KHOJ-2019 ANSWER KEY WITH SOLUTION CLASS - 10...UG – 1 & 2, Concorde Complex, Above OBC Bank. R.C....

UG – 1 & 2, Concorde Complex, Above OBC Bank. R.C. Dutt Road., Alkapuri Baroda. Ph. (0265) 6625979

KHOJ-2019

ANSWER KEY WITH SOLUTION CLASS - 10

PART - I PART - II PART - III

Q. No. Answer Q. No. Answer Q. No. Answers Q.

No. Answers

1 C 1 C 1 A 31 A 2 C 2 B 2 B 32 B 3 C 3 D 3 A 33 A 4 A 4 A 4 D 34 A 5 A 5 D 5 A 6 A 6 D 6 C 7 A 7 D 7 C 8 D 8 A 8 A 9 B 9 C 9 C PART - IV

10 B 10 D 10 C 1 C 11 B 11 D 11 C 2 C 12 A 12 C 12 D 3 D 13 D 13 B 13 D 4 A 14 D 14 C 14 C 5 B 15 B 15 C 15 A 6 D 16 B 16 D 16 B 7 D 17 A 17 C 17 C 8 B 18 B 18 C 18 B 9 A 19 A 19 C 19 A 10 C 20 B 20 D 20 D 11 B

21 C 21 B 12 B 22 B 22 A 13 C 23 A 23 D 14 B 24 A 24 C 15 C 25 D 25 D 16 A 26 D 26 D 27 D 27 C 28 A 28 B 29 B 29 C 30 A 30 D

DATE : 06/10/2019 KHOJ SOLUTIONS CLASS : 10

Page # 1UG–1 & 2, Concorde Complex, Above OBC Bank. R.C. Dutt Road., Alkapuri Baroda.IIT ASHRAM

PART - I1. (c)

A, B, C, D, E are five iron piece

A = 2B 1 9B 4 C C2 2

DC2

ED2

C < E < A So, A > B, B > C, D > C, E > Dfrom here we can conclude that C is lightest

2. (c)Since there are total seven pieces so P2 has 6 parts which are equal.Since small part is of 20 gms.

So, weight of 2P 20 6 120gm

P2

P1

1P 120gm

Original Cake = 120 + 120 = 240 gm

3. (c)Suppose distance b/w XY = d km

speed of T = x km/h

speed of S = 34

x km/h

X x

Tx km/h

S

K Z

d

Y

3d

52

d5

3x km/h4

Suppose at 4:00 pm T reach at Kdistance b/w XK = x km.

at time t both train reaches at Z.

dis. travelled by T = 35

d - x

dis. travelled by S = 25

d

time for both is same

So,

3 2/5dd x5 3x

x 4

3d 5x 8d5x 15 x

45d - 75x = 40d 5d = 75x

d 15hrsx

4. (a)Suppose present age of Barun = x yrs.

Arun’s = 0.4x yrsafter ‘t’ yrs -

Barun’s age = x + t yrs Arun’s = 0.4x + t yrs Arun’s = 12

Barun’s



0.4x + t = x t

2

0.8x + 2t = x + t t = .2x

So % increment in Barun’s age .2x 100x

20%

5. (a)

Day 1 work 1

120

Day 2 work 1 1

120 120

Day 3 work 1 1 1

120 120 120

Suppose no. of days nSo Day 1 work + Day 2 work + --------- + Day

n work = 1

1 2120 120

+ ---------- n 1

120

1120

1 2 ________ n 1 n(n 1) 120

2

n (n+1) = 240 n (n+1) = 15 16So, n = 15 days

6. (a)Speed of A = x1

B = x2

C = x3

time by A to travel 10 km 1 2

10 9x x

Similarly time taken by B 2 3

10 9x x

1 2

10 9 Kx x

1 210 9x , xK K

2 3

10 9 mx x

2 310 9x , xm m

So,9 10K m

9m K 9K10 m 10

Suppose A beat C by x km.

