in

one hour, i.e. (1/2)º in one minute.

(iii) The minute hand rotates through an angle of 6º in

one minute.

Ex.1 Express in radians 47º 25� 36�.

Sol. 47º 25�

'

6036

= 47º

'

53

25

= 47º

'

5128

= 47º

º

601

5128

=

º

7532

47

= º

75

3557

=

180753557 c

=

135003557

c.


50PAGE # 50

Ex.2 Express in degrees :

(a) c

15

2

(b) (� 2)c .

Sol. (a) ºc

180152

152

= 24º

(b) (� 2)c = º

1802

= º

)2(722

180

=

º

116

114

= � 114º

'

60

11

6 = � 114º

'

11

832

= � 114º 32�

''

60

118

= � 114º 32� 44�.

Ex.3 Express in radians 345g 25� 36�.Sol. 345g 25� 36� = 345.2536g

= 20000003452536

c = 1.726268 c

Ex.4 One angle of a triangle is 3x2

grades another is

2x3

degrees, whilst the third is 75

x radians ; express

all angles in degrees.

Sol.32

xg = 32

x 53

10º9

xº

And 75x

75xc

180º = 5

ºx12

But53

xº + 23

xº + 5

12 xº = 180º

6xº + 15xº + 24xº = 1800 45xº = 1800 x = 40º

Hence, three angles of the triangle are 24º, 60º and 96º.

Ex.5 The angles of a triangle are in A.P. and the number ofdegrees in the least is to the number of radians in thegreatest is 60 to c. Find the angles in degrees.

Sol. The three angles in A.P. ; if y is common difference, letthese angles be (x � y)º, xº and (x + y)º.

x + y + x + x � y= 180º

x = 60º

According to the question.

60

180)yx(

)yx(c

or (x � y) = (x + y) º180

× 60º

3 (x � y) = x + y

4y = 2x y = x/2

y = 2

º60 = 30º.

Hence three angles are 30º, 60º and 90º.

Ex.6 The angles in one regular polygon is to that in another

as 3 : 2, also the number of sides in the first is twice

that in the second ; how many sides do the polygons

have ?

Sol. Suppose the second regular polygon has number of

side = x.

The first regular polygon will have number of side = 2x.

Each angle of the first polygon = x2

)4x4( right angle

And each angle of the second polygon

= x

)4x2( right angle

x

4x2:

x24x4

= 3 : 2

x

12x6x

4x4

4x � 4 = 6x � 12

2x = 8 x = 4.

The number of sides in the first and second polygons

respectively are 8 and 4.

Ex.7 The radius of a certain circle is 30 cm, find the

approximately length of an arc of this circle ; if the length

of the chord of the arc be 30 cm.

Sol. Let ABC be the circle whose centre is O and AC is

chord.

In AOC, AO = OC = AC = 30 cm.

AOC = 60º = 3

Hence,

arc AC = radius × 3

= 30 × 3

= 10 = 31.4159 cm.

TRIGONOMETRY

Trigonometry means, the science which deals with

the measurement of triangles.

Trigonometric Ratios :

A right angled triangle is shown in Figure. B is of

90º. Side opposite to B is called hypotenuse. There

are two other angles i.e. A and C. If we consider

C as , then opposite side to this angle is called

perpendicular and side adjacent to is called base.

(i) Six Trigonometric Ratios are :

sin = HypotenuselarPerpendicu

= H

P=

ACAB

51PAGE # 51

cosec = larPerpendicuHypotenuse

= P

H=

ABAC

cos = Hypotenuse

Base = H

B =

ACBC

sec = Base

Hypotenuse = B

H =

BCAC

tan = Base

larPerpendicu = B

P =

BCAB

cot = larPerpendicu

Base =P

B =

ABBC

(ii) Interrelationship in Basic Trigonometric Ratios :

tan = cot

1 cot =

tan1

cos = sec

1 sec =

cos1

sin = eccos

1 cosec =

sin1

We also observe that

tan =

cossin

and cot =

sincos

Ex.8 If tan = nm

, then find sin

Sol. Let P = m and B = n

tan = B

P =

nm

H2 = P2 + B2

H2 = m22 + n22

H = 22 nm

sin = HP

= 22 nm

m

sin = 22 nm

m

.

Ex.9 If cosec A = 5

13, then prove that :

tan2 A � sin2 A = sin4A sec2 A.

Sol. We have cosec A = 5

13 =

larPerpendicuHypotenuse

.

So, we draw a right triangle ABC, right angled at C

such that Hypotenuse AB = 13 units and perpendicular

BC = 5 units

By pythagoras theorem,

AB2 = BC2 + AC2 (13)2 = (5)2 + AC2

AC2 = 169 � 25 = 144

AC = 144 = 12 units

tan A = ACBC

=125

and sin A = ABBC

= 135

and sec A = ACAB

= 1213

C

B

A

513

12

L.H.S tan2 A � sin2 A

= 2

125

�

2

135

= 14425

� 16925

= 169144

)144169(25

= 1691442525

R.H.S. sin4A × sec2 A

= 4

13

5

×

2

12

13

= 24

24

1213

135

= 22

4

1213

5

=

1691442525

So, L.H.S = R.H.S Hence Proved.

Ex.10 In ABC, right angled at B, AC + AB = 9 cm and

BC = 3 cm. Determine the value of cot C, cosec C, sec C.

Sol. In ABC, we have

(AC)2 = (AB)2 + BC2

(9 � AB)2 = AB2 + (3)2

[ AC + AB = 9cm AC = 9 � AB]

81 + AB2 � 18AB = AB2 + 9

72 � 18 AB = 0

AB = 1872

= 4 cm.

Now, AC + AB = 9 cm

AC = 9 � 4 = 5 cm

C

BA

5cm3cm

4cm

So, cot C = 43

ABBC

, cosec C = 45

ABAC

,

sec C = 35

BCAC

.

52PAGE # 52

TRIGONOMETRIC TABLE

Ex.11 Given that cos (A � B) = cos A cos B + sin A sinB, find

the value of cos15º.

Sol. Putting A = 45º and B = 30º

We get

cos (45º � 30º) = cos 45º cos 30º + sin 45º sin 30º

cos 15º = 23

2

1 + 2

1

2

1

cos 15º = 22

13 .

Ex.12 A Rhombus of side of 10 cm has two angles of 60º

each. Find the length of diagonals and also find its

area.

Sol. Let ABCD be a rhombus of side 10 cm and

BAD = BCD = 60º. Diagonals of parallelogram

bisect each other.

So, AO = OC and BO = OD

In right triangle AOB

BA30º

D C

O

sin 30º = ABOB

21

=10OB

OB = 5 cm

BD = 2(OB)

BD = 2 ( 5 )

BD = 10 cm

cos 30º =ABOA

23

=10OA

OA = 35

AC = 2(OA)

AC = 2 ( 35 ) = 310 cm

So, the length of diagonals AC = 310 cm & BD = 10 cm.

Area of Rhombus = 21

× AC × BD

= 21

× 310 × 10 = 350 cm2.

TRIGONOMETRIC RATIO S OF

COMPLEMENTARY ANGLES

sin (90 � ) = cos cos (90 � ) = sin

tan (90 � ) = cot cot (90 � ) = tan

sec (90 � ) = cosec cosec (90 � ) = sec

Ex.14 º70cotº20tan

º36tanº54cot � 2.

Sol.º70cotº20tan

º36tanº54cot � 2

= º36tan

)º36º90cot( +

º70cot)º70º90tan(

� 2

= º36tanº36tan

+ º70cotº70cot

� 2

[cot (90 � ) = tan and tan (90 � ) = cot ]

= 1 + 1 � 2 = 0.

Ex.15. º75tan5º15cot2

º22cosº68sin2

� 5

º70tanº50tanº40tanº20tanº45tan3.

Sol.º75tan5º15cot2

º22cosº68sin2

� 5

º70tanº50tanº40tanº20tanº45tan3

= º75tan5

)º75º90cot(2º22cos

)º22º90sin(2

� 5

)º50tanº40)(tanº70tanº20)(tan1(3

= º75tan5º75tan2

º22cosº22cos2

� 5

]º50tan)º50º90][tan(º70tan)º70º90[tan(3

= 2 � 53

52 (cot 70º tan 70º) (cot 50º tan 50º)

[ tan (90 º � ) = cot , cot (90 º � ) = tan &

sin (90º � ) = cos]

= 2 � 53

52 = 2 � 1 = 1.

Ex.16 If sin 3A = cos (A � 26º) where 3A is an acute angle,

find the value of A.

Sol. sin 3A = cos (A � 26º)

cos (90º � 3A) = cos (A � 26º)

[sin = cos(90º � ]

90º � 3A = A � 26º

4A = 116º

A = 29º

53PAGE # 53

AREA OF TRIANGLE

In a ABC, a & b are the length of 2 sides of triangle

and is the included angle between them.

Then, Area of triangle = 21

ab sin

Proof :

Const. : Draw a line from B perpendicular to AC,

i.e.BD AC

sin = a

BD BD = a sin

Area of triangle = 21 base height

= 21

b BD

= 21

b a sin = 21

ab sin

ANGLE OF ELEVATION

In order to see an object which is at a higher level

compared to the ground level we are to look up. The

line joining the object and the eye of the observer is

known as the line of sight and the angle which this line

of sight makes with the horizontal drawn through the

eye of the observer is known as the angle of elevation.

Therefore, the angle of elevation of an object

helps in finding out its height (Figure).

ANGLE OF DEPRESSION

When the object is at a lower level than the observer�s

eyes, he has to look downwards to have a view of the

object. In that case, the angle which the line of sight

makes with the horizontal through the observer�s eye

is known as the angle of depression (Figure).

Ex.17 A man is standing on the deck of a ship, which is 8 m

above water level. He observes the angle of elevation

of the top of a hill as 60º and the angle of depression of

the base of the hill as 30º. Calculate the distance of the

hill from the ship and the height of the hill.

Sol. Let x be distance of hill from man and h + 8 be height of

hill which is required.

In right triangle ACB,

tan 60º = xh

BCAC

3 = xh

In right triangle BCD,

tan 30º = x8

BCCD

x8

3

1 x = 8 3

Height of hill = h + 8

= 3 .x + 8 = 3 38 + 8 = 32 m.

Distance of ship from hill = x = 8 3 m.

Ex.18 A vertical tower stands on a horizontal plane and is

surmounted by a vertical flag staff of height 5 meters.

At a point on the plane, the angle of elevation of the

bottom and the top of the flag staff are respectively 30º

and 60º. Find the height of tower.

Sol. Let AB be the tower of height h metre and BC be the

height of flag staff surmounted on the tower.

Let the point on the plane be D at a distance x meter

from the foot of the tower

In ABD

tan 30º = ADAB

3

1 =

xh

x = 3 h .....(i)

54PAGE # 54

In ADC

tan 60º = ADAC

3 = x

h5

x = 3

h5 ......(ii)

From (i) and (ii)

3 h = 3

h5

3h = 5 + h 2h = 5

h = 25

= 2.5 m

So, the height of tower = 2.5 m

Ex.19 The angle of elevation of an aeroplane from a point

on the ground is 45º. After a flight of 15 sec, the elevation

changes to 30º. If the aeroplane is flying at a height of

3000 metres, find the speed of the aeroplane.

Sol. Let the point on the ground is E which is y metres from

point B and let after 15 sec flight it covers x metres

distance.

In AEB

tan 45º = EBAB

1 = y3000

y= 3000 m .......(i)

In CED

tan 30º = EDCD

3

1 = yx

3000

( AB = CD)

x + y = 3000 3 .......(ii)

From equation (i) and (ii)

x + 3000 = 3000 3

x = 3000 3 � 3000 x = 3000 ( 3 � 1)

x = 3000 × (1.732 � 1) x = 3000 × 0.732

x = 2196 m.

Speed of Aeroplane = takenTime

eredcovcetanDis

= 15

2196 m/sec

= 5

1815

2196 Km/hr

= 527.04 Km/hr

Hence, the speed of aeroplane is 527.04 Km/hr.

Ex.20 If the angle of elevation of a cloud from a point h

metres above a lake is and the angle of depression

of its reflection in the lake is , prove that the distance

of the cloud from the point of observation is

tantansech2

.

Sol. Let AB be the surface of the lake and let C be a point of

observation such that AC = h metres. Let D be the

position of the cloud and D� be its reflection in the lake.

Then BD = BD�.

In DCE

tan = CEDE

CE = tan

H......(i)

In CED�

tan = EC

'ED

55PAGE # 55

CE =

tanhHh

CE =

tanHh2

.......(ii)

From (i) & (ii)

tan

H =

tanHh2

H tan = 2h tan + H tan

H tan � H tan = 2h tan

H (tan � tan ) = 2h tan

H =

tantantanh2

........(iii)

In DCE

sin = CDDE

CD = sin

DE

CD = sin

H

Substituting the value of H from (iii)

CD =

sintantantanh2

CD =

sintantancossin

h2

CD =

tantansech2

Hence, the distance of the cloud from the point of

observation is

tantansech2

. Hence Proved.

1. Radian measure of 175º 45� is :

(A) 720700

(B) 720703

(C) 720705

(D) 720710

2. Degree measure of c

4

1

is :

(A) 15º 19�5� (B) 14º 19� 5�

(C) 15º 18º 6� (D) 14º 18º 6�

3. A horse is tied to a post by a rope. If the horse moves

along a circular path always keep the rope tight and

describes 88 metres when it has traced out 72º at the

centre, then the length of rope is :

(A) 60 m (B) 65 m

(C) 70 m (D) 72 m

4. Angle between the minute hand of a clock and hour

hand when the time is 7 : 20 am is :

(A) 80º (B) 100º

(C) 120º (D) 140º

5. Degree measure of 6

7 is :

(A) 210° (B) 240°

(C) 270° (D) None

6. If tan = ba

then value of is

cosb�sinacosbsina

=

(A) 22

22

b�a

ba (B) 22

22

ba

ba

�

(C) 22 ba

a

(D) 22 ba

1

7. If tan 15º = 2 � 3 , then the value of cot2 75º is :

(A) 7 + 3 (B) 7 � 2 3

(C) 7 � 4 3 (D) 7 + 4 3

8. If a = 34

cot2 30º + 3 sin2 60º � 3 cosec2 60º � 43

tan2 30º

and b = 3 tan2 45º + cos 0º � cot 90º then logb(a) is :

(A) 2 (B) �1

(C) 21

(D) 21

9. The angles of the triangles ABC and DEF are given as

follows : A = 900, B = 300, D = 900 and E = 300. If the side

BC is twice the side EF, which of the following statement

is true?

(A) Sin B = 2 Sin E (B) Sin E = 2 Sin B

(C) Sin B = Sin E (D) Sin A = Sin B

10. The value of the expression

34

cot2 30º + 3sin2 60º � 2cosec2 60º � 43

tan2 30º is :

(A) 1 (B) � 3

20

(C) 3

10(D) 5

11. The value of the expression

º45tanº60cosº30sin2º60tan4º45cosº30sin5 222

is :

(A) 4 (B) 9

(C) 1253

(D) 6

55

12. The value of tan 5º tan 10º tan 15º tan 20º...... tan 85º is :

(A) 1 (B) 2

(C) 3 (D) None

56PAGE # 56

13. If + = 2

and sin = 31

, then sin is :

(A) 32

(B) 3

22

(C) 32

(D) 43

14. If 7 sin = 24 cos ; 0 < < 2

, then value of

14 tan � 75 cos � 7 sec is equal to :

(A) 1 (B) 2

(C) 3 (D) 4

15. If tan = 4, then

cossincossin

tan3

is equal to :

(A) 0 (B) 2 2

(C) 2 (D) 1

16. Find the value of x from the equation :

6eccos

4eccos

4tan

3sec

6cot

4cos

6sinx

2

2

2

(A) 4 (B) 6

(C) � 2 (D) 0

17. The area of a triangle is 12 sq. cm. Two sides are 6 cm

and 12 cm. The included angle is :

(A) cos�1

31

(B) cos�1

61

(C) sin�1

61

(D) sin�1

31

18. If + = 90º and = 2, then cos2 + sin2 equals to :

(A) 21

(B) 0

(C) 1 (D) 2

19. The difference between two angles is 19º and their

sum is 9

890 grades. Find the greater angle.

(A) 63º (B) 35º

(C) 27º (D) 54º

20. If and are angles in the first quadrant, tan = 71

,

sin = 10

1, then using the formula sin (A + B) = sin AA

cos B + cos A sin B, one can find the value of ( + 2) to

be :

(A) 0º (B) 45º

(C) 60º (D) 90º

21. If each of , and is a positive acute angle such that

sin ( + � ) = 21

, cos( + � ) = 21

and

tan ( + � ) = 1, then the values of , and is :

(A) 45º, 45º & 90º (B) 60º, 45º & 75º

(C) 21

37 , 45° & 21

52 (D) none

22. If tan (A � B) = 3

1 and tan (A + B) = 3 ,

0º < A + B 90º, A > B. Then the value of A and B is :

(A) 45º, 30º (B) 45º, 15º

(C) 60º, 30º (D) none

23. If A, B, C are the interior angles of a triangle ABC, then

cos

2BA

equals to :

(A) cos2C

(B) sec2C

(C) cosec2C

(D) sin2C

24. An aeroplane when flying at a height 2500 m from theground, passes vertically above another aeroplane. Atan instant when the angles of elevation of the twoaeroplanes from the same point on the ground are 45º

and 30º respectively, then the vertical distance between

the two aeroplanes at that instant is :(A) 1158 m (B) 1058 m(C) 1008 m (D) none

25. The shadow of a tower is 30 metres when the sun�saltitude is 30º. When the sun�s altitude is 60º, then the

length of shadow will be :(A) 60 m (B) 15 m(C) 10 m (D) 5m

26. The angles of elevation of the top of a vertical towerfrom two points 30 metres apart, and on the samestraight line passing through the base of tower, are300 and 600 respectively. The height of the tower is :(A) 10 m (B) 15 m

(C) 315 m (D) 30 m

27. If the angle of elevation of a cloud from a point 200metres above a lake is 30º and the angle of depression

of its reflection in the lake is 60º, then the height of the

cloud (in metres) above the lake is :(A) 200 (B) 300(C) 500 (D) none

28. The angle of elevation of the top of tower from the topand bottom of a building h metre high are and , thenthe height of tower is :(A) h sin cos / sin ( + )(B) h cos cos / sin(�)(C) h cos sin / sin(�)(D) None of these

57PAGE # 57

29. When a eucalyptus tree is broken by strong wind, its

top strikes the ground at an angle of 30º to the ground

and at a distance of 15 m from the foot. What is the

height of the tree?