So,1 3

10 10 xx x

3

1

x 10 xx 10

910 xm

10 10K

9K 10 x 99 10 x

10m 10 10

81 10 x x 10 8.1 1.9km10



7. (a)71 = 772 = 4973 = 34374 = 2401

So, when 74 divided by 100 remainder is 1.

4700 1757 7 So,

7007100

Remainder is 1.

8. (d)no. of odd days upto April 1, 2002 =Dec. 31, 2001 = 1 odd day

Jan. = 3Feb. = 0March = 3April = 1 .

1 odd daySo, April 1st is Monday

2nd = TuesdaySo dates or, Tuesday = 2, 9, 16, 23, 30

9. (b)no. of odd days from Feb. 29, 2016 to

Feb. 29, 2020 = 5So, no. of odd days in 4 years = 5

To celebrate her birthday on Monday no. of odd days must be zero. By observing pattern,So, after 28 years her birthday would fall on (i.e. on 29 Feb. 2044) Monday.

Since she will live till 2099.So, no. of birthday on Monday.

2016 + 28 = 2044 2044 + 28 = 2072So in 2044 & 2072 her birthday will come on Monday, twice.

10. (b)Watch gain 5 sec. in 3 min.

So in 1 min. 5 sec.3

in 60 min. 5 603

100 sec.

from 7 : 00 am to 4:00 pmTotal hours = 9 hrs.

In 9 hrs watch will gain = 9 100 = 900 sec. = 15 min.

So, till 4:00 pm it will gain 15 min.hence it will show 4:15 pm when actual time is 4:00 pm.



11. (b)From the figure it is clear that the opposite faces are-

So when dice is closed either option (b) or (c) is correct (because opposite surface cannot cometogether).But when we move in anti clock wise direction (as shown)

should come in anticlock wise direction of

So only option (b) is correct.12. (a)

no. of cubes having no surface colored = 3(n 2) no. of cubes which are on surface of blank surface= 8 + 10 = 18 cubes

13. (d)no. of cubes have atleast red color = 16 + 16 = 32

14. (d)no. of cubes have atleast blue color = 16 + 16 = 32

15. (b)

Sohanji

Rohit Reeta

InduNeelam

So, Sohanji is Indu’s Grandfather.



16. (b)

2 3 2 5 6 22 4,2 8,2 16,2 32,2 64,3 9, 3 2 2 23 27,5 25,6 36,7 49

17. (a)The effective growth rate (4% –2%) is 2%So population after two years

20000 (1+0.02)2 = 20808

18. (b)Look at cases, by observing pattern

32 + 2 = 11, 332 + 22 = 1111,3332 + 222 = 111111

So sum of digit = 10

19. (a)Water Milk

Vessel I 1/3 2/3Vessel II 2/7 5/7The ratio of water to milk in the mixture-

1 1 2 4 2 1 5 4:3 5 7 5 3 5 7 5

= 31 : 74

20. (b)

Work for 16 hrs 16 224 3

part

Remaining work 2 113 3

part

13 part will be completed in

1 18 6 hrs.3

So total time = 16 + 6 = 22 hrs.



PART - II

1.(c) 3 3 3x 6 6 6 ... so 3x 6 x . Cube both sides to get 3x x 6 so 3x x 6 0 and

2x 2 x 2x 3 0. So x 2 . The other roots are complex.

2.(b) Let a 1991, so we have

21 11 11 1 1

aa aa a

a a a

1 1 a a

3.(d) By repeated application of the Pythagorean theorem (see fig.),we have:

2 2 2 41 BD BC CD

2 2 2 5 BF BD DF

2 2 2 54 BE BF EF2 2 2 45 AE BE AB .

So 45 3 5 AE .

A3 4

67

5B

C

DE FF

4.(a) Let s be the side length of the square (see fig.). We have 090 m BAM m BMA and m BMA0180 m AMN m NMC , so m BAM m NMC and BAM CMN , with ratio of similitude

AB:MC = 2:1. Since , ,2 4

s sBM CN so

3 ,4

sND CD CN and : 1 : 3CN ND

A s

s

s

s3s

2

2

44

B

CD

M

N



5.(d) Let P, Q be the midpoints of the longer sides of the rectangle. Let R, S be the points where thesemicircles meet. Let T be the point where PQ meets RS.

5 cm

P

Q

R S

4 cm

T

Each of the semicircles has radius 5 cm. Therefore PR, PS, QR and QS all have length 5 cm. ThereforePSQR is a rhombus. Hence the diagonals PQ and RS bisect each other at right angles. It follows thatPT and QT each have length 4cm. Let the common length of RT and ST be x cm.

We now apply Pythagoras’ Theorem to the right-angled triangle PTR. This gives 2 2 24 5 , x and

hence 2 2 25 4 25 16 9 x . Therefore x = 3.

It follows that both RT and ST have length 3 cm. Hence the length of RS is 6 cm. Therefore the widthof the overlap of the two semicircles is 6 cm.

6.(d) To avoid a lot of complicated arithmetic, we exploit the facts that 201.72.017100

and 101610.16100

.

Then we take out the common factor 201.7100 . This gives

201.7 10162.017 2016 10.16 201.7 2016 201.7100 100

201.7 (2016 1016)100

201.7 1000100

201.7 10 = 2017.

7. (d)Sol. Let I, J, K and L be the vertices of the sqaure. Let M be the midpoint of JK, let N be the point where the

diagonal IK meets LM. Let the line through N parallel to LK meet IL at R and JK at S. Let T be the footof the perpendicular from N to LK. Let the square have side length s.

I J

KL

MNR S

TIn the triangles INL and KNM, the opposite angles INL and KNM are equal. Also, as IL is parallelto JK, the alternate angles LIN and MKN are equal. Therefore the triangles INL and KNM are



similar. Hence 2. IN ILNK MK Similarly, the triangles INR and KNS are similar. Therefore 2.

NR INNS NK

So 23

NR s and 13

NS s .

Now, NTKS is square, because its angles are all right-angles, and 045 NKT . Therefore

13

NT NS s .

It follows that the area of the triangles LNK, INL and MNK are 2 21 1 1 1 2 1( ) , ( )2 3 6 2 3 3

s s s s s s and

21 1 1 12 2 3 12

s s s , respectively.

The area of the region P, is that of the triangle LNK, that is, 216

s . The area of the region Q is obtained

by subtracting the areas of the triangles LNK, INL and MNK from the area of the square. So region Q

has area 2 2 2 2 21 1 1 56 3 12 12

s s s s s . So the ratio of these areas is 2 21 5 1 5: : 2 : 56 12 6 12

s s .

8.(a) f(x) x = (x 1) (x 2) (x 3) (x 4) (x 5) f (0) = d = ( 1) ( 2) ( 3) ( 4) ( 5).

9.(c) The probaility of one getting the right answer and other two do not can be written as-1 1 1 1 1 1 1 1 1 111 1 1 1 1 14 3 2 3 4 2 2 4 3 24

10.(d) x = 3y + 32y = x 9 x = 2y + 9From equation (1) and (2), we get3y + 3 = 2y + 9 y = 6 and x = 21

11.(d) Let the speed of the boat be x km/hr, stream be y km/hr (x + y) 2 = 20 x + y = 10 ... (1)

5x y 202

x y 8 ... (2)From (1) and (2), we get y = 1

12.(c) Clearly, L.C.M. = (L.C.M. of p and p3) (L.C.M. of q2 and q) = p3q2

13.(b) 1 1 2 1 2 3 1 2 3 40, 2 3 , 0, 2 3 1 2 3 41 1 1 1

1 1 2 1 2 3 1 2 3 41 1 0 2 3 0 2 3 5 .