(A) m315 (B) m310

(C) 20 m (D) 10 m

30. A man at the top of a vertical lighthouse, observes a

boat coming directly towards it.If it takes 20 minutes

for the angle of depression to change from 30º to 60º,

the time taken by the boat to reach the lighthouse from

the point when the angle of depression was 30º, is :

(A) 30 minutes (B) 20 minutes

(C) 10 minutes (D) 5 minutes

31. In a triangle ABC, the internal bisector of the angle A

meets BC at D. If AB = 4, AC = 3 and A = 60º, then

the length of AD is :

(A) 32 (B) 7

312

(C) 8

315(D) None of these

32. The expression (1 � tan A + sec A) (1 � cot A + cosec

A) has value : [IJSO-2008](A) � 1 (B) 0

(C) + 1 (D) + 2

33. A person on the top of a tower observes a scooter

moving with uniform velocity towards the base of the

tower. He finds that the angle of depression changes

from 30º to 60º in 18 minutes. The scooter will reach

the base of the tower in next : [IJSO-2008](A) 9 minutes

(B) 18 / ( 3 � 1) minutes

(C) 6 3 minutes

(D) the time depends upon the height of the tower

34. In the diagram, PTR and QRS are straight lines. Given

that, tan xº = 34

and "T" is the midpoint of PR, calculate

the length of PQ, in cm. [NSTSE 2009]

S

TP

Q

R

xº

3 cm

6 cm

(A) 8 (B) 9

(C) 59 (D) 10

35. The trigonometric expression

cot2

sin11�sec

+ sec2

sec11�sin

has the value

[IJSO-2009](A) �1 (B) 0

(C) 1 (D) 2

36. (1 + tan2 ) / (1 + cot2 ) = [IJSO-2009](A) tan2 (B) cot2

(C) sec2 (D) cosec2

37. If cos + sin = 2 cos, then cos � sin = ?

[NSTSE 2009]

(A) 2 tan (B) 2 sin

(C) sincos

2(D) none of these.

38. The tops of two poles of heights 20m and 14m are

connected by a wire. If the wire makes an angle of 30º

with the horizontal, then the length of the wire is :

[NSTSE 2009](A) 40 m (B) 12 m

(C) 28 m (D) 68 m

39. If sin 2

2x1x

......34

.23

.12

= 1, 0º < x < 100º, then the

value of x is equal to : [NSTSE 2010](A) 91º (B) 80º

(C) 49º (D) 46º

40. If p = xsin1xsin�1

, q =

xcosxsin�1

, r = xsin1

xcos

, then

Which one of the following statement is correct ?

[NSTSE-2010](A) p = q r (B) q = r p

(C) r = p q (D) p = q = r

41. If sin + cosec = 2, then [sin8 + cosec8 ] will havethe value : [IJSO-2010]

(A) 2 (B) 24

(C) 26 (D) 28

42. An aeroplane is flying horizontally at a height of

3150 m above a horizontal plane ground. At a

particular instant it passes another aeroplane verti-

cally below it. At this instant, the angles of elevation of

the planes from a point on the ground are 30º and 60º.

Hence, the distance between the two planes at that

instant is :

[IJSO-2011]

(A) 1050 m. (B) 2100 m.

(C) 4200 m. (D) 5250 m.

PAGE # 59

PROTOPLASM

INTRODUCTION

All the living organisms are essentially formed ofnumerous coordinated compartments called as cells.Every cell basically formed of two functional regionsas plasma membrane and protoplasm. The groundsubstance of protoplasm, after removing nucleus, allthe cell organelles and cell inclusions, is calledhyaloplasm/ cytoplasm. It consists of high watercontents containing various compounds of biologicalimportance, some of which are soluble in water e.g.glucose, amino acids, minerals etc. while some ofthese are insoluble in water e.g. lipids.

� Physical Properties :

(i) Protoplasm is a polyphasic colloidal system.

(ii) Its specific gravity is slightly above that of water.

(iii) Its viscosity has been found to be like that ofglycerin.

(iv) It has power of responding to external stimuli, likeheat, electric shocks, application of chemicalsetc.This property of protoplasm is called irritability.

(v) It exhibits streaming movement e.g. rotatorymovements in the leaves of aquatic plants like Hydrillaand Vallisneria.

(vi) Amoeboid movement of the protoplasm can alsobe noticed in myxomycetes and Amoeba.

(vii) In general, the pH of cytoplasm is slightly acidici.e. 6.8, however pH of the nucleoplasm is 7.6 to 7.8.

� Chemical Properties :

The collection of various types of biomolecules of acell collectively form cellular pool. Elements do notoccur in free form but combines to form organicmolecules and inorganic molecules. Organic andinorganic compounds occur in a ratio of 9 : 1. Cellularpool is mainly constituted by :

� Inorganic materials include salts, minerals and water.These materials generally occur in aqueous phasewhich contains molecules and ions dissolved in water.

� Organic compounds as carbohydrates, lipids, aminoacids, proteins, nucleic acids and vitamins.These molecules usually occur in aqueous and non�aqueous phase.

(a) Inorganic Compounds :

� The inorganic substances include salts, minerals andwater.

(A) Minerals: These occur in ionic state and form only1-3 % of cellular pool. Cellular functions fail to occurin the absence of proper ionic balance in the cellcytoplasm and extracellular fluid.

� A compound which releases H+ ion when dissolvedin water is called as an acid, e.g., HCl, H

2SO

4 etc. and

base releases OH- ion, e.g., NaOH, KOH etc.

� Salt is a compound formed, when an acid and a basereact with each other.

� A cell has many salts of Na+, K+, Ca++ and Cl-, HCO3

-,PO

4-3 etc.

� A large amount of minerals also occur as hard depositsas crystals within the cell.

� The salt concentration in cells and in body fluids is ofgreat importance for normal cell functioning.

(B) Gases: Oxygen, carbon dioxide , nitrogen andother gases are also present in protoplasm.

(C) Water : Water is not an organic molecule becauseit does not contain carbon. The bonding properties ofwater account for some of its characteristics, whichare very important to living organisms.

(i) Water is the main component of cell contents andbody fluid.

(ii) It is neutral with pH 7. It ionises to H+ and OH- ion.Phospholipids, nucleic acids and proteins byaccepting or donating H+ ions from water attain specificionic state.

(iii) It forms an average 55 to 60% of living material.

(iv) Water dissolves more substances in it than anyother liquid due to it's highest known dielectricconstant (the measure of capacity to neutralize theattraction between electric charges).Only polarmolecules dissolve in water.

(v) It is generally non toxic to the cell. Colloids likestarch, glycogen and protein remain dispersed inwater in cell cytoplasm.

(vi) It is a medium of heat exchange and transfer.

(vii) It participates in chemical reactions both as areactant and a product. It forms an ideal medium forchemical reactions, because dissolved molecules canmake intimate contact.

(viii) It acts as a lubricating and protective fluid.

(b) Organic Compounds :

A large series of covalent compounds are formed withthe help of carbon, hydrogen and some otherelements. These are called as organic compounds.Their special properties distinguish them frominorganic compounds found in living bodies.

(i) Carbohydrates :

� Carbohydrates can be chiefly composed of carbon,hydrogen and oxygen. In this hydrogen and oxygenatoms are present in a ratio of 2 : 1. As in carbohydrateshydrogen and oxygen are present therefore these arealso termed as hydrates of carbon.


PAGE # 60

� Carbohydrates have general formula as Cn H2n On.

� Carbohydrates are widely distributed in plant tissuesand in animal tissues. In animals they are in the formof glucose and glycogen. In plants they are in the formof cellulose and starch.

� One gram of carbohydrate yields about 4 kilocaloriesof energy.

� Carbohydrates can be defined chemically as aldehydeor ketone derivatives or the poly hydric (more than one� OH group) alcohol and their derivatives.

Carbohydrates can be further divided as :(A) Monosaccharides (B) Disaccharides(C) Polysaccharides

(A) Monosaccharides : These sugars cannot behydrolyzed into simpler forms. They have the generalformula Cn H2n On. The simplest types ofmonosaccharides are glyceraldehyde anddihydroxyacetone. Depending upon the number ofcarbon atoms present, these can be further subdividedinto trioses (e.g. Glyceraldehyde), tetrose(e.g. Erythrose), pentoses (e.g. Ribose, Deoxyribose )etc. Generally if free �H is present at carbon 1 the

sugar is an aldose but if a CH2OH group is substituted,

the sugar is a ketose. They have reducing propertydue to the presence of aldehyde or ketone grouppresent in them. Some examples are as follows :

� Glucose : It is a hexose. It 's formula is C6 H

12 O

6 . It is

normally found in fruit juice and formed in the body bythe hydrolysis of starch, cane sugar, maltose andlactose. Glucose is said to be the sugar of the body. Itis a principal sugar in blood, serving the tissue as amajor metabolic fuel. Normal level of blood glucoseis 80 �120 mg / 100 ml of blood. When the blood

sugar level exceeds the threshold value i.e. 180 mg /100 ml, glucose begins to appear in the urine. Thiscondition is called as glycosuria.

O IIC � HI

H � C � OH I

HO � C � HI

H � C � OHI

H � C � OH

ICH OH2

Structure of glucose

Glucose simple chain

H

OHOH

H

H

CH OH2

OH

OH

OH

H

Ring structure of glucose

� Note : The simple ring structure of glucose is given byHaworth.

� Fructose : Fructose or fruit sugar is also known aslevulose. Similar to glucose it is a ketohexose andless readily absorbed by tissue cells. It is obtained bythe hydrolysis of cane sugar.

� Galactose : It is found in milk sugar or lactose alongwith glucose. It is synthesized in the mammary glandsand combines with glucose to make the lactose ofmilk. In the liver it can be changed to glucose and thusused in the body. It is a part of glycolipids andglycoproteins.

(B) Disaccharides : The disaccharides are sugarscomposed of two molecules of the same or differentmonosaccharides, united by a glycosidic linkage.They have a general formula Cn (H2O)n�1 . Theseinclude maltose, lactose , sucrose.

� Maltose: They consist of two glucose residues. It'soccurrence have been reported in germinating cerealsand malt. It is the major product of enzymatichydrolysis of starch.

� Lactose: It is found in milk to the extent of about 5%.Upon hydrolysis it yields a mixture of galactose andglucose.

� Sucrose : A single molecule of sucrose consists ofone glucose and one fructose molecule. It is thecommon sugar of commerce and the kitchen. It isderived commercially from either cane sugar or occursin varying amount in a variety of fruits, seeds, leaves ,flowers, roots and in maple sugar.

� On hydrolysis it yields an equimolar mixture of glucoseand fructose.

(C) Polysaccharides : Polysaccharides (Glycans)are those which yield more than six molecules ofmonosaccharides on hydrolysis. It's general formulaeis (C6 H10 O5)n. It's examples are as follows :

� Cellulose : It is the chief constituent of the frame workof plants, constituting 50% or more of all carbon invegetation. It is a linear and unbranchedhomopolysaccharide of about 6000 to 10,000 - DGlucose molecules.

� Mammals do not have cellulase enzyme and thereforecannot digest wood & vegetable fibers.

� Purest form of cellulose is found in cotton which isabout 90%.

� Starch : It is the most important food source ofcarbohydrates and is found in cereals, potatoes,legumes and other vegetables. Chemically, the starchis formed of two glucose polymers : -Amylose (anunbranched but spiral chain of about 200-2000 -Glucose molecules) and Amylopectin (a branchedchain of about 2000-20,000 -Glucose molecules.).Natural starch is insoluble in water and gives a bluecolour with iodine solution.

� Glycogen : The counterpart of starch in the animalbody is glycogen thats why it is also called as animalstarch, which occurs in significant amount in liver andmuscles. Glycogen is non � reducing sugar which

gives red colour with iodine. It is a branchedhomopolysaccharide formed of about 30,000 -D-Glucose molecules.

� Note : Glycosidic linkage: The linkage between the

hydroxyl groups of two monosaccharide molecules

with the release of one molecule of water.

PAGE # 61

� Biological significance of carbohydrates :

� Carbohydrates serve as an important structuralmaterial in some animals and in all plants, wherethey constitute the cellulose framework.

� Carbohydrates are essential for life. Almost all animalsuse them as respiratory fuel. In animal cells,carbohydrates are in the form of glucose and glycogen,which serve as an important source of energy for thevital activities.

� Carbohydrates play a key role in the metabolism ofamino acids and fatty acids.

� Some carbohydrates have highly specific functionse.g. ribose in the nucleoprotein of the cells, galactosein certain lipids and the lactose of milk.

(ii) Protein : The name protein is derived from theGreek word proteios, which means"Of the firstrank". This was coined by Berzelius in 1838. Proteinsare the complex nitrogenous substances found in thecells of animals and plants. Chemically proteins arepolymers of molecular units called as amino acids.These polymers contain carbon, oxygen, nitrogen andhydrogen atoms. Usually sulphur atoms are alsopresent .Certain proteins contain phosphorus or sometrace metal elements, such as copper, iron etc. inaddition to the other elements. The proteins have highmolecular weight. One gram of protein yields 4kilocalories of energy. The amino acids found in amolecule of protein are linked together by peptidebonds. The general structure of a amino acid isrepresented by the following formula :

The R group is variable in different amino acids.Amino acids can react with acid and base both, this isdue to the presence of carboxyl and amino groups inthem. There are about 20 amino acids that take partin the formation of proteins. The 20 amino acids arefurther divided into three groups :

� Essential amino acids : They are 8 in number. Theyare not synthesized in a human body and are obtainedfrom food etc. are called as essential amino acids.

� Non � essential amino acids : They are 10 in number.They are synthesized in a human body and are termedas non � essential amino acids.

� Semi�essential amino acids : They are two in numberand needed by growing children and lactating andpregnant women.

Essential Amino acids

Non-Essential Amino acids

Semi-Essential Amino acids

Isoleucine Alanine Arginine

Leucine Aspargine His tidine

Methionine Aspartic acid �

Phenylalanine Cys teine �

Threonine Glutam ic acid �

Tryptophan Glutam ine �

Valine Glycine �

Lys ine Proline �

� Serine �

� Tyros ine �

� Note : Peptide bond : Polypeptide and simple proteinconsist entirely of long chain of amino acids linkedtogether by peptide bonds formed between thecarboxyl group of one amino acid and the amino groupof other amino acid. A molecule of water is releasedout during bond formation.

H � N � C � C � OH

HI

HI

R1

I

OI I

H � N � C � C � OH

HI

HI

R2

I

OI I

H � N � C � C �

HI

HI

R1

I

OI I

N � C � C � OH

HI

HI

R2

I

OI I

Peptide bond formation

� Biological significance of proteins :

� They act as a structural components of cell. They areessential for growth and repair of the body.

� All the enzymes are made up of proteins. They help tocatalyze various reactions occurring in our body.

� They play important roles as hormones, antibodies, etc.

� Haemoglobin , the respiratory pigment of animals isa conjugated protein composed of colourless basicprotein the globin and haem.

(iii) Lipid : Term lipid was coined by Bloor. Fats andtheir derivatives are collectively known as lipids (Ingreek Lipas = fat). The principal componentassociated with most lipids are the fatty acids. Thelipids are a heterogenous groups of substances whichhave the common property of being relatively insolublein water and soluble in non � polar solvents such as

ether, chloroform and benzene. They consist ofcomparatively less oxygen. One gram of fat yields 9kilocalories of energy. Similar or different fatty acidsparticipate in the composition of a fat molecule. Thelipids include fats, oils, ghee, waxes and relatedcompounds.

PAGE # 62

� Note : Lipids generally consist of a single moleculeof glycerol and three molecules of fatty acids joinedtogether by ester bonds. Therefore these are alsotermed as triglycerides. Three molecules of waterare released during the formation of triglycerides.

� Biological significance of lipids :

� They takes part in the synthesis of steroids, hormones,vitamin D, bile salts etc.

� They act as a solvent for fat soluble vitamins i.e. vitaminA , D, E and K.

� They act as storage compounds in animals, in thefruits and seeds of plants and in other organism.

� They act as structural cellular components particularlyin cell membranes. They are found in the form ofphospholipids, glycolipids and sterols.

� They act as insulators. They provide electrical andthermal insulation. They are deposited beneath theskin and other internal organs to reduce the heat loss.They also work as shock absorbers and othermechanical impacts.

(iv) Nucleic acids : These are the hereditary materialsof living organisms. There are two types of nucleicacids :

(A) DNA (Deoxyribose nucleic acids) : DNA is coiledmacromolecule made of two antiparallel chains heldtogether by hydrogen bonds. DNA has diameter of 20Å. One turn of spiral has a distance of 34 Å and distance

between two adjacent nucleotides is 3.4 Å.

� Nucleotides : A single nucleotide consist of followingparts :

� Pentose sugar : It is a 5 - carbon containing sugarwhich is ribose is RNA and deoxyribose in DNA.

� Nitrogen bases : There are two types of purines whichinclude adenine (A), and guanine (G) and pyrimidineswhich include thymine (T), uracil (U) and cytosine (C).In DNA adenine, thymine, guanine and cytosinepresent while in RNA uracil is present in place ofthymine.

� Phosphate group : PO4

-3 group in the form of H3PO

4

(B) RNA : (Ribonucleic acid) Structure of RNA isfundamentally the same as DNA but there are somedifferences. The differences are as follows.

� In place of deoxyribose sugar of DNA, there is apresence of ribose sugar in RNA.

� In place of nitrogen base, thymine present in DNAthere is a nitrogen base uracil in RNA.

� RNA is made up of only one polynucleotide chain i.e.RNA is single stranded.

� In RNA, polynucleotide chain runs in 3' 5' direction.

� Exception : RNA found in Reo�virus is double

stranded i.e. it has two polynucleotide chains.

� Types of RNA- A cell contains three types of RNA :

1. Ribosomal RNA (r � RNA) : This RNA is 80% of thecell's total RNA. It is the most stable form of RNA. It isfound in ribosomes and it is produced in nucleolus.They are present as 80 � S type of ribosome in

eukaryotic cells and 70 � S type of ribosome in

prokaryotic cells. It is the site of protein synthesis.

2. Transfer RNA (t � RNA) :

� It is 10 � 15% of total RNA

� It is synthesized in the nucleus by DNA.

� It is also known as soluble RNA.

� It is also known as adapter RNA.

� It is the smallest RNA. At the time of protein synthesisit acts as a carrier of amino acids.

� It has the most complex structure.

3. Messenger RNA (m � RNA) : The m � RNA is 1 � 5 %

of the cells total RNA. The m � RNA is produced by

genetic DNA in the nucleus. This process is called astranscription, m � RNA is also called as templateRNA. It acts as the template for protein synthesis.

(v) Enzymes : Enzymes are protein catalysts forbiochemical reactions in the living cells. Thesubstance which increases reaction rate is called ascatalyst and the phenomenon is called as catalysis.The term enzyme is derived from Greek word whichmeans 'in yeast' because the yeast cells were thefirst to reveal enzyme activity in living organisms.Enzyme was first introduced by W. Kuhne in 1878.Berzelius was the first to define and recognize thenature of catalyst. In 1926 J.B Sumner isolated theenzyme urease as a crystalline protein for the firsttime. Enzymes could be intracellular and extracellularenzymes. When the enzymes remain and functioninside the cells, they are called as endoenzymes orintracellular enzymes. The enzymes which leave thecell and function outside the cell are called extracellular enzymes.

� General properties of enzymes :

� They remain unaltered at the end.

� They are required in small quantities.

� They accelerate the rate of reaction.

� They are proteinaceous in nature.

� Enzymes are highly specific towards substrate.

� Certain enzymes exhibit the property of reversibility.

(vi) Pigments :

The coloured substance found in the living being iscalled as pigment. The beauty of nature is due toanimals, birds and flowers having different pigments.The living beings depend on sun for energy. The greenpigment in nature is called as chlorophyll, can only

PAGE # 63

store light energy obtained from the sun, in the form ofchemical energy. Thus, chlorophyll is the nutritionalbasis of life on earth. The colour of our skin is due tothe pigment melanin. Haemoglobin & haemocyaninpigments play an important role in transportation ofoxygen in the body of living beings. Pigments belongto the group carotenoid are found in both plants andanimals.

BIOCHEMICAL REACTIONS

The reactions undergoing inside a living cell to sustainlife are called as biochemical reactions. The biologicalsystem can't use heat liberated in biological reactionsdirectly as they are isothermic so the biologicalsystems use chemical energy (ATP) to perform variousliving processes. Biochemical reactions are catabolic(breakdown/exergonic reactions) and anabolic(synthetic reactions), collectively they are called asmetabolic reactions.

EXERCISE

1. Which of the following is a disaccharide ?(A) Galactose (B) Fructose(C) Maltose (D) Dextrin

2. Nucleic acids are made up of(A) amino acids (B) pentose sugars(C) nucleosides (D) nucleotides

3. Which of the following is not a carbohydrate ?(A) Starch (B) Glycogen(C) Wax (D) Glucose

4. To get quick energy one should use(A) carbohydrates (B) fats(C) vitamins (D) proteins

5. Circular and double stranded DNA occurs in(A) golgi body (B) mitochondria(C) nucleus (D) cytoplasm

6. The most abundant protein in human body is(A) collagen (B) myosin(C) actin (D) albumin

7. Which is not a polysaccharide ?(A) Sucrose (B) Starch(C) Glycogen (D) Cellulose

8. The decreasing order of the amount of organiccompounds, present in an animal body is(A) carbohydrates, proteins, fats, and nucleic acid(B) proteins, fats, nucleic acid and carbohydrates(C) proteins, fats, carbohydrates and nucleic acid(D) carbohydrates, fats, proteins and nucleic acid

9. Term protoplasm was introduced by(A) Purkinje (B) Schultze(C) Sutton and Boveri (D) Van Mohl

10. Which of the following is a monosaccharide ?(A) pentose sugar (B) hexose sugar(C) glucose (D) all of the above

11. The process of m-RNA synthesis on a DNA templateis known as(A) translation (B) transcription(C) transduction (D) transformation

12. Which amino acid is non essential for a human body ?(A) Glycine (B) Phenylalanine(C) Arginine (D) Methionine

13. Double helix model of DNA was proposed by(A) Watson and Crick(B) Schleiden and Schwann(C) Singer and Nicholson(D) Kornberg and Khurana

14. Which of the following nitrogen base is not found inDNA ?(A) Thymine (B) Cytosine(C) Guanine (D) Uracil

15. Glycogen is a / an(A) polymer of amino acids(B) polymer of fatty acids(C) unsaturated fat(D) polymer of glucose units

16. Carbohydrate is a(A) polymer of fatty acids(B) polymer of amino acids(C) polyhydric aldehyde or ketone(D) none of the above

17. In which form, food stored in animal body ?(A) Glucose (B) Glycogen(C) Cellulose (D) ATP

18. Chemically enzymes are(A) fats (B) carbohydrates(C) hydrocarbons (D) proteins

19. Long chain molecules of fatty acids are obtained by(A) polymerisation of two carbon compounds(B) decomposition of fats(C) polymerisation of glycogen(D) conversion of glycogen

20. The amino acids which are not synthesized in thebody are called as(A) non�essential

(B) essential(C) deaminated(D) all of them are correct

21. Fats in the body are formed when(A) glycogen is formed from glucose(B) sugar level becomes stable in blood(C) extra glycogen storage in liver and muscles isstopped(D) all of the above

PAGE # 64

22. Which element is not found in nitrogen base ?(A) Nitrogen (B) Hydrogen(C) Carbon (D) Phosphorous

23. Proteins are the polymers of(A) amino acids (B) natural protein(C) enzymes (D) nucleic acids

24. DNA polymerase is needed for(A) replication of DNA (B) synthesis of DNA(C) elongation of DNA (D) all of the above

25. Duplication of DNA is called as(A) replication (B) transduction(C) transcription (D) translation

26. Ligase enzyme is used for(A) denaturation of DNA(B) splitting of DNA into small fragments(C) joining fragments of DNA(D) digestion of lipids

27. Orange juice contains plenty of(A) vitamin C (B) vitamin A(C) vitamin D (D) vitamin E

28. Sucrose is composed of(A) glucose & fructose(B) glucose & glycogen(C) two molecules of glucose(D) glycogen & fructose

29. Similarity in DNA and RNA is(A) both are polymers of nucleotides(B) both have similar pyrimidine(C) both have similar sugar(D) None of the above

30. In which stage of cell cycle, DNA replication occurs(A) G

1�phase (B) S � Phase

(C) G2 � phase (D) M � phase

c31. Bacteria cannot survive in a highly salted picklebecause [IJSO-Stage-I/2011](A) they become plasmolysed and consequently die.(B) they do anaerobic respiration.(C) water is not available to them.(D) of all the reasons mentioned above.

32. Maximum vitamin A content is likely to be found inthe extract of [IJSO-Stage-I/2011](A) sprout of pulse (B) cod liver(C) white muscles (D) rose petals

33. The ointment prescribed for burns usually contains,among other ingredients, [IJSO-Stage-I/2011](A) vitamin A (B) vitamin B(C) vitamin D (D) vitamin E

34. Unsaturated fatty acids contain [IJSO-Stage-I/2012](A) atleast one double bond(B) two double bonds(C) more than two double bonds(D) no double bond

PAGE # 65

SERIES COMPLETION

Series completion problems deals with numbers,alphabets and both together. While attempting to

solve the question, you have to check the pattern ofthe series. Series moves with certain mathematicaloperations. You have to check the pattern.

Type of questions asked in the examination :(i) Find the missing term(s).(ii) Find the wrong term(s).

NUMBER SERIES

(a) Some Important Patterns :

(i) a, a ± d, a ± 2d, a ± 3d.......(Arithmetic Progression)

(ii) a, ak, ak2, ak3, ................(Geometric Progression)

(iii) a, ka

, 2k

a, 3k

a, .............(Geometric Progression)

(iv) Series of prime numbers � i.e. 2, 3, 5, 7, 11, ......

(v) Series of composite numbers �i.e. 4, 6, 8, 9, 10, 12, .................

Directions : (1 to 10) Find the missing numbers :

Ex 1. 16, 19, 22, 25, ?(A) 27 (B) 28

(C) 29 (D) 25Sol. (B) As per series a, a + d, a + 2d,.........

a = 16

d = 3a + 4d = 16 + 4 × 3

Ex 2. 9, 18, 36, ?, 144(A) 70 (B) 56

(C) 54 (D) 72Sol. (D) As per series, a, ak, ak2, ak3, ........

a = 9, k = 2

ak3 = 9 × 23 = 72

Ex 3. 2, 6, 14, 26, ?(A) 92 (B) 54(C) 44 (D) 42

Sol. (D) The pattern is +4, +8, +12, +16, .......

Ex 4. 240, ? , 120, 40, 10, 2(A) 120 (B) 240(C) 40 (D) 10

Sol. (B) The pattern is ×1, ×21

, ×31

, ×41

, ×51

missing term = 240 × 1 = 240

Ex 5. 8, 12, 21, 46, 95, ?(A) 188 (B) 214(C) 148 (D) 216

Sol. (D) The pattern is + 22, + 32, + 52, + 72, .......missing number = 95 + 112 = 216

Ex 6. 3, 9, 36, 180, ?(A) 1080 (B) 900(C) 720 (D) None of these

Sol. (A) Each term is multiplied by 3, 4, 5 and so onrespectively. Therefore, the next term would be180 × 6 = 1080.

(b) Multiple Series :

A multiple series is a mixture of more than oneseries :

Ex 7. 4, 7, 3, 6, 2, 5, ?(A) 0 (B) 1(C) 2 (D) 3

Sol. (B) The sequence is a combination of two seriesI 4, 3, 2, ?II 7, 6, 5The pattern followed in I is � 1, � 1, � 1

missing number = 2 � 1 = 1

Ex 8. 14, 15, 12, 16, 9, 18, 4, 21, ?(A) 2 (B) 3(C) � 3 (D) � 5

Sol. (C) The sequence is a combination of two series. 14, 12, 9, 4, (....) and 15. 16, 18, 21The pattern followed in is � 2, � 3, � 5, .......

missing number = 4 � 7 = � 3

Ex 9. 1, 1, 4, 8, 9, ? ,16, 64(A) 21 (B) 27(C) 25 (D) 28

Sol. (B) (i) 1, 4, 9, 16 [12, 13, 22, 23, 32, 33.............](ii) 1, 8, __, 64 mixed combination

Ex 10. 3, 6, 24, 30, 63, 72, ?, 132(A) 58 (B) 42(C) 90 (D) 120

Sol. (D) The difference between the terms is givenbelow as :

3 6 24 30 63 72 ? 132

3 18 6 33 9 48 ?

3 15 3 15 ?

Difference

Difference

Therefore, alternate difference between thedifference is 3 and 15 respectively.Hence, the next term would be 72 + 48 = 120.


PAGE # 66

Directions : (11 to 12) Find the wrong term(s) �

Ex 11. 5, 8, 10, 12, 15, 18, 20, 23(A) 8 (B) 12(C) 15 (D) 18

Sol. (B)

Therefore, number 12 is wrong and should bereplaced by 13.

Ex 12. 1, 3, 8, 19, 42, 88, 184(A) 3 (B) 8(C) 19 (D) 88

Sol. (D)

1 1848 19 42 893

2 5 11 23 47 95

3 6 12 24 48

Hence, number 88 is wrong and should bereplaced by 89.or 1 × 2 + 1, 3 × 2 + 2, 8 × 2 +3, 19 × 2 + 4, 42 × 2 + 5,

89 × 2 + 6

Directions : (13 to 14) In each of the following questions, anumber series is given. After the series, below it inthe next line, a number is given followed by (P), (Q),(R), (S) and (T). You have to complete the seriesstarting with the number given following thesequence of the given series. Then answer thequestion given below it.

Ex 13. 12 28 64 14037 (P) (Q) (R) (S) (T)Which number will come in place of (T) ?(A) 1412 (B) 164(C) 696 (D) 78

Sol. (A)

Similarly (P) (Q) (R) (S) (T)

37 78 164 340 696 1412

×2+4 ×2+8 ×2+12 ×2+16 ×2+20

Therefore, the number 1412 will come in place of (T).

Ex 14. 2 9 57 3373 (P) (Q) (R) (S) (T)Which number will come in place of (Q) ?(A) 113 (B) 17(C) 3912 (D) 8065

Sol. (A)

Similarly, (P) (Q) (R) (S)

3 17 113 673

×8�7 ×7�6 ×6�5 ×5�4

3361

Therefore, the number 113 will come in place of (Q).

ALPHABET SERIES

(a) Pattern of Alphabets Show Variation Based on :

(i) Position of the letters (ii) Difference between the alphabets

(i) Position of alphabets :

Alphabets in order :

Alphabets in reverse order :

Directions : (15 to 24) Find the missing term(s) :

Ex 15. B, E, H, ?(A) K (B) L(C) J (D) M

Sol. (A) In the given series, every letter is moved threesteps forward to obtain the corresponding lettersof the next term. So, the missing term is K.

Ex 16. Q, N, K, ?, E(A) H (B) I(C) J (D) G

Sol. (A) In the given series, every letter is moved threesteps backward to obtain the corresponding lettersof the next term. So, the missing term is H.

Ex 17. A, Y, D, W, G, U, J, ?(A) R (B) T(C) S (D) P

Sol. (C) The given sequence consists of two series :. A, D, G, J in which each letter is moved threesteps forward to obtain the next term. Y, W, U, ? in which each letter is moved two stepsbackward to obtain the next term.So, the missing term would be S.

Ex 18. AG, LR, WC, HN, ?(A) SX (B) RY(C) SY (D) TX

PAGE # 67

Sol. (C) The first letter of each group and the secondletter of each group differs by 11 letters betweenthem.

Alphabeticalpositions

Difference inAlphabeticalpositions

A 1

L 12

W 23

H 8

11 11 11

Similarly,

Alphabeticalpositions

Difference inAlphabeticalpositions

G 7

R 18

C 3

N 14

11 11 11

Therefore, the next group of letter would be SY.

11

H SAnd

11

N Y

Ex 19. AD, EI, JO, PV, ?(A) VD (B) WC(C) WD (D) VE

Sol. (C) The first letter of subsequent groups have adifference of 4, 5 and 6 places respectively, whereasthe second letter of the subsequent groups has adifference of 5, 6, and 7 places respectively,Therefore, on following the same pattern, we get�WD� as the next term which would replace the

question mark.

Ex 20. AB, BA, ABD, DBA, PQRS, ?(A) SRQP (B) SRPQ(C) SQRP (D) RSQP

Sol. (A) The first term is reversed to get second term.The third term is reversed to get the fourth term.Similarly, to get the sixth term, we reverse the fifthterm. So, the missing term would be SRQP.

Ex 21. HEJ, JGL, LIN, NKP, ?(A) MOR (B) PNS(C) PMR (D) NPT

Sol. (C) First letter of each group differs by 2 letters.Second letter of each group differs by 2 letters.Third letter of each group differs by 2 letters. All theletters differ in the forward direction. Hence, thenext choice would be PMR.

Ex 22. XYQ, ZAR, BCS, DET, ?(A) GFU (B) FUG(C) FZU (D) FGU

Sol. (D) Here, first two terms of every group of lettersare in continuation, like XY, ZA, BC, DE, and thethird letter of each group is again in forwardcontinuation,i.e. Q, R, S, T. Hence, the term replacing thequestion mark would be FGU.

Ex 23. 17Z5, 15X4, 13V3, ?, 9R1(A) 11S2 (B) 11T2(C) 11U2 (D) 11T3

Sol. (B) The first number & second letter of every termis moved two steps backward & the third numberof every term is moved one step backward. So, themissing term would be 11T2.

Ex 24. (ABC) � 6, (DEF) � 15, (GHI) � 24, ?

(A) (IJK) � 33 (B) (JKM) � 33

(C) (IJK) � 32 (D) (JKL) � 33

Sol. (D) In a given seriesLet A = 1, B = 2, C = 3, D = 4, E = 5, F = 6, and so on

6CBA321

, 15FED654

, 24IHG

987

So, the missing term would be 33LKJ121110

Directions : (25 to 27) Find the wrong term (s) :

Ex 25. ABD, DGK, HMS, NTB, SBL, ZKW

(A) NTB (B) DGK(C) SBL (D) ZKW

Sol. (A) First letter of first, second, third,.........terms ismoved three, four, five, ........steps forwardrespectively. Similarly, second letter is moved five,six, seven,......steps forward respectively and thirdletter is moved seven, eight, nine,........stepsforward respectively. Hence, NTB is the wrong termand should be replaced by MTB.

Ex 26. EPV, FQW, GRX, HTY, ITZ(A) FQW (B) GRX(C) HTY (D) ITZ

Sol. (C) In every term, first second and third letter is inalphabetical order to its next term respectively.Fourth term is not following the same rule. Hence,HTY is the wrong term and should be replaced byHSY.

Ex 27. D4V, G10T, J20R, M43P, P90N

(A) P90N (B) G10T

(C) J20R (D) D4V

Sol. (B) First letter of every term is moved three steps

forward in each next term. Second number of every

term of the pattern × 2 + 1, × 2 + 2, × 2 + 3,............and

third letter of every term is moved two steps

backward. Hence, G10T is the wrong term and

should be replaced by G9T.

LETTER REPEATING SERIES

Pattern of such questions is that some letters insequence are missing.

(i) The letters may be in cyclic order (clockwise oranti-clockwise).(ii) To solve a problem, we have to select one of thealternative from the given alternatives. Thealternative which gives a sequence form of lettersis the choice.


Ex 28. a a _ b a a _ b b b _ a(A) baa (B) abb(C) bab (D) aab

PAGE # 68

Sol. (A) we proceed step by step to solve the aboveseries:Steps :

1. The first blank space should be filled in by 'b' sothat we have two a's followed by two b's.

2. Second blank space should be filled in by 'a' sothat we have three a's followed by three b' s.

3. The last blank space must be filled in by 'a' to keepthe series in sequence.

Ex 29. _ bca _ ca _ c _ b _(A) aabbc (B) abbbc(C) aabcc (D) abbac

Sol. (D)

Series is abc/ abc/ abc/ abc. So, pattern abc isrepeated.

Ex. 30 a _ abb _ aa _ ba _ a _ b(A) ababa (B) aabba(C) aabab (D) aaabb

Sol- (C) Series is aaabb/ aaabb/ aaabb. So, patternaaabb is repeated.

Ex 31. a _ c _ abb _ ca _ a(A) baca (B) bbca(C) bacc (D) bacb

Sol- (A) Series is abc/ aabbcc/ aaa

Ex 32. bc _ b _ c _ b _ ccb(A) cbcb (B) bbcb(C) cbbc (D) bcbc

Sol- (A) Series is bccb / bccb / bccb. So, pattern bccb isrepeated

Directions : (33 to 34) The question given below is basedon the letter series, In series, some letters aremissing. Select the correct alternative. If more thanfive letters are missing, select the last five lettersof the series.

Ex 33. xyzu _ yz _ v _ _ uv _ _ _ _ _ _ _(A) uvxyz (B) vuzyx(C) uvzyx (D) vuxyz

Sol. (A) The series is x y z u v / y z u v x/ z u v x y/u v x y zThus the letters are written in a cyclic order.