14. (c) 2f(x) x 4x 3

4 3

2 2

Soz 1 1 2 2 2

2 22 4 2 3Soz 2 2

3

Soz = 16/3

Poz = 161 1 1 1 3

15. (c) For infinite many solutions, 1 1 1

2 2 2

a b ca b c

2 1 3 53 ( 1) 2

mn

17 11,4 5

m n .

16. (d) A x

2x

B

EFG

CDIn BDC, GE DC

then GE BECD BC

i.e.GE 42x 7

8xGE7

Similarly, In 3xADB, GF7

3x 8x 11xEF GF GE7 7 7

7EF K AB

11x7 k x7

k 11



17. (c) If 2 21 5 ta n 4 se c 2 3

2 215 tan 4 1 tan 23

2tan 1

tan 1

045

20 0 2 0sec45 cosec45 sin 45

152

18.(c) centroid, G is , 0, 03 3

a b c a b c

i.e. 0 03

a b c a b c

then 3 3 3 2 2 23 ( ) ( ). a b c abc a b c a b c ab ab ca3 3 3 3 0 a b c abc3 3 3 3 a b c abc

3 3 33

a b cabc abc abc

2 2 23

a b cbc ac ab

Ans.

19. (c)

Sol. Let D be the midpoint of the side BC. By Apollonius theorem, 2 2 2 2AB AC 2(BD AD ) .

Hence 2

2 2 21 1 81BD (10 5 ) 62 2 4

Thus 9BD2

and BC = 9. The sides of the triangle are therefore, a = 9, b = 5, c = 10.

The semi-perimeter s = 12 and the area ABC is given by

( ) ( ) ( ) 6 14s s a s b s c

OR



A

B CD

510

6 12

By Appolloneous theorem, 2 2 2 2AB AC 2AD 2 CD

we get 9 9CD , BD2 2

By Herons formula, area of 2ADC 3 14 cm

and area of 2ABD 3 14 cm

ar ABC ar ADC ar ABD 6 14

6 x 6 14

x 14

20. (d)

Sol. From the given conditions we have

3 32 2 28 28a ba b ab

a b

The possibilities are

(a, b) = (1, 28), (2, 14), (4, 7), (7, 4), (14, 2), (28, 1)

21. (c)

Sol. 5 4 3 2 2 3 22 5 2 8 (2 5 2 8)m m m m m m m m

(3 1) (2 (3 1) 5(3 1) 2 8)m m m m m 2(3 1) (6 2 15 5 2 8)m m m m m 2(3 1)(6 15 3)m m m

(3 1) (6(3 1) 15 3) m m m

(3 1) (3 9)m m 29 30 9m m

9(3 1) 30 9 3m m m



Hence5 4 3 2

2

2 5 2 8 3 11 3

m m m m mm m

22. (b) 2 2 2Sin 90 88 Sin 90 86 Sin 90 84 ...............22 + ................ 2Sin 902 0 2 0 2 0cos 88 cos 86 cos 84 .............................+ 1

it has total of 45 terms 1 + 1 + 1 + 1 + ........................22 times + 1 23 Ans.

23. (a) P L.C.M.(2,4,6,8,10) 120, Q LCM (1,3,5,7,9) 315

Also, L.C.M. of (P, Q) = 2520 option (a) satisfies the relation between P, Q and L.

24. (a) 2sin 3cos 2

2 2(2sin 3cos ) (2)

2 24sin 9 cos 12 sin cos 4 (1)Let3sin 2 cos x

squaring both sides2 2 29sin 4 cos 12 sin cos x (2)

Adding (1) & (2)

2 2 213 sin cos 4 x

x 3

25. (d) AB : BD : DC 3 : 1 : 3then AB = 3x, BD = x, CD = 3xIn ABC , by pythagoras theorem.