EX 34. abcd _ bc _ e _ _ de _ _ _ _ _ _ _(A) deabc (B) edcba(C) decba (D) edabc

Sol. (A) The series is a b c d e / b c d e a / c d e a b / b e a b cThus the letters are written in a cyclic order.

Direction : (35 to 36) There is a letter series in the first rowand a number series in the second row. Eachnumber in the number series stands for a letter inthe letter series. Since in each of that series someterm are missing you have to find out as to whatthose terms are, and answer the questions basedon these as given below in the series.

Ex 35. a _ h _ _ c _ n e _ h _ e a c _ _ _ _ _2 1 _ 4 3 _ 5 _ _ 2 5 4 _ _ _ _ _ _ _ _The last five terms in the series are(A) 32524 (B) 43215(C) 25314 (D) 32541

Sol. (B) By taking a = 2, c = 1, n = 4, h = 5 and e = 3, thenumbers series runs as 21543 15432 5432143215. If first digit of a group of five digits is placedas the last digit, we obtain the second group of fivedigits and so on.

Ex 36. _ m y e _ _ y l x _ y l m _ _ l _ _ _ _4 6 _ 5 8 6 _ _ _ 5 7 _ 6 5 8 _ _ _ _ _The last five terms of the number series are(A) 46758 (B) 74658(C) 76485 (D) 46785

Sol. (D) By taking e = 5, l = 4, m = 6, y = 7 and x = 8 thenumber series runs as 46758 67485 74658 46785.By taking the digits in the groups of five, we findthat first digit of the first group (i.e. 4) is the thirddigit of the second group and the last two digitshave interchanged their positions. The same ruleapplies in others groups also.

Direction : (37) In the following question, three sequencesof letter/numbers are given which correspond toeach other in some way. In the given question, youhave to find out the letter/numerals that come inthe vacant places marked by (?). These are givenas one of the four alternatives under the question.Mark your answer as instructed.

Ex 37. C B _ _ D _ B A B C C B_ _ 2 3 5 4 _ _ ? ? ? ?p _ p q _ r _ q _ _ _ _(A) 4 5 5 4 (B) 4 3 3 4(C) 4 2 2 4 (D) 2 5 5 2

Sol. (C) Comparing the positions of the capital letters,numbers and small letters, we find p correspondsto C and 2 corresponds to p. So, p and 2 correspondto C. q corresponds to A and 3 corresponds to q.So, q and 3 corresponds to A. Also, 5 correspondsto D. So, the remaining number i.e., 4 correspondsto B. So, BCCB corresponds to 4, 2, 2, 4.

MISSING TERMS IN FIGURES

Directions : (38 to 47) Find the missing number(s) :

Ex 38.

6 9 15

8 12 20

4 6 ?

(A) 5 (B) 10(C) 15 (D) 21

Sol. (B) In the first row, 6 + 9 = 15In the second row, 8 + 12 = 20 In the third row, missing number = 4 + 6 = 10.

PAGE # 69

Ex 39.

(A) 11 (B) 6(C) 3 (D) 2

Sol. (C) Clearly, in the column, 83

46

In the column, 272

318

We take x in place of ?

Similarly in the column, 95

15

x , x 315

59

Ex 40.

3C 27D 9E

7I 21K 3M

4D ? 7J

(A) 11E (B) 28G(C) 35I (D) 48F

Sol. (B) The letters in the first row form a series C, D, E(a series consecutive letters). The letters in thesecond row form a series I, K, M (a series ofalternate letters). Similarly, the letters in the thirdrow will form the series D, G, J (a series in whicheach letter is three steps ahead of the previousone). So, the missing letter is G. Also, thenumber in the second column is equal to theproduct of the numbers in the first and thirdcolumns. So, missing number is (4 × 7) i.e. 28.

Thus, the answer is 28G.

Ex 41.

6

4 5

41

? 5

1

2

7

(A) 16 (B) 9(C) 85 (D) 112

Sol. (C) Hint ; 42 + 52 = 16 + 25 = 4112 + 22 = 1 + 4 = 562 + 72 = 36 + 49 = 85

Ex 42.84 81 8814 18 ?12 9 11

(A) 16 (B) 21(C) 61 (D) 81

Sol. (A) In first figure, 2

1412 = 84.

In second figure, 2

189 = 81.

Let the missing number In third figure be x.

Then, 2x

11 = 88 or x = 11

288 = 16.

Ex 43.

3 5 5

10 30 ?2 3 56 9 6

4 5 2

(A) 15 (B) 20(C) 25 (D) 40

Sol. (B) ClearlyIn first figure] 6 × 3 � 4 × 2 = 18 � 8 = 10

In second figure] 9 × 5 � 5 × 3 = 45 � 15 = 30

In third figure] 6 × 5 � 2 × 5 = 30 � 10 = 20

Ex 44. 26 ? 294 3 35 6 8

6 4 5

(A) 32 (B) 22(C) 18 (D) 27

Sol. (B) In first figure] 5 × 4 + 6 = 26

In second figure] 8 × 3 + 5 = 29

missing number in third figure] 6 × 3 + 4 = 22

Ex 45.174 336 ?

8 5 3

3 2 9

6 7 5 3

2 7 9

2 5

6 4 5(A) 140 (B) 150(C) 200 (D) 180

Sol. (B) In first figure] 8 × 5 × 3 + 3 × 2 × 9 = 120 + 54 =

174In second figure] 6 × 7 × 5 + 2 × 7 × 9= 210 + 126

= 336

missing number in third figure] 3 × 2 × 5 + 6 × 4 × 5 = 30 + 120 = 150

Ex 46.

11 9 15 7 25 21

40 176 ?

(A) 184 (B) 210(C) 241 (D) 425

Sol. (A) The number at the bottom is the difference ofsquares of two numbers given at topIn first figure] 112 � 92 = 121 � 81 = 40

In second figure] 152 � 72 = 225 � 49 = 176

In third figure] 252 � 212 = 625 � 441 = 184

Ex 47. 33

3 5

6 3

7

48

5 4

4 3 5

45

?

(A) 47 (B) 45(C) 37 (D) 35

Sol. (D) In first figure, 6 × 3 + 3 × 5 = 33

In second figure, 5 × 4 + 4 × 7 = 48

In third figure, 5 × 4 + 3 × 5 = 35

PAGE # 70

EXERCISE-1

Directions : (1 to 25) Find the missing numbers :

1. 2, 8, 18, 32, ?(A) 62 (B) 60(C) 50 (D) 46

2. 16, 54, 195, ?(A) 780 (B) 802(C) 816 (D) 824

3. 14, 316, 536, 764, ?(A) 981 (B) 1048(C) 8110 (D) 9100

4. 8, 11, 15, 22, 33, 51, ?, 127, 203(A) 80 (B) 53(C) 58 (D) 69

5. 2, 3, 6, 18, ?, 1944(A) 154 (B) 180(C) 108 (D) 452

6. 7,19, 55, 163, ?(A) 387 (B) 329(C) 527 (D) 487

7. 1, 2, 9, 4, 25, 6, ?(A) 51 (B) 49(C) 50 (D) 47

8. 16, 33, 67, 135, ?(A) 371 (B) 175(C) 271 (D) 287

9. 8, 24, 16, ?, 7, 14, 6, 18, 12, 5, 5, 10(A) 14 (B) 10(C) 7 (D) 5

10. 2, 12, 36, 80, 150, ?(A) 194 (B) 210(C) 252 (D) 258

11. 101, 100, ?, 87, 71, 46(A) 92 (B) 88(C) 89 (D) 96

12. 100, 50, 52, 26, 28, ? 16, 8(A) 30 (B) 36(C) 14 (D) 32

13. 6, 24, 60, 120, 210, 336, ?, 720(A) 496 (B) 502(C) 504 (D) 498

14. 3, 1, 4, 5, 9, 14, 23, ?(A) 32 (B) 37(C) 41 (D) 28

15. 3, 6, 18, 72, 360, ?(A) 720 (B) 1080(C) 1600 (D) 2160

16. 78, 79, 81, ?, 92, 103, 119(A) 88 (B) 85(C) 84 (D) 83

17. 0, 6, 20, 42, 72, ?(A) 106 (B) 112(C) 110 (D) 108

18. 2, 9, 28, 65, ?(A) 121 (B) 195(C) 126 (D) 103

19. 1, 11, ?, 11, 11, 11, 16, 11(A) 1 (B) 11(C) 6 (D) 192

20. 137, 248, 359, 470, ?(A) 582 (B) 581(C) 571 (D) 481

21. 3, 15, 35, ?, 99, 143(A) 63 (B) 77(C) 69 (D) 81

22. 9, 16, 30, 58, ?(A) 104 (B) 114(C) 116 (D) 118

23. 3, 12, 27, 48, 75, 108, ?(A) 192 (B) 183(C) 162 (D) 147

24. 1, 4, 12, 30, ?(A) 60 (B) 62(C) 64 (D) 68

25. 94, 166, 258, ?, 4912(A) 3610 (B) 1644(C) 1026 (D) 516

Directions : (26 to 28) In each of the following questions, anumber series is given. After the series, below it inthe next line, a number is given followed by (P), (Q),(R), (S) and (T). You have to complete the seriesstarting with the number given following thesequence of the given series. Then answer thequestion given below it.

26. 2 3 8 275 (P) (Q) (R) (S) (T)Which of the following numbers will come in placeof (T) ?(A) 184 (B) 6(C) 925 (D) 45

27. 5 18 48 1127 (P) (Q) (R) (S) (T)Which number will come in place of (S) ?(A) 172 (B) 276(C) 270 (D) 376

28. 15 159 259 3237 (P) (Q) (R) (S) (T)Which of the following numbers will come in placeof (R) ?(A) 251 (B) 315(C) 176 (D) 151

PAGE # 71

Directions : (29 to 35) Find the wrong term(s) �

29. 9, 11, 15, 23, 39, 70, 135(A) 23 (B) 39(C) 70 (D) 135

30. 3, 9, 36, 72, 216, 864, 1728, 3468(A) 3468 (B) 1728(C) 864 (D) 216

31. 2, 5, 11, 20, 30, 47, 65(A) 5 (B) 20(C) 30 (D) 47

32. 121, 143, 165, 186, 209(A) 143 (B) 165(C) 186 (D) 209

33. 9, 15, 24, 34, 51, 69, 90

(A) 15 (B) 24

(C*) 34 (D) 51(A) 15 (B) 24(C) 34 (D) 51

34. 9, 13, 21, 37, 69, 132, 261

(A) 21 (B) 37

(C) 69 (D) 132

35. 105, 85, 60, 30, 0, � 45, � 90

(A) 85 (B) � 45

(C) 105 (D) 0

EXERCISE-2


1. X, U, S, P, N, K, I, ?(A) J (B) K(C) M (D) F

2. Z, X, U, Q, L, ?(A) F (B) K(C) G (D) E

3. A, H, N, S, W, ?(A) A (B) Y(C) B (D) Z

4. Q, T, V, Y, A, ?(A) B (B) C(C) D (D) F

5. X, A, D, G, J, ?(A) N (B) O(C) M (D) P

6. Z, L, X, J, V, H, T, F, ?, ?(A) R, D (B) R, E(C) S, E (D) Q, D

7. AZ, YB, CX, WD, ?(A) VE (B) UE(C) EU (D) EV

8. DFK, FEL, HDM, JCN, ?(A) KBN (B) KBO(C) LBO (D) LBN

9. JXG, HTJ, FPN, ?, BHY(A) EKS (B) ELS(C) DLR (D) DLS

10. CYD, FTH, IOL, LJP, ?(A) PET (B) OET(C) OEY (D) PEV

11. ZGL, XHN, VIQ, TJU, ?(A) RKX (B) RKY(C) RLZ (D) RKZ

12. MTH, QRK, UPN, YNQ, ?(A) CKT (B) ELT(C) CLT (D) EKT

13. ZSD, YTC, XUB, WVA, ?(A) VZZ (B) ZVX(C) VWZ (D) VZX

14. RML, VIJ, ZFH, DDF, ?(A) HDC (B) CHI(C) HCD (D) DIC

15. LRX, DJP, VBH, NTZ, ?(A) ELS (B) FMR(C) GKS (D) FLR

16. MAD, OBE, SCH, YDM, ?(A) HET (B) HES(C) GET (D) UAE

17. 2B, 4C, 8E, 14H, ?(A) 22L (B) 24L(C) 22K (D) 2M

18. 1 BR, 2 EO, 6 HL, 15 KI, ?(A) 22 NF (B) 31 NF(C) 31 NE (D) 28 NF

19. P3C, R5F, T8I, V12L, ?(A) Y17O (B) X17M(C) X17O (D) X16O

20. Z 15 A, W 13 C, ?, Q 9 G, N 7 I(A) T 12 E (B) R 11F(C) T 11E (D) R 13 D

21. B3M, E7J, H15G, K31D, ?(A) N65A (B) O63A(C) N63A (D) N63Z

22. 5X9, 8U12, 11R15, 14O18, ?(A) 17L21 (B) 17K21(C) 17M21 (D) 17L23

23. 6C7, 8F10, 11J14, 15O19, ?(A) 19U24 (B) 20U25(C) 19U25 (D) 20U24

24. B2E, D5H, F12K, H27N, ?(A) J58Q (B) J56Q(C) J57Q (D) J56P

PAGE # 72

Directions : (25 to 30) Find the wrong term(s) :

25. ECA, JHF, OMK, TQP, YWU(A) ECA (B) JHF(C) TQP (D) YWU

26. DKY, FJW, HIT, JHS, LGQ(A) FJW (B) LGQ(C) JHJ (D) HIT

27. DVG, FSI, HPK, JNM, LJO(A) DVG (B) JNM(C) HPK (D) LJO

28. CDF, DEG, EFH, FHI(A) CDF (B) DEG(C) FHI (D) EFH

29. ZLA, BMY, CNW, FOU, HPS(A) ZLA (B) BMY(C) FOU (D) CNW

30. G4T, J10R, M20P, P43N, S90L

(A) G4T (B) J10R

(C) M20P (D) P43N

EXERCISE-3

Directions : (1 to 15) Which sequence of letters when placedat the blanks one after the other will complete thegiven letter series ?

1. a _ b a a _ a a _ _ a b(A) a a a a (B) b a a a(C) b b a a (D) a b b a

2. _ a a b b _ a _ a b _ b(A) b b a a (B) b a b a(C) b a a b (D) a b a b

3. a a b _ a a a _ b b a _(A) b a a (B) a b b(C) b a b (D) a a b

4. a _ _ b _ a _ a b _ a a(A) a b a a b (B) b b a b a(C) b b a b b (D) b a a b a

5. abc _ d _ bc _ d _ b _ cda(A) bacdc (B) cdabc(C) dacab (D) dccbd

6. a _ bbc _ aab _ cca _ bbcc(A) bacb (B) acba(C) abba (D) caba

7. _ b c _ _ b b _ a a b c(A) acac (B) babc(C) abab (D) aacc

8. _ b c c _ ac _ a a b b _ a b _ c c(A) aabca (B) abaca(C) bacab (D) bcaca

9. a _ cab _ a _ c _ b c(A) bbac (B) abab(C) abba (D) bcba

10. ba _ cb _ b _ bab _(A) acbb (B) bcaa(C) cabb (D) bacc

11. a _ bc _ a _ bcda _ ccd _ bcd _(A) abddbd (B) acbdbb(C) adbbad (D) bbbddd

12. cc _ ccdd _ d _ cc _ ccdd _ dd(A) dcdcc (B) dcddc(C) dccdd (D) None of these

13. a _ b a a _ b a a _ b a(A) a a b (B) b a b(C) b b a (D) b b b

14. b a b b b _ b _ b _ b b(A) b b a (B) b a a(C) a b a (D) a a a

15. m _ l _ ml _ m _ llm(A) lmmm (B) lmlm(C) lmml (D) mllm

Directions : (16 to 19) The questions given below are basedon the letter series, In each of these series, someletters are missing. Select the correct alternative. Ifmore than five letters are missing, select the lastfive letters of the series.

16. _ _ r _ ttp _ _ s _ tp _ _ _ s _ _ _(A) rstqp (B) tsrqp(C) rstpq (D) None

17. _ _ x _ zbxazyxabyz _ _ _ _ _(A) abxzy (B) abzxy(C) abxyz (D) bxayz

18. x _ xxy _ x _ xy _ yxx _ _ yy _ y(A) xyyyy (B) xxyyx(C) yxxyx (D) xyxyx

19. _ _ r _ tqrptsrpqst _ _ _ _ _(A) pqrts (B) pqtrs(C) pqrst (D) qrpst

Directions : (20 to 23) There is a letter series in the first rowand a number series in the second row. Eachnumber in the number series stands for a letter inthe letter series. Since in each of that series someterm are missing you have to find out as to whatthose terms are, and answer the questions basedon these as given below in the series.

20. a b _ c d _ a _ a b d _ d b a _1 _ 3 _ 3 2 _ 1 _ _ _ 4 _ _ _ _The last four terms in the series are(A) 1234 (B) 3112(C) 3211 (D) 4312

PAGE # 73

21. _ b n t _ _ n a m _ n a b _ _ a _ _ _ _1 3 _ 2 5 3 _ _ 5 2 4 _ 3 2 5 _ _ _ _ _The last five terms in the series are(A) 13425 (B) 41325(C) 34125 (D) 13452

22. n _ g f _ t _ f h t n _ _ t _ b _ f1 3 _ 2 4 5 0 _ 4 _ _ 3 _ _ _ _ _ _The last five terms of the number series are(A) 50123 (B) 40321(C) 40231 (D) 51302

23. _ m i a x _ i r x a _ _ m a _ _ _ _ _ _4 _ 5 _ 7 3 _ _ _ 6 _ _ _ _ _ _ _ _ _ _The last five term of the letter series are(A) r m x i a (B) x m r a i(C) x r m a i (D) r m i x a

Directions : (24 to 26) In each of the following questions,three sequences of letter/numbers are given whichcorrespond to each other in some way. In eachquestion, you have to find out the letter/numeralsthat come in the vacant places marked by (?). Theseare given as one of the four alternatives under thequestion. Mark your answer as instructed.

24. _ A C _ B D _ C D C D2 _ 4 1 _ 1 4 _ _ _ _r s _ q r _ p ? ? ? ?(A) p q p q (B) p r p r(C) r q r q (D) r s r s

25. A _ B A C _ D _ B C D C_ 4 _ 3 _ 2 _ 5 ? ? ? ?d c _ _ b a c b _ _ _ _(A) 2 4 5 4 (B) 2 5 4 5(C) 3 4 5 4 (D) 4 5 2 5

26. _ A D A C B _ _ B D C C2 4 _ _ 2 3 5 3 _ _ _ _p _ _ q _ _ r s ? ? ? ?(A) p r s s (B) p s r r(C) r p s s (D) s r p p

EXERCISE-4

Directions : (1 to 39) Find the missing term in the givenfigures

1.