2 2 2AC AB BC

2 2 2(20) 9x 16x

2400 25x

x 4

Again In ABD,

2 2 2AD AB BD

2 2 2AD 12 4

AD 160 160

AD 4 10



26. (d) mx2 + nx + 1 = 0nsin cosm

(1)

1sin cosm

Squaring both sides eqn (1)2

2 22

nsin cos 2sin cosm

2

22 n1m m

2 2m n 2m

or 2 2n m 2m Ans.

27. (d) Let Son’s present age be ‘x’ years.then Acc. to Ques.

Father’s present age = 30 x 20 7 (1)Also, Father’s present age = (x + 31) (2)

30 x 20 7 x 31

30x 600 7 x 31

29x 638

638x29

x = 22 Sum of their ages = 22 + 53 = 75 Ans.

28. (a) By div. algo.{let remainder = ax + b}

51 2x x 3x 2 q(x) ax b

51x x 2 x 1 q(x) ax b

at x 2

512 2a b ----- (1)at x = 1

511 a b

1 = a + b ----- (2)from (1) and (2)

51

51

a 2 1

b 2 2

51 51So, remainder is ax b 2 1 x 2 2



29. (b) 3 3 3a b c 3abc {when a + b + c = 0}.

then, a + b + c = 1/33(abc) . Ans.

30. (a) By 2nnS a l

and ( 1)l a n d

888 = 222 + (n-1) 2666 = (n-1) 2

334 n

334334 222 8882

S

334 11102

= 185370



PART - III1.Sol.(a)

Since 1 1 1 1 1 1f v u v u f

Putting the sign convention properly.

1 1 1 1 1 1( ) ( ) ( )

v u f v u f

Comparing this equation with y = mx + c

Slope 0tan 1 135 m or 045 and intercept 1

Cf

- 4501350

l/v

l/u

1 C

f

2.Sol. (b)

Since a parallel beam will reflect parallel and seen to meet at infinity.3.Sol.(a)

All the conductors have equal lengths. Area of cross-section of A is 2 2 23 2

a a a

Similarly area of cross-section of B = Area of cross-section of 2C a

Hence according to formula lRA

; resistances of all the conductors are equal i.e., A a CR R R

4.Sol.(d)

2 2222 31

1 2 31 2 3

: : : : l lllR R R R

m m m m 1 2 39 4 1: : : : 27:6:11 2 3

R R R

5.Sol. (a)

Direction of induced current can be find out of fleming’s right hand rule.6.Sol. (c)

lR

A

2mRA m

l m

7.Sol.(c)

18 1962.5 10 1.6 10 101

nei amperet



8.Sol.(a)

Total internal reflection occurs when i > c

9.Sol.(c)

As r > i therefore 1 2n n

10.Sol.(c)

In both the cases initial verticle velocity is zero, so time taken by them to reach the ground is same ratio1 : 1

11.Sol.(c)

For last second of motionv = 0 m/st = 1sa = – g m/s2

v = u + atO = u – g u = gv2 – u2 = 2as0 – g2 = 2 (-g)s

s = g2

12.Sol.(d)

F = mal = 100 × a

21a 0.01 cm /s100

13.Sol.(d)

2GMg

R h

So, when height increase, acceleration due to gravity will decreases



14.Sol.(c)

KE = 2P

2m

P is increased 50%

So, 3P ' P2

Now,22 pP' 9 9KE' KE

2m 4 2m 4

9KE' KE4

change in 2 21 1

9 1 1KE mv mv4 2 2

21

1 9mv 12 4

21

5 1 mv4 2

15.Sol.(a)

m1 = m2

v1 = v2

F = 2mv

rSo,

21 1

1 i 22

1 22 1

2

m vF r r

m vF rr

16.Sol.(b)

v2 – u2 = 2as

22v v 2g 75

16

2–15v 2g 7516

75m

V

v4

2 2 75 16 9v15

2v 10 16g

v2 = 1600 v = 40 m/s

v2 – u2 = 2as –1600 = – 2 × 10 × s –1600s 80m

–20

17.Sol.(c)

Q = i2Rt



18.Sol.(b)

At – 36ºC bromine will be liquid

19.Sol.(a)

Atoms & molecules are constituents of matter.

20.Sol.(d)

Evaporation leads to cooling in perspiration in human body and earthen put. Plant keep themseviescool by transpiration of leaves.

21Sol.(b)

Sodium amalgam is mixture

22.Sol.(a)

Milk is fat dispersed in water

23.Sol.(d) Gun powder is heterogenuns mixture

24.Sol.(c)

Mass of solution - 100 + 34.7 = 134.7g

Volume of solution = mass of solution

density 3134.7 103.61m

1.3

(m/v) % = Mass of solute 100

Volume of solution =

34.7 100 33.49%103.61

25.Sol.(d)

four coloured dues were used to colour four sweets



26.Sol.(d)

2nd & 3rd period consist of both S & P block elements where 1st period consist only S block.

27.Sol. (c)

Ca consist of – 20C consist of – 6O consist of – 8CaC03 consists of = 20 + 6 + 24 = 50 protons100 gram consists of = 50 × 6.022 × 1023

10 gram consist of = 50 × 6.022 × 1022

= 3.011 × 1024 protons

28. (b)Sol. X is HCl and B is Na2CO3

2 3 2 2Na CO HCl 2NaCl CO H O

29. (c)Sol. The volume of sulphuric acid require to produce colour change for any indicator is same.

30. (d)

Sol. The conjugate base of 12 4H PO is 2

4HPO

31.Sol.(a)

Soda Ash is Na2CO3 H2O32.Sol.(b)

Malchite is an ore of copper.

33. (a)Sol. Zigzag line separetes metal from non-metal.

34. (a)Sol. P, B, Al have valency 3.

P, Ar, Al have 3 shells.



PART - IV1. (c)

P in the given concept map is Chloroplast.

2. (c) the following characteristics are applicable to cockroach .

(i) Bilaterally symmetrical and segmented body

(iii) Open circulatory system

3. (d)

The table given below shows a list of organisms and its method of reproduction.Organism Method of reproductionAmoeba P Binary fissionHydra Q Budding

Mucor R Sporulation

4. (a)Sperm is a haploid cell .

5. (b)

The diagram below shows how both X and Y are transported in a plant.

X starch Y water

6. (d)Retina is part of the human eye where images are formed.

7. (d)

Most of the photosynthesis (80%) which takes place on this earth is carried out by algae present inocean and fresh water sources.

8. (b)

Bile juice has no digestive enzyme

9. (a)Dark reaction and light reaction of photosynthesis takes place in stroma and grana of chloroplastrespectively



10. (c)

The most important function of villi in the small intestine is to provide increased surface area for absorptionof digested food.

11. (b)The final product of digestion of carbohydrates and proteins are glucose and amino acids respectively

12. (b)

In a closed circulatory system, blood is completely enclosed with in vessels

13. (c) (ii) and (iv)Left ventricle pumps oxygenated blood to different body parts while right ventricle pumps deoxygenatedblood to lungs and Right atrium receives deoxygenated blood from different parts of the body while leftventricle pumps oxygenated blood to different parts of the body

14. (b)

Wavelength of visible light is 400 - 700 nm

15. (c)Nitrogen is used in the synthesis of proteins?

16. (a)

If the tip of a seeding is cut off, growth as well as bending ceases because it hampers perception of lightstimulus

KHOJ-2019 ANSWER KEY WITH SOLUTION CLASS - 10...UG – 1 & 2, Concorde Complex, Above OBC Bank. R.C....

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Transcript of KHOJ-2019 ANSWER KEY WITH SOLUTION CLASS - 10...UG – 1 & 2, Concorde Complex, Above OBC Bank. R.C....