(A) 36 (B) 9(C) 25 (D) 64

2.

(A) 14 (B) 18(C) 11 (D) 13

3.

(A) 112 (B) 92(C) 82 (D) 102

4.

(A) 235 (B) 141(C) 144 (D) 188

5.

18 3012

6

3216 40

8

36 18 27

?

(A) 18 (B) 12(C) 9 (D) 6

6. 12 21 ?6 7 85 6 4

4 5 10

(A) 14 (B) 22(C) 32 (D) 320

7.

5 9 8

5 15 ?

3 5 6

(A) 12 (B) 11(C) 16 (D) 26

8.

(A) 72 (B) 18(C) 9 (D) 19

9.

(A) 1 (B) 18(C) 90 (D) 225

10.

(A) 20 (B) 22(C) 24 (D) 12

PAGE # 74

11.

7 11 4912 8 5415 4 ?

(A) 36 (B) 7(C) 25 (D) 0

12.

18 24 3212 14 163 ? 472 112 128

(A) 2 (B) 3(C) 4 (D) 5

13. C26

5

3

2

4 H70

4

5

4

10 J90

6

?

8

6

(A) 1 (B) 3(C) 4 (D) 5

14. 80 70 80

29 29 5927 30 40

33 31 10

43 44 2045 43 39

39 42 ?

(A) 69 (B) 49(C) 50 (D) 60

15.

101 48

35 5615 184

43 3438 ?

(A) 127 (B) 142(C) 158 (D) 198

16.

1 7 63 3 ?5 4 8

35 74 104

(A) 1 (B) 2(C) 3 (D) 4

17.

(A) 33 (B) 145(C) 135 (D) 18

18.

(A) 28 (B) 36(C) 81 (D) 49

19. 72 6 140

5

8

4 10 7

?

2

6 2

4 314 12 3

6

6 8

(A) 16 (B) 14(C) 20 (D) 22

20.5 3 94 8 4

20 24 ?9 11 13

(A) 117 (B) 36(C) 32 (D) 26

21.

(A) 26 (B) 25(C) 27 (D) 30

22.

12 16 3630 40 3418 32 18

30 44 ?

(A) 48 (B) 9(C) 44 (D) 64

23.

5

16 109 2

6

21

22 53 19

15

51

17 ? 48

13

(A) 25 (B) 129(C) 7 (D) 49

24.

4 5 4

33 54 ?3 4 32 2 5

2 3 6

(A) 78 (B) 82(C) 94 (D) 86

PAGE # 75

25.

2

4 5

3

28

5

7 4

3

38

1

2 3

7

?

(A) 14 (B) 18(C) 11 (D) 26

26.

(A) 9 (B) 11(C) 10 (D) 12

27. BIG - 792 HCA - 138 FED - 456 E?H - 87?

(A) G, 6 (B) I, 9(C) G, 5 (D) I, 5

28. 26 21 ?

36 9 25

25 16 36

64 25 14449 81 64

(A) 19 (B) 23(C) 25 (D) 31

29. 2 1 ?8 6 83 4 12

2 3 6

6 8 4(A) 3 (B) 4(C) 5 (D) 6

30. 80 70 80

29 29 5927 30 40

33 31 10

43 44 2045 43 39

39 42 ?

(A) 69 (B) 49(C) 50 (D) 60

31.

(A) 0 (B) 2(C) 3 (D) 1

32.

(A) 12 (B) 9(C) 14 (D) 10

33 Find the missing letters from left to right.

(A) JSN (B) JNS(C) JRS (D) KRS

34.3 8 10 2 ? 1

6 56 90 2 20 0

(A) 0 (B) 3(C) 5 (D) 7

35. 80 65 ?15 9 132 7 16

5 4 116 6 8

(A) 48 (B) 72(C) 35 (D) 120

36.

(A) 38 (B) 64(C) 4 (D) 16

37.

101 48

35 5615 184

43 3438 ?

(A) 127 (B) 142(C) 158 (D) 198

38.

18 32 1830 40 2712 16 36

6 8 ?(A) 18 (B) 12(C) 9 (D) 6

39.4 9

6

9 16

12

16 ?

20

(A) 60 (B) 50(C) 21 (D) 25

40. Find the value of X in the following figure :

15 4

33 2

27 2

36 8

32 X

18 9

22 11

12 3

(A) 3 (B) 4(C) 8 (D) 12

PAGE # 76

Directions : (1 to 5) Read the following information carefullyand answer the questions given below it.(i). Five professors (Dr. Joshi, Dr. Davar, Dr.Natrajan, Dr. Choudhary and Dr. Zia) teach fivedifferent subjects (zoology, physics, botany, geologyand history) in four universities ( Delhi, Gujarat,Mumbai, and Osmania). Do not assume anyspecific order.(ii). Dr. Choudhary teaches zoology in MumbaiUniversity .(iii). Dr. Natrajan is neither in Osmania Universitynor in Delhi University and he teaches neithergeology nor history.(iv). Dr. Zia teaches physics but neither in MumbaiUniversity nor in Osmania University.(v). Dr. Joshi teaches history in Delhi University.(vi). Two professors are from Gujarat University.(vii). One professor teaches only one subject andin one University only.

Ex 1. Who teaches geology ?(A) Dr Natrajan (B) Dr. Zia(C) Dr. Davar (D) Dr. Joshi

Ex 2. Which university is Dr. Zia from ?(A) Gujarat (B) Mumbai(C) Delhi (D) Osmania

Ex 3. Who teaches botany ?(A) Dr. Zia (B) Dr. Davar(C) Dr. Joshi (D) Dr. Natrajan

Ex 4. Who is from Osmania University ?(A) Dr. Natrajan (B) Dr. Davar(C) Dr. Joshi (D) Dr. Zia

Ex 5. Which of the following combinations is correct ?(A) Delhi University - Dr. Zia(B) Dr. Choudhary - geology(C) Dr. Davar - Mumbai University(D) Dr. Natranjan - Gujarat University

Sol. : (1 to 5)From the given information in the question :From II, we get Dr. Choudhary teaches zoology inMumbai University.

From III, We get Dr. Natrajan is neither in Osmanianor in Delhi University. Therefore, he will be eitherat Mumbai or Gujarat University. Similarly, as heteaches neither geology nor history, therefore, hemust be teaching physics or botany. ..........(1)From IV, Dr. Zia Physics but as he is not teachingin either Mumbai or Osmania University, he mustbe teaching either in Delhi or Gujarat University...(2)Form V, we get Dr Joshi teaches history in DelhiUniversity Form (1) and (2), we conclude that DrNatarajan teaches botany. And from (1), (2) and VI,we get both Natarajan and Zia teach in GujaratUniversity. Finally, On summarisation we canprepare the following table.

PUZZLE TEST

Names University SubjectDr. Joshi Delhi HistoryDr. Davar Osmania GeologyDr. Natrajan Gujarat BotanyDr. Choudhary Mumbai Zoology

Dr. Zia Gujarat Physics

On the basis of the above table, rest of the questionscan be solved very easily.

1. (C) Dr. Davar teaches geology.

2. (A) Dr. Zia is from Gujarat university.

3. (D) Dr. Natrajan teaches botany.

4. (B) Dr. Davar is from Osmania University.

5. (D) Dr. Natranjan - Gujarat University is the correctcombination.

Ex 6. Ramesh is taller than Vinay who is not as tall asKaran. Sanjay is taller than Anupam but shorterthan Vinay. Who among them is the tallest ?(A) Ramesh (B) Karan(C) Vinay (D) Cannot be determined

Sol. (D) In this question ranking of Karan is not defined.Consequently, either Ram or Karan occupies thetop position with regard to height. Hence,option (d) is the correct choice.

Directions : (7 to 11) Read the following information carefullyand answer the questions given below it :There are five men A, B, C, D and E and six womenP, Q, R, S, T and U. A, B and R are advocates; C, D,P, Q and S are doctors and the rest are teachers.Some teams are to be selected from amongstthese eleven persons subject to the followingconditions :A, P and U have to be together.B cannot go with D or R.E and Q have to be together.C and T have to be together.D and P cannot go together.C cannot go with Q.

Ex 7. If the team is to consist of two male advocates, twolady doctors and one teacher, the members of theteam are(A) A B P Q U (B) A B P U S(C) A P R S U (D) B E Q R S

Sol. (B) The male advocates are A and B, lady doctorsare P, Q and S ; teachers are E, T and U.Now, A and B will be selected.A, P and U have to be together. Now, we have toselect one lady doctor more. It can be Q or S. But Qand E have to be together. Since E is not selected,so S will be selected. Thus, the team is A B P U S.


PAGE # 77

Ex 8. If the team is to consist of one advocate, two

doctors, three teachers and C may not go with T,

the members of the team are :

(A) A E P Q S U (B) A E P Q T U

(C) B E Q S T U (D) E Q R S T U

Sol. (B) The advocates are A, B and R ; doctors are

C, D, P, Q, S ; teachers are E, T and U. The team

consists of 3 teachers i.e. E, T, U. Now, A, P and U

have to be together. E and Q have to be together.

Thus, the team is A E P Q T U.

Ex 9. If the team is to consist of one male advocate, one

male doctor, one lady doctor and two teachers, the

members of the team are :

(A) A C P T U (B) A D E P T

(C) A D E P U (D) B C E Q U

Sol. (A) The male advocates are A and B ; male doctors

are C and D ; lady doctors are P, Q and S ; teachers

are E, T and U. If A is selected, P and U will be

selected. D and P cannot go together. So, a male

doctor C will be selected. C and T have to be

together. Thus, the team is A C P T U. If B is

selected, D will not be selected. So, male doctor C

will be chosen. C and T have to be together. Now,

the second teacher to be selected is E or U. But, U

cannot go without A. So, E will be selected. E and Q

have to be together. Thus, the team can also be

B C E Q T.

Ex 10. If the team is to consist of one advocate, three

doctors and one male teacher, the members of

the team are:

(A) A D P S U (B) C D R S T

(C) D E Q R S (D) D E Q R T

Sol. (C) The advocates are A, B and R ; the doctors are

C, D, P, Q and S ; male teacher is E. Clearly, E will

be selected. E and Q have to be together. C and Q

cannot be together. So, C will not be selected. P

also cannot be selected because U is not selected.

So, two other doctors D and S will be selected. P is

not selected, so A will not be selected. D is

selected, so B cannot be selected. Thus, the team

is D E Q R S.

Ex 11. If the team is to consist of two advocates, two

doctors, two teachers and not more than three

ladies, the members of the team are :

(A) A B C P T U (B) A C P R T U

(C) A E P Q R T (D) B C E Q R T

Sol. (A) A C P R T U and A E P Q R T are wrong because

each of these combinations consists of four ladies.

B C E Q R T is incorrect because B and R cannot

go together.

Directions : (12 to 15) Read the following paragraphcarefully :Four women A, B, C and D and three men E, F andG play bridge, a game for four players.(i) The group consists of three married couplesand a widow.(ii) Spouses are never partners in a game.(iii) No more than one married couple ever plays inthe same game.(iv) One day they played four games as follows.

A and E versus B and F.A and G versus D and F.B and C versus F and G.C and E versus D and G.

Ex 12. Whom is E married to ?(A) A (B) B(C) C (D) D

Ex 13. Whom is F married to ?(A) A (B) B(C) C (D) D

Ex 14. Whom is G married to ?(A) A (B) B(C) C (D) D

Ex 15. Which of the following is a widow ?(A) A (B) B(C) C (D) D

Sol. : (12 to 15)From (iv), is married either to A or to C. If F is marriedto A, then G is married to B or to C. If G is married toB, then E is married to D ; if G is married to C, thenE is married to B or to D. If F is married to C, then Gis married to B ; then E is married to D. Hence, themarried couples are : FA, GB, ED or FA, GC, EB orFA, GC, ED or FC, GB, ED. Of these, only FA, GB,ED does not contradict any of the statements.

12. (D) E is married to D.

13. (A) F is married to A.

14. (B) G is married to B.

15. (C) C is a widow.

Ex 16. A vagabond runs out of cigarettes. He searches forthe stubs, having learnt that 7 stubs can make anew cigarette, good enough to be smoked, hegathers 49 stubs, If he smokes 1 cigarette everythree - quarters of an hour, how long will his supplylast ?(A) 5.25 hr (B) 6 hr(C) 4.5 hr (D) 3 hr

Sol. (B) He has got = 7749

cigarettes.

The duration of time he will take to smoke these

7 cigarettes = 43

7 hr = 5.25 hr (i.e. 5 hr and 15

min). Now note that after he has smoked these 7cigarettes, he will collect 7 more stubs (one formeach), form which he will be able to make another

cigarette. This will take him another 43

hr (45 min)

to smoke. Therefore, total time taken = 6hr.

PAGE # 78

Directions : (17 to 18) Read the following information andanswer the questions that follow.There are 70 clerks working with M/s. Jha LalKhanna & Co. chartered accountants, of which 30are female.(i) 30 clerks are married.(ii) 24 clerks are above 25 years of age(iii) 19 Married clerks are above 25 years of age;among them 7 are males.(iv) 12 males are above 25 years of age(v) 15 males are married.

Ex 17. How many unmarried girls are there ?(A) 12 (B) 15(C) 18 (D) 10

Ex 18. How many of these unmarried girls are above 25 ?(A) 12 (B) 15(C) 4 (D) 0

Sol. (17 to 18) : From the given data, we can make thefollowing table with the help of which rest of thequestions can be solved very easily.

Male (40) Female (30)

Above 25

Married 7 12

Unmarried 5 0

Below 25

married 8 3

unmarried 20 15

Total 40 30

17. There are 15 unmarried girls.

18. In these 15 unmarried girls no one is above 25.

EXERCISE

Directions : (1 to 5) Study the following information carefullyand answer the questions given below it :There are five friends A, B, C, D and E. Two of themare businessmen while the other three belong todifferent occupations viz. medical, engineer andlegal. One businessman and the lawyer stay inthe same locality S, while the other three stay inthree different localities P, Q and R. Two of thesefive persons are Hindus while the remaining threecome from three different communities viz. Muslim,Christian and Shikh. The lawyer is the oldest inage while one of the businessmen who runs afactory is the youngest. The other businessman isa cloth merchant and agewise lies between thedoctor and the lawyer. D is a cloth merchant andstays in locality S while E is a Muslim and stays inlocality R. The doctor is a Christian and stays inlocality P, B is a Shikh while A is a Hindu and runsa factory.

1. Who stays in locality Q ?(A) A (B) B(C) C (D) E

2. What is E�s occupation ?(A) Business (B) Engineer(C) Lawyer (D) Doctor

3. Agewise who among the following lies between Aand C ?(A) Lawyer (B) Doctor(C) Cloth merchant (D) Engineer

4. What is B�s occupation ?(A) Business (B) Engineer(C) Lawyer (D) Doctor

5. What is C�s occupation ?(A) Doctor (B) Lawyer(C) Engineer (D) Business

Directions : (6 to 10) Study the given information carefullyand answer the questions that follow.There are four people sitting in a row : one eachfrom India, Japan, USA and Germany, but not inthat order,. They are wearing caps of different colours - green,yellow, red and white, not necessarily in that order.II. One is wearing a kurta and one a T-shirt.III. The Indian is wearing a green cap and a jacket.IV. The American is not seated at either end.V. The persons with kurta and T-shirt are sittingnext to each other.VI. The persons with kurta wears a red cap andsits next to the Japanese.VII. The Japanese wears a shirt and is not seatedat either end.VIII. The man with white cap wears T-shirt and isseated at one end.

6. Who wears the T-shirt ?(A) Indian (B) Japanese(C) American (D) German

7. Who is wearing a kurta ?(A) Indian (B) Japanese(C) American (D) German

8. What is the colour of the cap worn by the Japanese?(A) Red (B) Green(C) Yellow (D) White

9. Who precedes the man wearing T-shirt ?(A) Indian (B) Japanese(C) American (D) German

10. Who precedes the man wearing jacket ?(A) Indian (B) German(C) Japanese (D) Cannot say

PAGE # 79

Directions : (11 to 15) Read the following informationcarefully and answer the questions that follow.I. There are six students ( A, B, C, D, E and F) in agroup. Each student can opt for only three choicesout of the six which are music, reading, painting,badminton, cricket and tennis.II. A, C and F like reading.III. D does not like badminton, but likes music.IV. Both B and E like painting and music.V. A and D do not like painting, but they like cricket.VI. All student except one like badminton.VII. Two students like tennis.VIII. F does not like cricket, music and tennis.

11. Which pair of students has the same combinationof choices ?(A) A and C (B) C and D(C) B and E (D) D and F

12. Who among the following students likes bothtennis and cricket ?(A) A and B (B) C(C) B and D (D) D

13. How many students like painting and badminton ?(A) 1 (B) 2(C) 3 (D) 4

14. Who among the following do not like music ?(A) A , C and D (B) A, B and C(C) A, C and F (D) B, D and F

15. Which of the following is the most popular choice?(A) Tennis (B) Badminton(C) Reading (D) Painting

16. R earns more than H but not as much as T, Mearns more than R. Who earns least amongthem?(A) R (B) T(C) H (D) M

17. Harish is taller than Manish but shorter thanSuresh. Manish is shorter than Anil but taller thanRaghu. Who among them is the shortest havingregard to height ?(A) Anil (B) Manish(C) Raghu (D) Cannot be determined

Direction : (18) Examine the following statements :I. Either A and B are of the same age or A is olderthan B.II. Either C and D are of the same age or D is olderthan C.III. B is older than C.

18. Which one of the following conclusions can bedrawn from the above statements ?(A) A is older than B(B) B and D are of the same age(C) D is older than C(D) A is older than C

Directions : (19 to 23) Read the information given belowand answer the questions.The age and height of six children in a class are asfollows :(i) A is taller and older than B but shorter andyounger than C.(ii) D is taller than E who is not as tall as B.(iii) The oldest is the shortest.(iv) The youngest would be fourth if the childrenstood in a line according to their height and onestarted counting from the tallest.(v) D is younger than F but older than E who isolder than C.

19. Who among them is the tallest ?(A) B (B) E(C) C (D) Data inadequate

20. Who is older than B but younger than C ?(A) F (B) D(C) A (D) Data inadequate

21. Which of the following statements is definitely true?(A) D is the most old person(B) B has the max. height(C) A is older than D(D) F is the shortest

22. Which of the following is the correct order of heightin descending order?(A) A, C, D, B, E, F (B) F, D, E, C, A, B(C) D, C, A, B, E, F (D) C, D, A, B, E, F

23. Whose Rank in height cannot be positioneddefinitely ?(A) B (B) D(C) C (D) E

Directions : (24 to 28) Study the information given belowand answer the questions that follow.(i) Six Plays P, Q, R, S, T and U are to be organisedfrom Monday to Saturday i.e. 10 to 15 one play eachday.(ii) There are two plays between R and S and oneplay between P and R.(iii) There is one play between U and T and T is tobe organised before U.(iv) Q is to be organised before P, not necessarilyimmediately.(v) The organisation does not start with Q.

24. The organisation would start from which play ?(A) P (B) S(C) T (D) None

25. On which date is play T to be organised ?(A) 10th (B) 11th

(C) 12th (D) None

26. The organisation would end with which play ?(A) P (B) Q(C) S (D) None

PAGE # 80

27. Which day is play Q organised ?(A) Tuesday (B) Wednesday

(C) Thursday (D) None

28. Which of the following is the correct sequence oforganising plays ?(A) PTRUQS (B) QSTURP

(C) SUTRQP (D) None

Directions : (29 to 30) Read the following informationcarefully and answer the questions given below it.I. Seven books are placed one above the other in a

particular way .II. The history book is placed directly above thecivics book.

III. The geography book is fourth from the bottomand the English book is fifth from the top.IV. There are two books in between the civics and

economics books.

29. To find the number of books between the civicsand the science books, which other extra piece ofinformation is required, from the following ?

(A) There are two books between the geographyand the science books.(B) There are two books between the mathematics

and the geography books .(C) There is one book between the English andthe science books.

(D) The civics book is placed before two booksabove the economics book.

30. To know which three books are kept above theEnglish book, which of the following additional

pieces of information, if any, is required?(A) The economics book is between the Englishand the science books.

(B) There are two books between the English andthe history books.(C) The geography book is above the English book.

(D) No other information is required.

Directions : (31 to 32) A five-member team that includesRama, Shamma, Henna, Reena, and Tina, isplanning to go to a science fair but each of them

put up certain conditions for going .They are asfollows.I. If Rama goes, then at least one amongst

Shamma and Henna must go.II. If Shamma goes, then Reena will not go.III. If Henna will go, then Tina must go.

IV. If Reena goes, then - Henna must go.V. If Tina goes, then Rama must go but Shammacannot go.

VI. If Reena plans not to go the fair, then Rama willalso not go.

31. If it is sure that Henna will go to the fair, then who

among the following will definitely go ?

(A) Rama (B) Shamma

(C) Reena (D) Rama and Reena

32. If Tina does not go to the fair, which of the following

statements must be true ?

(i) Henna cannot go

(ii) Shamma cannot go

(iii) Reena cannot go

(iv) Rama cannot go

(A) (i) and (ii) (B) (iii) and (iv)

(C) (i), (iii) and (iv) (D) (i) and (iv)

Directions : (33 to 37) Read the following paragraph

carefully and choose the correct alternative.

The office staff of XYZ corporation presently

consists of three females A, B, C and five males D,

E, F, G and H. The management is planning to

open a new office in another city using three males

and two females of the present staff. To do so they

plan to separate certain individuals who do not

function well together. The following guidelines

were established

I. Females A and C are not to be together

II. C and E should be separated

III. D and G should be separated

IV. D and F should not be part of a team.

33. If A is chosen to be moved, which of the following

cannot be a team ?

(A) ABDEH (B) ABDGH

(C) ABEFH (D) ABEGH

34. If C and F are to be moved to the new office, how

many combinations are possible ?

(A) 1 (B) 2

(C) 3 (D) 4

35. If C is chosen to the new office, which number of

the staff cannot be chosen to go with C ?

(A) B (B) D

(C) F (D) G

36. Under the guidelines, which of the following must

be chosen to go to the new office ?

(A) B (B) D

(C) E (D) G

37. If D goes to the new office, which of the following

is/are true ?

I. C cannot be chosen

II. A cannot be chosen

III. H must be chosen.

(A) I only (B) II only

(C) I and II only (D) I and III only

PAGE # 81

Directions : (38 to 42) Study the following information

carefully and answer the questions that follow :

A team of five is to be selected from amongst five

boys A, B, C, D and E and four girls P, Q, R and S.

Some criteria for selection are :

A and S have to be together

P cannot be put with R.

D and Q cannot go together.

C and E have to be together.

R cannot be put with B.

Unless otherwise stated, these criteria are

applicable to all the questions below :

38. If two of the members have to be boys, the team

will consist of :

(A) A B S P Q (B) A D S Q R

(C) B D S R Q (D) C E S P Q

39. If R be one of the members, the other members of

the team are :

(A) P S A D (B) Q S A D

(C) Q S C E (D) S A C E

40. If two of the members are girls and D is one of the

members, the members of the team other than D

are :

(A) P Q B C (B) P Q C E

(C) P S A B (D) P S C E

41. If A and C are members, the other members of the

team cannot be :

(A) B E S (B) D E S

(C) E S P (D) P Q E

42. If including P at least three members are girls, the

members of the team other than P are :

(A) Q S A B (B) Q S B D

(C) Q S C E (D) R S A D

Directions : (43 to 44) Read the given information carefully

and answer the questions that follow :

Ratan, Anil, Pinku and Gaurav are brothers of Rakhi,

Sangeeta, Pooja and Saroj, not necessarily in that

order. Each boy has one sister and the names of

bothers and sisters do not begin with the same

letter. Pinku and Gaurav are not Saroj �s or

Sangeeta�s brothers. Saroj is not Ratan�s sister.

43. Pooja�s brother is

(A) Ratan (B) Anil

(C) Pinku (D) Gaurav

44. Which of the following are brother and sister ?

(A) Ratan and Pooja (B) Anil and Saroj

(C) Pinku and Sangeeta (D) Gaurav and Rakhi

Directions : (45 to 49) Read the following informationcarefully and answer the questions given below.(i) There is a family of six persons- L, M, N, O, Pand Q. They are professor, businessman,chartered account, bank manager, engineer andmedical representative, not necessarily in thatorder.(ii) There are two married couples in the family.(iii) O, the bank manager is married to the ladyprofessor.(iv) Q, the medical representative, is the son of Mand brother of P.(v) N, the chartered accountant, is the daughter - inlaw of L.(vi) The businessman is married to the charteredacconuntant.(vii) P is an unmarried engineer.(viii) L is the grandmother of Q

45. How is P related to Q.(A) Brother (B) Sister(C) Cousin (D) Either brother or sister

46. Which of the following is the profession of M ?(A) Professor(B) Chartered accountant(C) Businessman(D) Medical representative

47. Which of the following is the profession of L ?(A) Professor (B) Charted accountant(C) Businessman (D) Engineer

48. Which of the following is one of the couples ?(A) QO (B) OM(C) PL (D) None of these

49. How is O related to Q?(A) Father (B) Grandfather(C) Uncle (D) Brother

Directions : (50 to 54)I. There is a group of six persons P,Q, R, S, T and Ufrom a family. They are Psychologist, Manager,Lawyer, Jeweller, Doctor and Engineer.II. The Doctor is grandfather of U, who is aPsychologist.III. The Manager S is married to P.IV. R, the Jeweller is married to the Lawyer.V. Q is the mother of U and T.VI. There are two married couples in the family.

50. What is the profession of T ?(A) Doctor (B) Jeweller(C) Manager (D) None of these

51. How is P related to T ?(A) Brother (B) Uncle(C) Father (D) Grandfather

52. How many male members are their in the family ?(A) One (B) Three(C) Four (D) Data inadequate

PAGE # 82

53. What is the profession of P ?(A) Doctor (B) Lawyer(C) Jeweller (D) Manager

54. Which of the following is one of the pairs of couplesin the family ?(A) PQ (B) PR(C) PS (D) Cannot be determined

Direction : (55) The ages of Mandar, Shivku, Pawan andChandra are 32, 21, 35 and 29 years, not in order,Whenever asked they lie of their own age but tellthe truth abut others.(i) Pawan says, �My age is 32 and Mandar�s age isnot 35�(ii) Shivku says, �My age is not 2 9 and Pawan�sage in not 21�(iii) Mandar says, �My age is 32.�

55. What is Chandra�s age ?(A) 32 years (B) 35 years(C) 29 years (D) 21 years

Directions : (56 to 57) Answer the questions on the basisof the information given below. 5 friends Nitin,Reema, Jai, Deepti and Ashutosh are playing agame of crossing the roads. In the beginning, Nitin,Reema and Ashutosh are on the one side of theroad and Deepti and Jai are on the other side. Atthe end of the game, it was found that Reema andDeepti are on the one side and Nitin, Jai andAshutosh are on the other side of the road. Rulesof the game are as follows :I. One �movement� means only one person crossesthe road from any side to the other side.II. No two persons can cross the roadsimultaneously from any side to the other side.

III. Two persons from the same side of the roads

cannot move in consecutive �movements�.

IV. If one person crosses the road in a particular

movement, he or she cannot immediately move

back to the other side.

V. Jai and Reema did not take part in first 3

movements.

56. What is the minimum possible number of

movements that took place in the entire game ?

(A) 3 (B) 4

(C) 5 (D) 6

57. If number of movements are minimised in the

game, then which of the following combination of

friends can never be together on one particular

side of the road during the course of the game ?

(A) Nitin, Reema amd Deepti

(B) Nitin, Jai and Deepti

(C) Deepti, Jai and Ashutosh

(D) Ashutosh, Nitin and Deepti

58. You have 12 similar looking coins. 11 of them weigh

the same. One of them has a different weight, but

you don�t know whether it is heavier or lighter. You

also have a scale. You can put coins on both sides

of the scale and it�ll tell you which side is heavier or

will stay in the middle if both sides weigh the same.

What is the minimum number of weighing required

to find out the odd coin.

(A) 3 (B) 4

(C) 5 (D) 6

PAGE # 83

CALENDAR AND CLOCK TEST

We are to find the day of the week on a mentioneddate. Certain concepts are defined as under.

An ordinary year has 365 days.

In an ordinary year, first and last day of the year aresame.

A leap year has 366 days. Every year which isdivisible by 4 is called a leap year. For example1200, 1600, 1992, 2004, etc. are all leap years.

For a leap year, if first day is Monday than last daywill be Tuesday for the same year.

In a leap year, February is of 29 days but in anordinary year, it has only 28 days.

Year ending in 00's but not divisiable by 400 is notconsidered a leap year. e.g., 900, 1000, 1100, 1300,1400, 1500, 1700, 1800, 1900, 2100 are not leapyears.

The day on which calendar started (or the very firstday ) i.e., 1 Jan, 0001 was Monday.

Calendar year is from 1 Jan to 31 Dec. Financialyear is from 1 April to 31 March.

ODD DAYS

The no. of days exceeding the complete no. ofweeks in a duration is the no. of odd days duringthat duration.

COUNTING OF ODD DAYS

Every ordinary year has 365 days = 52 weeks +1 day. Ordinary year has 1 odd day.

Every leap year 366 days = 52 weeks + 2 days. Leap year has 2 odd days.

Odd days of 100 years = 5,Odd days of 200 years = 3,Odd days of 300 years = 1,Odd days of 400 years = 0.

Explanation : 100 years = 76 ordinary years + 24 leap years

( The year 100 is not a leap year)= 76 odd days + 2 × 24 odd days = 124 odd days.

Odd days = 7

124 = 5 odd days.

Similarly, 200 years = 10 odd days = 03 odd days

300 years = 7

15= 1 odd day..

400 years = 7

120 = 0 odd day (1 is added as 400

is a leap year)Similarly, 800, 1200, 1600, 2000, 2400 yearscontain 0 odd days.

After counting the odd days, we find the dayaccording to the number of odd days.

Sunday for 0 odd day, Monday for 1 odd day and soon as shown in the following table.

Table : 1 (Odd days for week days)

0 1 2 3 4 5 6

SaturdayMondaySunday Tuesday WednesdayDays

Odd Days

Thursday Friday

Table : 2 (Odd days for months in a year)

Ordinary Year

DaysOdd Days

Leap year DaysOdd

Days

January 31 3 January 31 3

February 28 0 February 29 1

March 31 3 March 31 3

April 30 2 April 30 2

May 31 3 May 31 3

June 30 2 June 30 2

Total 181 days 6 Total 182 days 0

July 31 3 July 31 3

August 31 3 August 31 3

September 30 2 September 30 2

October 31 3 October 31 3

November 30 2 November 30 2

December 31 3 December 31 3

Total 184 days 1 Total 184 days 2

Table : 3 (Odd days for every quarter)

Ivth three months1 Oct. to31 Dec.

Total year1 Jan to 31 Dec.

Ist three months1 Jan to

31 March

Monthsof

years

IInd three months1 Apr to30 June

IIIrd three months1 July to 30 Sep.

Total days90 / 91

Ord. / Leap91 92 92

365 / 366Ord. / Leap

1Odd day

1 / 2 Ord. / Leap

Odd days6 / 0

Ord. / Leap0

Odd day1

Odd day

Ex 1. If it was Saturday on 17th December 1982 whatwill be the day on 22nd December 1984 ?

Sol. Total number of odd days between 17 Dec.1982to 17 Dec.1984 the number of odd days = 1+2 = 3.From 17 to 22 Dec. number of odd days = 5 3 + 5 = 8 odd days = 1 odd day. Saturday + 1 odd day = Sunday.


PAGE # 84

Ex 2. Find the day of the week on 16 January, 1969.Sol. 1600 years have �0� odd day. .....................(A)

300 years have �1� odd day. ......................(B)68 years have 17 leap years and 51 ordinary years.Thus = (17 × 2 + 51 × 1 ) = 85 odd days

' 01' odd day ...(C)

16 January has = ' 02' odd days..(D)Adding (A) + (B) +(C) +(D),We get, 0 + 01 +01 +02 = 04 odd days

Ans. Thursday

Ex 3. Find the day of the week on 18 July, 1776 (leapyear).

Sol. Here 1600 years have �0� odd day.....................(A)100 years have �5� odd days..............................(B)75 years = (18 leap years + 57 ordinary years)= (18 × 2 + 57 × 1)

= 93 odd days= (7 × 13 + 2) = �2� odd days.............................(C)Now, the no. of days from 1st January to 18 July,1776= 182 + 18 = 200 days= (28 × 7 + 4) days = �4� odd days.....................(D)Adding (A) + (B) +(C) +(D),We get, 0 + 5 + 2 + 4 = 04 odd days

Ans. Thursday

Ex 4. On what dates of October, 1975 did Tuesday fall ?Sol. For determining the dates, we find the day on 1st

Oct, 1975.1600 years have �0� odd days.....................(A).300 years have �01� odd days.....................(B).74 years have (18 leap years + 56 ordinary years)2 × 18 + 1 × 56 = 92 odd days

= �01� odd days.............(C)Days from 1st January to 1st Oct., 19751st Jan � 30 June + 1st July to 1st Oct.181 + 31 + 31 + 30 + 1 = 274 days= �01� odd days......(D) (274/7= 01 days)Adding (A) + (B) +(C) +(D) = 0 + 01 +01 +01= '03' odd days

Ans. Wednesday( 1st Oct), hence 7,14,21,28 Oct. willTuesday fall.

Ex 5. Calendar for 1995 will serve for 2006, prove ?Sol. The Calendar for 1995 and 2006 will be the same

,if day on 1st January of both the years is the same.This is possible only if the total odd days between31st Dec. 1994 and 31st Dec.2005 is 0. [one daybefore both the years as we want to know the dayon 1st January of both the years i.e. same]During this period, we have3 leap years(1996, 2000, 2004) and08 ordinary years(1995,1997,1998,1999, 2001, 2002, 2003,2005)Total odd days = (2 × 3 + 1 × 8) = 14 = 0 odd

days (Thus Proved)

Ex 6. The year next to 1996 having the same Calendar

will be -

Sol. 1996 1997 1998 1999 2000 2001 2002 2003

2 1 1 1 2

Total = 2 + 1 + 1 + 1 + 2 = 7= 0 odd days

Hence, year 2001 will have the same calendar as

year 1996.

Ex 7. Prove that last day of a century cannot be Tuesday,

Thursday or Saturday.

Sol. 100 years have = 5 odd days

Last day of st century is Friday

200 years have = 10 odd days

Last day of IInd century is Wednesday

= 3 odd days

300 years have = 15 odd days

Last day of rd century is Monday

= 01 odd day

400 years have = (5 × 4 + 1)

Last day of 4th century is Sunday

= 21 odd days

= 0 odd days

Since the order keeps on cycling, we see that the

last day of the century cannot be Tuesday, Thursday

or Saturday.

Important Notes :

Minute hand and hour hand coincides once in every

hour. They coincide 11 times in 12 hours and 22

times in 24 hours.

They coincide only one time between 11 to 1 O�

clock. at 12 O� clock.

Minute hand and hour hand are opposite once in

every hour. They do it 11 times in 12 hours and 22

times in 24 hours.

They opposite only one time between 5 to 7 O�

clock. at 6 O� clock.

Both hands (minute and hour) are perpendicular

twice in every hour. 22 times in 12 hours and 44

times in 24 hours.

In one minute, hour hand moves 1/2º and minute

hand moves 6º. In one hour, hour hand moves 30º

and minute hand moves 360º.

In an hour, minute hand moves 55 minutes ahead

of hour hand.

PAGE # 85

HANDS COINCIDE

Ex.8 At what time between 3 O�Clock and 4 O�Clock will

the two hands coincide ?Sol. At 3 O�clock the distance between the two hands is

15 minutes when they coincide with each other thedistance between the two hands will be 0 min.So, the time taken (15 + 0 ) = 15 minutes.

Minute hand is 55 min. ahead of hour hand in 60 min. Minute hand is 1 min. ahead of hour hand in

5560 min.

Minute hand is 15 min. ahead of hour hand in

55

1560 =

11180

= 114

16 min.

Hence the right time is 114

16 minute past 3.

HANDS ARE OPPOSITE

Ex.9 At what time between 2 O�clock and 3 O�clock will

the two hands be opposite ?Sol. At 2 O�clock the distance between the two hands is

10 minutes. When they are at 30 minutes distance,they are opposite to each other. The time taken(30 + 10 ) = 40 min.

Minute hand is 55 min. ahead of hour hand in 60 min. Minute hand is 1 min. ahead of hour hand in

5560

min.

Minute hand is 40 minutes ahead of hour hand

in 55

4060 =

11480

= 117

43 min.

Hence, the right time is 117

43 min. past 2.

HANDS ARE PERPENDICULAR

Ex.10 At what time between 4 O�clock and 5 O�clock will

the hands are perpendicular ?Sol. At 4 O�clock the distance between the two hands is

20 min. When they are at 15 minutes distance,they are perpendicular to each other.

Case-I When the time taken (20 � 15) = 5 min.

Minute hand is 55 min. ahead of hour hand in 60 min.Minute hand is 5 min. ahead of hour hand in

55

560 =

1160

= 115

5 min.


5 min. past 4.

Case-II When the time taken (20 + 15) = 35 min.


60 min.


55

3560 =

11420

= 112

38 min.


38 min. past 4.

MIRROR IMAGE OF CLOCK

If the time is between 1 O�clock to 11 O�clock, then

to find the mirror image, time is subtracted from

11 : 60.

If the time is between 11 O�clock to 1 O�clock, then

to find the mirror image, time is subtracted from

23 : 60.

Ex.11 The time in the clock is 4 : 46, what is the mirror image ?

Sol. (11 : 60) � (4 : 46) = 7 : 14.

Ex.12 The time in the clock is 12 : 35, then find its mirror

image.

Sol. (23 : 60) � (12 : 35) = 11 : 25.

TO FIND THE ANGLE BETWEEN TWO HANDS

Angle are of two types :

Positive angle : It is obtained by moving from hour

hand to minute hand moving in clockwise direction.

Negative angle : It is obtained by moving from

minute hand to hour hand.

Both types of angles are 360º in total. If one angle

is known, other can be obtained by subtracting from

360º.

Ex.13 At 4 : 30, what is the angle formed between hour

hand and minute hand ?

Sol. At 4 O� clock angle between hour and min. hand is

of 120º.

In 30 min. minute hand make an angle of 180º.

So, the resultant angle is 180º � 120º = 60º.

But in 30 min. hour hand will also cover an angle of 15º.

Hence, the final angle between both hands is

60º � 15º = 45º.

Short trick

PAGE # 86

Ex.14 A bus for Delhi leaves every thirty minutes from a busstand. An enquiry clerk told a passenger that the bushad already left ten minutes ago and the next bus willleave at 9.35 A.M. At what time did the enquiry clerkgive this information to the passenger ?

Sol. Bus leaves after every 30 minutes.The next bus will leave at 9 : 35 A.M.The last bus left at 9 : 35 � 0 : 30 = 9 : 05 A.M.

but clerk said that bus had left 10 minutes earlier.9 : 05 + 0 : 10 = 9 : 15 A.M.

EXERCISE

1. Find the day of the week on 26 January, 1950.(A) Tuesday (B) Friday(C) Wednesday (D) Thursday

2. Which two months in a year have the same

calendar ?(A) June, October (B) April, November(C) April, July (D) October, December

3. Are the years 900 and 1000 leap years ?

(A) Yes (B) No(C) Can't say (D) None of these

4. If it was Saturday on 17th November, 1962 whatwill be the day on 22nd November, 1964 ?

(A) Monday (B) Tuesday(C) Wednesday (D) Sunday

5. Sangeeta remembers that her father's birthdaywas certainly after eighth but before thirteenth of

December. Her sister Natasha remembers thattheir father's birthday was definitely after ninth butbefore fourteenth of December. On which date of

December was their father's birthday ?(A) 10th (B) 11th(C) 12th (D) Data inadequate

6. Find the day of the week on 15 August, 1947.

(A) Tuesday (B) Friday(C) Wednesday (D) Thursday

7. Karan was born on Saturday 22nd March 1982. Onwhat day of the week was he 14 years 7 months

and 8 days of age ?(A) Sunday (B) Tuesday(C) Wednesday (D) Monday

8. If on 14th day after 5th March be Wednesday, what

day of the week will fall on 10th Dec. of the sameyear ?(A) Friday (B) Wednesday

(C) Thursday (D) Tuesday

9. If the day before yesterday was Saturday, what day

will fall on the day after tomorrow ?

(A) Friday (B) Thursday

(C) Wednesday (D) Tuesday

10. If February 1, 1996 is Wednesday, what day is March

10, 1996 ?

(A) Monday (B) Sunday

(C) Saturday (D) Friday

11. If the seventh day of a month is three days earlier

than Friday, what day will it be on the nineteenth

day of the month ?

(A) Sunday (B) Monday

(C) Wednesday (D) Friday

12. Mohini went to the movies nine days ago. She goes

to the movies only on Thursday. What day of the

week is today ?

(A) Thursday (B) Saturday

(C) Sunday (D) Tuesday

13. At what time are the hands of a clock together

between 5 and 6 ?

(A) 33113

min. past 5 (B) 28113

min. past 5

(C) 27113

min. past 5 (D) 26113

min. past 5

14. At what time between 9 and 10 will the hands of a

clock be in the straight line, but not together ?

(A) 16 minutes past 9

(B) 114

16 minutes past 9

(C) 116

16 minutes past 9

(D) 119

16 minutes past 9

15. At what time between 5 & 5 : 30 will the hands of a

clock be at right angle ?

(A) 1110

10 minutes past 5

(B) 115

11 minutes past 5

(C) 1110

9 minutes past 5

(D) 119

10 minutes past 5

16. Ajay left home for the bus stop 15 minutes earlier

than usual. It takes 10 minutes to reach the stop.

He reached the stop at 8.40 a.m. What time does

he usually leave home for the bus stop ?

(A) 8.30 a.m. (B) 8.45 a.m.

(C) 8.55 a.m. (D) Data inadequate

PAGE # 87

17. The priest told the devotee, "The temple bell isrung at regular intervals of 45 minutes. The lastbell was rung five minutes ago. The next bell isdue to be rung at 7.45 a.m." At what time did thepriest give this information to the devotee ?(A) 7.40 a.m. (B) 7.05 a.m.(C) 6.55 a.m. (D) None of these

18. There are twenty people working in an office. Thefirst group of five works between 8.00 A.M. and 2.00P.M. The second group of ten works between 10.00A.M. and 4.00 P.M. And the third group of five worksbetween 12 noon and 6.00 P.M. There are threecomputers in the office which all the employeesfrequently use. During which of the following hoursthe computers are likely to be used most ?(A) 10.00 A.M. �� 12 noon

(B) 12 noon �� 2.00 P.M.

(C) 1.00 P.M. �� 3.00 P.M.

(D) 2.00 P.M. �� 4.00 P.M.

19. A tired worker slept at 7.45 p.m.. If he rose at 12noon, for how many hours did he sleep ?(A) 5 hours 15 min. (B) 16 hours 15 min.(C) 12 hours (D) 6 hours 45 min.

20. How many times are the hands of a clocksperpendicular in a day ?(A) 42 (B) 48(C) 44 (D) 46

21. If a clock shows 04: 28 then its mirror image willbe ?(A) 07: 42 (B) 07: 32(C) 08: 32 (D) 08: 42

22. A watch, which gains uniformly, is 3 minutes slowat noon on Monday and is 3 minutes 48 secondsfast at 2 p.m. on the following Monday. What time itwas correct ?(A) 2 p.m. On Tuesday(B) 2 p.m. On Wednesday(C) 3 p.m. On Thursday(D) 1 p.m. On Friday.

23. How many times are the hands of a clocks coincidein a day ?(A) 10 (B) 11(C) 12 (D) 22

24. At what time between 2 and 3 O� clock the hands of

a clock will make an angle of 160º ?

(A) 20 minutes past 2 (B) 30 minutes past 2 (C) 40 minutes past 2 (D) 50 minutes past 2

25. Ashish leaves his house at 20 minutes to seven inthe morning, reaches Kunal�s house in 25 minutes,

they finish their breakfast in another 15 minutesand leave for their office which takes another 35minutes. At what time do they leave Kunal�s house

to reach their office ?(A) 7.40 am (B) 7.20 am(C) 7.45 am (D) 8.15 am

26. The train for Lucknow leaves every two and a halfhours from New Delhi Railway Station. Anannouncement was made at the station that thetrain for Lucknow had left 40 minutes ago and thenext train will leave at 18. 00 hrs. At what time wasthe announcement made ?(A) 15.30 hrs (B) 17.10 hrs(C) 16.00 hrs (D) None of these

27. A monkey climbs 30 feet at the beginning of eachhour and rests for a while when he slips back 20feet before he again starts climbing in thebeginning of the next hour. If he begins his ascentat 8.00 a.m., at what time will he first touch a flag at120 feet from the ground ?(A) 4 p.m. (B) 5 p.m.(C) 6 p.m. (D) None of these

28. If the two incorrect watches are set at 12 : 00 noonat correct time, when will both the watches showthe correct time for the first time given that the firstwatch gains 1 min in 1 hour and second watchloses 4 min in 2 hours :(A) 6 pm, 25 days later(B) 12 : 00 noon, 30 days later(C) 12 noon, 15 days later(D) 6 am 45 days later

29. Rajeev and Sanjeev are too close friends Rajeev'swatch gains 1 minute in an hour and Sanjeev'swatch loses 2 minutes in an hour. Once they setboth the watches at 12 : 00 noon, with my correctwatch. When will the two incorrect watches ofRajeev and Sanjeev show the same time together?(A) 8 days later (B) 10 days later(C) 6 days later (D) can't be determined

30. At a railway station a 24 hour watch loses 3 minutesin 4 hours. If it is set correctly on Sunday noonwhen will the watch show the correct time ?(A) 6 pm after 40 days(B) 12 noon after 75 days(C) 12 pm after 100 days(D) 12 noon after 80 days

31. A swiss watch is being shown in a museum whichhas a very peculiar property. It gains as much inthe day as it loses during night between 8 pm to 8am. In a week how many times will the clock showthe correct time ?(A) 6 times (B) 14 times(C) 7 times (D) 8 times

32. A wrist watch which is running 12 minutes late ona Sunday noon is 16 minutes ahead of the correcttime at 12 noon on the next Sunday. When is theclock 8 minutes ahead of time ?(A) Thursday 10 am (B) Friday noon(C) Friday 8 pm (D) Tuesday noon

PAGE # 88

33. A clock loses 2 minutes in a hour and another clockgains 2 minutes in every 2 hours. Both these clocksare set correctly at a certain time on Sunday andboth the clocks stop simultaneously on the nextday with the time shown being 9 am and 10 : 06AM. What is the correct time at which they stopped?(A) 9 : 54 am (B) 9 : 44 pm(C) 9 : 46 am (D) 9 : 44 am

34. David sets his watch at 6 : 10 am on Sunday, whichgains 12 minutes in a day. On Wednesday if thiswatch is showing 2 : 50 pm. What is the correcttime ?(A) 1 : 50 pm (B) 2 : 10 pm(C) 2 : 30 pm (D) 3 : 30 pm

35. Ramu purchased a second hand Swiss watchwhich is very costly. In this watch the minute-hand

and hour hand coincide after every 113

65 minutes.

How much time does the watch lose or gain perday ?(A) 4 min (B) 5 min(C) 4 min, 20 sec (D) none of these

36 My watch was 8 minutes behind at 8 pm on Sundaybut within a week at 8 pm on Wednesday it was 7minutes ahead of time. During this period at whichtime this watch has shown the correct time :(A) Tuesday 10 : 24 am(B) Wednesday 9 : 16 pm(C) It cannot show the correct time during this period(D) None of the above

37. Out of the following four choices which does notshow the coinciding of the hour hand and minute-hand :(A) 3 : 16 : 2 (B) 6 : 32 : 43(C) 9 : 59 : 05 (D) 5 : 27 : 16

38. Kumbhakarna starts sleeping between 1 pm and

2 pm and he wakes up when his watch shows

such a time that the two hands (i.e., hour-hand

and minute-hand) interchange the respective

places. He wakes up between 2 pm and 3 PM on

the same night. How long does he sleep ?

(A) 135

55 min (B) 1310

110 min

(C) 136

54 min (D) None of these

39. A clock loses 3% time during the first week and

then gains 2% time during the next one week. If the

clock was set right at 12 noon on a Sunday, what

will be the time that the clock will show exactly 14

days from the time it was set right ?

(A) 1 : 36 : 48 (B) 1 : 40 : 48

(C) 1 : 41 : 24 (D) 10 : 19 : 12

Direction : (40 to 41) A 12 dial clock has its minute hand

defective. Whenever it touches dial 12, it

immediately falls down to 6 instead of running

smoothly (the hour hand remains unaffected during

that fall). It was set right at 12 �O� clock in the noon.

40. What was the actual time when the minute hand of

the clock touched dial 9 for the 5th time?

(A) 2 : 15 (B) 3 : 00

(C) 5 : 15 (D) 6 : 45

41. If the actual time is 10 : 10, what is the position of

the hour hand in that defective clock ?

(A) Between 2 and 3 (B) Between 4 and 5

(C) Between 10 and 11 (D) Between 3 and 4

PAGE # 89

CUBE AND DICE-TEST

CUBES

A cube is three dimensional figure, having 8corners, 6 surfaces and 12 edges. If a cube ispainted on all of its surfaces with any colour andfurther divided into various smaller cubes, we getfollowing results. Smaller cubes with threesurfaces painted will be present on the corners ofthe big cube.

3 33

3

33

3

3 3

3

3

3

1 1

1 1

11

11

11 1

1

2 2

2 2

2

22

2

22

2

2

2

2

22

22

222

22 2

Smaller cubes with two surface painted will bepresent on the edges of the big cube. Smallercubes with one surface painted will be present onthe surfaces of the big cube. Smaller cubes withno surface painted will be present inside the bigcube.

If a cube is painted on all of its surfaces with acolour and then divided into smaller cubes of equalsize then after separation, number of smaller cubesso obtained will be calculated as under :Number of smaller cubes with three surfacespainted = 8Number of smaller cubes with two surfacespainted = (n � 2) × 12

Number of smaller cubes with one surfacespainted = (n � 2)2 × 6

Number of smaller cubes with no surfaces painted= (n � 2)3

Where n = No of divisions on the surfaces of thebigger cube

= cubesmalleroneofedgeoflengthcubebigofedgeoflength

TYPE I

If a cube is painted on all of its surfaces with singlecolour and then divided into various smaller cubesof equal size.

Directions : ( 1 to 4) A cube of side 4 cm is painted black onall of its surfaces and then divided into varioussmaller cubes of side 1 cm each. The smallercubes so obtained are separated.

Total cubes of obtained = 64111

444

Here n = 41

4

cubesmallofside

cubebigofside

Ex 1. How many smaller cubes have three surfaces

painted ?

(A) 4 (B) 8

(C) 16 (D) 24

Sol. (B) Number of smaller cubes with three surfaces

painted = 8

Ex 2. How many smaller cubes have two surfaces

painted ?

(A) 4 (B) 8

(C) 16 (D) 24

Sol. (D) Number of smaller cubes with two surfaces

painted = (n � 2) × 12 = (4 � 2) × 12 = 24

Ex 3. How many smaller cubes have only one surface

painted ?

(A) 8 (B) 16

(C) 24 (D) 32

Sol. (C) Number of smaller cubes with one surface

painted = (n � 2)2 × 6 = (4 � 2)2 × 6 = 4 × 6 = 24

Ex 4. How many smaller cubes will have no side painted ?

(A) 18 (B) 16

(C) 22 (D) 8

Sol. (D) Number of smaller cubes with no surface

painted = (n � 2)3 = (4 � 2)3 = (2)3 = 8

TYPE II

If a cube is painted on all of its surfaces with

different colours and then divided into various

smaller cubes of equal size.

Directions : ( 5 to 7 ) A cube of side 4 cm is painted black on

the pair of one opposite surfaces, blue on the pair

of another opposite surfaces and red on remaining

pair of opposite surfaces. The cube is now divided

into smaller cubes of equal side of 1 cm each.


PAGE # 90

Ex 5. How many smaller cubes have three surfacespainted ?(A) 4 (B) 8(C) 16 (D) 24

Sol. (B) Number of smaller cubes with three surfacespainted = 8(These smaller cubes will have all three surfacespainted with different colour blue, black and red.)

Ex 6. How many smaller cubes have two surfacespainted ?(A) 4 (B) 8(C) 16 (D) 24

Sol. (D) Number of smaller cubes with two surfacespainted = 24. And out of this -(a) Number of cubes with two surfaces paintedwith black and blue colour = 8.(b) Number of cubes with two surfaces paintedwith blue and red colour = 8.(c) Number of cubes with two surfaces paintedwith black and red color = 8.

Ex 7. How many smaller cubes have only one surfacepainted ?(A) 8 (B) 16(C) 24 (D) 32

Sol. (C) Number of smaller cubes with one surfacepainted = 24. And out of this -(a) Number of cubes with one surface paintedwith black colour =8.(b) Number of cubes with one surface paintedwith blue colour = 8.(c) Number of cubes with one surface paintedwith red colour = 8.

TYPE III

If a cube is painted on its surfaces in such a waythat one pair of opposite surfaces is left unpainted.

Directions : ( 8 to 11 ) A cube of side 4 cm is painted red onthe pair of one opposite surfaces, green on thepair of another opposite surfaces and one pair ofopposite surfaces is left unpainted. Now the cubeis divided into 64 smaller cubes of side 1 cm each.


Sol. (A) Number of smaller cubes with three surfacespainted = 0 (Because each smaller cube at thecorner is attached to a surface which is unpainted.)


Sol. (C) Number of smaller cubes with two surfacespainted = Number of cubes present at the corners+ Numbers of cubes present at 4 edges= 8 + (n � 2) × 4 = 8 + 8 = 16


Sol. (D) Number of smaller cubes with one surfacepainted = Number of cubes present at the 8 edges+ number of cubes present at the four surfaces=(n � 2) × 8 + (n � 2)2 × 4

= 2 × 8 + 4 × 4 = 16 + 16 = 32

Ex 11. How many smaller cubes will have no side painted?(A) 18 (B) 16(C) 22 (D) 8

Sol. (B) Number of smaller cubes with no side painted= Number of cubes on the two unpainted surfaces +number of cubes present inside the cube.= (n � 2)2 × 2 + (n � 2)3 = 4 × 2 + (2)3 = 8 + 8 = 16.

TYPE IV

If a cube is painted on its surfaces in such a waythat one pair of adjacent surfaces is left unpainted.

Directions : (12 to 15 )A cube of side 4 cm is painted red onthe pair of one adjacent surfaces, green on thepair of other adjacent surfaces and two adjacentsurfaces are left unpainted. Now the cube is dividedinto 64 smaller cubes of side 1 cm each.


PAGE # 91

Sol. (A) Number of smaller cubes with three surfacespainted = Number of smaller cubes at two corners= 2


Sol. (D) Number of smaller cubes with two surfacespainted = Number of smaller cubes at four corners+ Number of smaller cubes at 5 edges.= 4 + (n � 2) × 5 = 4 + 2 × 5

= 4 + 10 = 14


Sol. (D) Number of smaller cubes with one surfacepainted = Number of smaller cubes at foursurfaces + Number of smaller cubes at 6 edges +Number of smaller cubes at two corners.= (n � 2)2 × 4 + (n � 2) × 6 + 2

= 4 × 4 + 2 × 6 + 2 = 16 + 12 = 28 + 2 = 30

Ex 15. How many smaller cubes will have no side painted?(A) 18 (B) 16(C) 22 (D) 8

Sol. (A) Number of smaller cubes with no surfacespainted = Number of smaller cubes from insidethe big cube + Number of cubes at two surfaces +Number of cubes at one edge.= (n � 2)3 + (n � 2)2 × 2 + (n � 2)

= (2)3 + (2)2 × + 2

= 8 + 8 + 2 = 18

DICES

Type-I

General Dice : In a general dice the sum of numberson the any two adjacent faces is �7�.Standard Dice : In a standard dice the sum ofnumbers on the opposite faces is '7'.

Ex 16. Which number is opposite 4 in a standard dicegiven below ?

41

5

(A) 1 (B) 3(C) 5 (D) Can�t be determined

Sol. Clearly , from the standard dice the sum ofnumbers on the opposite faces is '7', so numberopposite to 4 is 3.

Type-II

Ex 17. The figures given below show the two differentpositions of a dice. Which number will appearopposite to number 2 ?.

(A) 3 (B) 4(C) 5 (D) 6

Sol. (C) The above question,where only two positions ofa dice are given, can easilybe solved with thefollowing method.

Step I. The dice, when unfolded, will appear as shown inthe figure given on the right side.

Step II. Write the common number to both the dice in themiddle block. Since common number is 4, hencenumber 4 will appear in the central block.

Step III. Consider the figure (i) and write the first number inthe anti-clockwise direction of number 4,(common number) in block I and second numberin block II. Therefore, numbers 3 and 2 being thefirst and second number to 4 in anticlockwisedirections respectively, will appear in block I & IIrespectively.

Step IV. Consider figure (ii) and wire first and secondnumber in the anticlock-wise direction to number4, (common number) in block (III) & (IV). Hencenumbers 6 and 5 will appear in the blocks III and IVrespectively.

Step V. Write remaining number in the remaining block.Therefore, number 1 will come in the remainingblock. Now, from the unfolded figures we find thatnumber opposite to 6 is 3, number opposite to 2 is5 and number opposite to 4 is 1. Therefore, option(C) is our answer.( Short Trick : From the given dice, we will take thecommon number as the base and then in itsrespect move clockwise direction and write asfollows : 4 � 2 � 3

4 � 5 � 6.Here,we find that number opposite to 6 is 3, numberopposite to 2 is 5 and number opposite to 4 isremaining number 1.Therefore, option (C) is our answer. )

Ex 18. On the basis of two figures of dice, you have to tellwhat number will be on the opposite face of number5 ?

(A) 1 (B) 2(C) 4 (D) 6

PAGE # 92

Sol. (D) The above question where only two positionsof a dice are given, can easily be solved with thefollowing method :If in the given dice, there are two numbers common,then uncommon numbers will always be oppositeof each other.Therefore, option (D) is our answer.

Type-III

Ex 19. From the following figures of dice, find whichnumber will come in place of �?�

(A) 4 (B) 5(C) 2 (D) 3

Sol. (D) If the above dice is unfolded, it will look like asthe figure (i) given below.

Figure (i)

Now the number in place of �?� can be obtained by

making a slight change in the figure as given here.Now comparing figure (ii) with third dice as above,we get that number in place of ? is 3.

Figure (ii)

Type-IV

Ex 20. A dice has been thrown four times and producesfollowing results.

Which number will appear opposite to the number3 ?(A) 4 (B) 5(C) 6 (D) 1

Sol. (A) From the figures (i), (ii) and (iv) we find thatnumbers 6, 1, 5 and 2 appear on the adjacentsurfaces to the number 3. Therefore, number 4will be opposite to number 3.

Type-V

Ex 21. Which of the following dices is identical to theunfolded figure as shown here ?

(X)

(A) (B)

(C) (D)

Sol. (A) From the unfolded figure of dice, we find thatnumber opposite to 2 is 4, for 5 it is 3 and for 1 it is6. From this result we can definitely say that figure(B), (C) and (D) can not be the answer figure asnumbers lying on the opposite pair of surfaces arepresent on the adjacent surfaces.

EXERCISE

Directions : (1 to 5) A cube is coloured orange on one face,pink on the opposite face, brown on one face andsilver on a face adjacent to the brown face. Theother two faces are left uncoloured. It is then cutinto 125 smaller cubes of equal size. Now answerthe following questions based on the abovestatements.

1. How many cubes have at least one face colouredpink ?(A) 1 (B) 9(C) 16 (D) 25

2. How many cubes have all the faces uncoloured ?(A) 24 (B) 36(C) 48 (D) 64

3. How many cubes have at least two faces coloured ?(A) 19 (B) 20(C) 21 (D) 23

4. How many cubes are coloured orange on one faceand have the remaining faces uncoloured ?(A) 8 (B) 12(C) 14 (D) 16

5. How many cubes one coloured silver on one face,orange or pink on another face and have fouruncoloured faces ?(A) 8 (B) 10(C) 12 (D) 16

PAGE # 93

Directions : (6 to 11) A cube is painted red on two adjacentsurfaces and black on the surfaces opposite tored surfaces and green on the remaining faces.Now the cube is cut into sixty four smaller cubes ofequal size.

6. How many smaller cubes have only one surfacepainted ?(A) 8 (B) 16(C) 24 (D) 32

7. How many smaller cubes will have no surfacepainted ?(A) 0 (B) 4(C) 8 (D) 16

8. How many smaller cubes have less than threesurfaces painted ?(A) 8 (B) 24(C) 28 (D) 48

9. How many smaller cubes have three surfacespainted ?(A) 4 (B) 8(C) 16 (D) 24

10. How many smaller cubes with two surfacespainted have one face green and one of theadjacent faces black or red ?(A) 8 (B) 16(C) 24 (D) 28

11. How many smaller cubes have at least one surfacepainted with green colour ?(A) 8 (B) 24(C) 32 (D) 56

Directions : (12 to 16) A cube of 4 cm has been painted onits surfaces in such a way that two oppositesurfaces have been painted blue and two adjacentsurfaces have been painted red. Two remainingsurfaces have been left unpainted. Now the cubeis cut into smaller cubes of side 1 cm each.

12. How many cubes will have no side painted ?(A) 18 (B) 16(C) 22 (D) 8

13. How many cubes will have at least red colour onits surfaces ?(A) 20 (B) 22(C) 28 (D) 32

14. How many cubes will have at least blue colour onits surfaces ?(A) 20 (B) 8(C) 24 (D) 32

15. How many cubes will have only two surfacespainted with red and blue colour respectively ?(A) 8 (B) 12(C) 24 (D) 30

16. How many cubes will have three surfaces coloured ?(A) 3 (B) 4(C) 2 (D) 16

Directions : (17 to 21) The outer border of width 1 cm of a

cube with side 5 cm is painted yellow on each side

and the remaining space enclosed by this 1 cm

path is painted pink. This cube is now cut into 125

smaller cubes of each side 1 cm. The smaller

cubes so obtained are now seperated.

17. How many smaller cubes have all the surfaces

uncoloured ?

(A) 0 (B) 9

(C) 18 (D) 27

18. How many smaller cubes have three surfaces

coloured ?

(A) 2 (B) 4

(C) 8 (D) 10

19. How many cubes have at least two surfaces

coloured yellow ?

(A) 24 (B) 44

(C) 48 (D) 96

20. How many cubes have one face coloured pink and

an adjacent face yellow ?

(A) 0 (B) 1

(C) 2 (D) 4

21. How many cubes have at least one face coloured ?

(A) 27 (B) 98

(C) 48 (D) 121

Directions : (22 to 31) A solid cube has been painted yellow,blue and black on pairs of opposite faces. Thecube is then cut into 36 smaller cubes such that32 cubes are of the same size while 4 others areof bigger sizes. Also no faces of any of the biggercubes is painted blue.

22. How many cubes have at least one face paintedblue ?(A) 0 (B) 8(C) 16 (D) 32

23. How many cubes have only one faces painted ?(A) 24 (B) 20(C) 8 (D) 12

24. How many cubes have only two faces painted ?(A) 24 (B) 20(C) 16 (D) 8

25. How many cubes have atleast two faces painted ?(A) 36 (B) 34(C) 28 (D) 24

26. How many cubes have only three faces painted ?(A) 8 (B) 4(C) 2 (D) 0

PAGE # 94

27. How many cubes do not have any of their facespainted yellow ?(A) 0 (B) 4(C) 8 (D) 16

28. How many cubes have at least one of their facespainted black ?(A) 0 (B) 8(C) 16 (D) 20

29. How many cubes have at least one of their facespainted yellow or blue ?(A) 36 (B) 32(C) 16 (D) 0

30. How many cubes have no face painted ?(A) 8 (B) 4(C) 1 (D) 0

31. How many cubes have two faces painted yellowand black respectively ?(A) 0 (B) 8(C) 12 (D) 16

Directions : (32 to 35) Some equalcubes are arranged in theform of a solid block asshown in the adjacentfigure. All the visiblesufaces of the block (exceptthe bottom) are thenpainted.

32. How many cubes do not have any of the facespainted ?(A) 27 (B) 8(C) 10 (D) 12

33. How many cubes have one face painted ?(A) 9 (B) 24(C) 22 (D) 20

34. How many cubes have only two faces painted ?(A) 0 (B) 16(C) 20 (D) 24

35. How many cubes have only three faces painted ?(A) 4 (B) 12(C) 6 (D) 20

Directions : (36 to 40) A cuboid of dimensions(6 cm 4 cm 1 cm) is painted black on both thesurfaces of dimensions (4 cm 1 cm), green on thesurfaces of dimensions (6 cm 4 cm). and red onthe surfaces of dimensions (6 cm 1 cm). Now theblock is divided into various smaller cubes of side1 cm. each. The smaller cubes so obtained areseparated.

36. How many cubes will have all three colours black,green and red each at least on one side?(A) 16 (B) 12(C) 10 (D) 8

37. How many cubes will be formed?(A) 6 (B) 12(C) 16 (D) 24

38. If cubes having only black as well as green colourare removed then how many cubes will be left?(A) 4 (B) 8(C) 16 (D) 30

39. How many cubes will have 4 coloured sides and2 sides without colour?(A) 8 (B) 4(C) 16 (D) 10

40. How many cubes will have two sides with greencolour and remaining sides without any colour?(A) 12 (B) 10(C) 8 (D) 4

41. Which alphabet is opposite D ?

(A) E (B) C(C) F (D) A

42. What should be the number opposite 4 ?

(i) (ii) (iii)

(A) 5 (B) 1(C) 3 (D) 2

43.

(i) (ii)

(iii) (iv)Which letter will be opposite to letter D ?(A) A (B) B(C) E (D) F

Directions : (44 to 45) The figure (X) given below is theunfolded position of a cubical dice. In each of thefollowing questions this unfolded figure is followedby four different figures of dice. You have to selectthe figure which is identical to the figure (X).

PAGE # 95

44. (X)

(A) (B)

(B) (D)

45. (X)

(A) (B)

(C) (D)

Directions : (46 to 48) In each of the following questions,select the correct option for the question asked.

(i) (ii)

46. Which number will come opposite to number 2?(A) 5 (B) 1(C) 6 (D) 3



49. On the basis of two figures of dice, you have to tell whatnumber will be on the opposite face of number 5?

(i) (ii)

(A) 1 (B) 2(C) 4 (D) 6

50. Which symbol will appear on the opposite surfaceto the symbol x?

(A) (B) =

(C) (D) O

51. Three positions of the same dice are given below.Observe the figures carefully and tell which numberwill come in place of �?�

(i)

16 3

(ii)

35 4

(iii)

42 ?

(A) 1 (B) 6(C) 3 (D) 5

52. On the basis of the following figures you have totell which number will come in place of �?�

(i)

36 1

(ii)

42 6

(iii)

?1 5

(A) 2 (B) 3(C) 6 (D) 4

Directions : (53 to 55) Choose from the alternatives, theboxes that will be formed when figure (X) is folded:

53. (X)

(A) (B)

(C) (D)

54. (X)

+

(A) (B) +

(C) + (D)

PAGE # 96

55. (X)

(A) (B)

(C) (D)

Direction : (56) The six faces of a cube have been markedwith numbers 1, 2, 3, 4, 5 and 6 respectively. Thiscube is rolled down three times. The threepositions are given. Choose the figure that will beformed when the cube is unfolded.

56.

(A) (B)

(C) (D)

57. Which number is opposite 3 in a standard dicegiven below ?

(A) 1 (B) 4(C) 5 (D) Can�t be determined

58. Which number is opposite 4 ?

(A) 5 (B) 3(C) 2 (D) 1

Directions : (59) In the following question four positions ofthe same dice have been shown. You have to seethese figures and select the number opposite tothe number as asked in each question.

59.

(i) (ii)

(iii) (iv)Which number is opposite to number 5?(A) 6 (B) 5(C) 1 (D) 3

Directions : (60 to 64) Choose the cube from the optionsthat will unfold to give the figure on the left

60.X

M

M

M MX

X

(A) (B) (C) (D) (E)

61.

4 1 8

3

7

9

(A) (B) (C) (D) (E)

981 14

7

7 87 48

7

62.

D

8

(A) (B) (C) (D) (E)

8 8 D

63.

B

(A) (B) (C) (D) (E)

B

PAGE # 97

64.

J

(A) (B) (C) (D) (E)

J J

Directions : (65 to 68) In each of the following questions, adiagram has been given which can be folded intoa cube. The entries given in the squares indicatethe entries on the face of the cube. In each questiona number or a letter has been given . Of the fouralternatives given below it, you have to find the onethat would appear on the face opposite to it in thecube.

65. Which letter is opposite Q ?

Q

O P LNM

(A) L (B) M(C) N (D) P

66. Which number/letter is opposite 2 ?

3 I CAB2

(A) A (B) C(C) 1 (D) 3

67. Which number/letter is opposite O?

N M 2

L

I O

(A) L (B) M(C) N (D) 2

68. Which letter is opposite R?

Q R

S P

U T

(A) P (B) S(C) T (D) U

98PAGE # 98

ANSWER KEY

FORCE AND NEWTON�S LAW OF MOTION(PHYSICS)

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans. B C C B A B D A ACD B C C C C CQue. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Ans. B D A A A C C D B B B B A D CQue. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

Ans. B C B A A C B D A C A B D A AQue. 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

Ans. D D CD D A D D C B A,C C A,B,C B B BQue. 61 62 63 64 65 66 67 68 69Ans. B A C C D C B B C

CARBON(CHEMISTRY)

Ques. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Ans. C C B B B A D A C B C D C B B

Ques. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30Ans. B A C B C B B A C C D C A B C

Ques. 31 32 33 34 35 36 37 38 39 40 41Ans. C B C A B D C B C A A

NUMBER SYSTEM(MATHEMATICS)

Q. 1 2 3 4 5 6 7 8 9 10

Ans. B A B A D A D C B A

Q. 11 12 13 14 15 16 17 18 19 20

Ans. A A D A B A D B C C

Q. 21 22 23 24 25 26 27 28 29 30

Ans. B C A C C A B D B B

Q. 31 32 33 34 35 36 37 38 39 40

Ans. C B A A B D C C D C

Q. 41 42 43 44 45 46 47 48 49 50

Ans. C A D D A A C C C D

Q. 51 52 53 54 55 56 57 58 59 60

Ans. C D C C C A C B D D

Q. 61 62 63 64 65 66 67 68 69 70

Ans. A B B A A A B B B C

Q. 71 72 73 74 75 76 77 78 79 80

Ans. D A B C A A,D B D B A

Q. 81 82 83 84 85 86 87 88 89 90

Ans. D A B B B B B D D B

Q. 91

Ans. C


99PAGE # 99

TRIGONOMETRY(MATHEMATICS)

Q. 1 2 3 4 5 6 7 8 9 10

Ans. B B C B A A C C C C

Q. 11 12 13 14 15 16 17 18 19 20

Ans. D A B B D B D A D B

Q. 21 22 23 24 25 26 27 28 29 30

Ans. C B D B C C D C A A

Q. 31 32 33 34 35 36 37 38 39 40

Ans. B D A D B A B B D D

Q. 41 42

Ans. A B

PROTOPLASM (BIOLOGY)

Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15A. C D C A B A A C A D B A A D DQ. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30A. C B D B B C D A D A C A A A BQ. 31 32 33 34A. A B D A

SERIES COMPLETION(MENTAL ABILITY)

EXERCISE-1 (Number Series)

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans. C D D A C D B C C C D C C B DQue. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Ans. B C C C B A B D D A C B B C AQue. 31 32 33 34 35

Ans. C C C D D

EXERCISE- 2 (Alphabet Series)

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans. D A D C C A D C D B D C C C DQue. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Ans. C A B C C C A B A C D B C D B

EXERCISE- 3 (Letter Repeating Series)

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans. D D A A C B A C D D C B D C BQue. 16 17 18 19 20 21 22 23 24 25 26

Ans. C A A A C D D D A B D

100PAGE # 100

EXERCISE- 4 (Missing Term In Figure)

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans. B D B D C C C D A D D B C A BQue. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Ans. B C A B B A C A D D A C D B A

Que. 31 32 33 34 35 36 37 38 39 40

Ans. B B A C A C B C D B

PUZZLE-TEST(MENTAL ABILITY)

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Ans. A B D C A D C C C C C D C C B

Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30Ans. C C D D C D D B B C A A D C DQue. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45Ans. D C B A B A D A D C D A C B DQue. 46 47 48 49 50 51 52 53 54 55 56 57 58Ans. C A D B D D D A C A A D B

CALENDAR AND CLOCK-TEST(MENTAL ABILITY)

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans. D C B D D B C B C C A B C B A

Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30Ans. B B B B C B C D C B D C B B DQue. 31 32 33 34 35 36 37 38 39 40 41Ans. D B D B A A C A D A C

CUBE AND DICE TEST(MENTAL ABILITY)

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans. D C C D A C C D B B C A C D BQue. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Ans. C D C B A B D D A D C A D C BQue. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45Ans. C D C D C A D C B C B B A D BQue. 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60Ans. D A B C D A B D B D C B A C CQue. 61 62 63 64 65 66 67 68Ans. A D E D C A B B

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