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PROBATIONARY OFFICERQUANTITATIVE APTITUDE NOTES..............................................................................

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SUBJECT : NUMBERS

Numbers-Key Notes

Divisibility1. A number is divisible by 2 if it is an even number.

2. A number is divisible by 3 if the sum of the digits is divisible by 3.

3. A number is divisible by 4 if the number formed by the last two digits is

divisible by 4.4. A number is divisible by 5 if the units digit is either 5 or 0.

5. A number is divisible by 6 if the number is divisible by both 2 and 3.

6. A number is divisible by 8 if the number formed by the last three digits

is divisible by 8.

7. A number is divisible by 9 if the sum of the digits is divisible by 9.

8. A number is divisible by 10 if the units digit is 0.

9. A number is divisible by 11 if the difference of the sum of its digits at

odd

places and the sum of its digits at even places, is divisible by 11.

Important formulas

i. ( a + b )( a - b ) = ( a^ 2 b^2 )

ii. ( a + b ) ^2 = ( a2 + b2 + 2 ab )iii. ( a - b ) 2 = ( a2 + b2 - 2 ab )

iv. ( a + b + c )^ 2 = a2+ b2 + c2 + 2 ( ab + bc + ca )

v. ( a 3 + b^3 ) = ( a + b )( a2 - ab + b2 )

vi. ( a^3 b^3 ) = ( a - b )( a^2 + ab + b^2)

vii. Sum of natural numbers from 1 to n

n(n+1)/2

viii. Sum of squares of first n natural numbers is =

n(n+1)(2n+1)/6

ix. Sum of cubes of first n natural numbers is A

( n(n+1)/2)^2

x. HCF= (HCF of the numerators)/(LCM of the denominators)

xi. LCM= (LCM of the numerators)/HCF of the denominatorsxii. Product of two numbers = Product of their H.C.F. and L.C.M

Note: When a number N is raised to any integral power m, the digit in the

units

place of the resulting value can be determined without actually evaluating

the

power. The digits when raised to powers will give values in which the

digits in

the units place follow a cylindrical pattern. Following is the pattern to

calculate

the digit in the units place of any derived power.

HCF models:-

If N is a composite number such that N = ap . bq . cr .. where a, b, c are primefactors of N and

p,q,r .. are positive integers, then

(a) The number of factors of N is given by the expression (p + 1) (q + 1) (r +

1)

(b) It can be expressed as the product of two factors in 1/2 {(p + 1) (q + 1) (r

+ 1)..} ways

(c) If N is a perfect square, it can be expressed

(i)as a product of two DIFFERENT factors in 1/2 {(p + 1) (q + 1) (r + 1).. -1}

ways

(ii)as a product of two factors in 1/2 {(p + 1) (q + 1) (r + 1) .+1} ways

(d) Sum of all factors of N = (ap+1 1 / a 1) . (bq+1 1 / b 1) . (cr+1 1

/ c 1)..

(e) The number of co-primes of N (< N), (N) = N(1 1/a) (1 1/b) (1 1/c)

.

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(f) Sum of the numbers in (e) = N/2 . (N)

(g) It can be expressed as a product of two factors in 2n-1, where n is the

number of different

prime factors of the given number N.

Exercise Questions

1. 117 * 117 + 83 * 83 = ?

a) 20698b) 20578

c) 21698

d) 21268

2. (1/4)3 + (3/4)3 + 3(1/4)(3/4)(1/4 + 3/4) =?

a) 1/64

b)27/64

c) 49/64

d)1

3. The difference of two numbers is 1365. On dividing the larger number by

the smaller, we get 6 as

quotient and 15 as reminder. What is the smaller number ?

a) 240b) 270

c) 295

d) 360

5. H.C.F. of two numbers is 16. Which one of the following can never be

their L.C.M

a) 32

b) 80

c) 64

d) 60

6. What is the remainder when 9 + 92 + 93 + .... + 98 is divided by 6?

a) 3

b) 2c) 0

d)5

7. The sum of the first 100 natural numbers is divisible by

a) 2, 4 and 8

b) 2 and 4

c)2 only

d)none of these

8. For what value of 'n' will the remainder of 351n and 352n be the same

when divided by 7?

a) 2

b)3

c)6d)4

9. Let n be the number of different 5 digit numbers, divisible by 4 with

the digits 1, 2, 3, 4, 5 and 6, no digit being repeated in the numbers.

What is the value of n?

a) 144

b) 168

c)192

d)none of these

10. Find the greatest number of five digits, which is exactly divisible by 7,

10, 15, 21 and 28.

a) 99840

b) 99900

c)99960

d) 99990Answer Key:1.B; 2.D; 3.B;5.D; 6.C; 7.C; 8.B; 9.C; 10.C

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Ratio & PropertionRatio:1. Duplicate ratio of a : b is a2 : b22. Sub-duplicate ratio of a : b is sqrt of(a) : sqrt of (b)3. Triplicate ratio of a : b is a3: b34. Sub-triplicate ratio of a : b is cuberoot of (a) : cuberoot of (b)

If a/b = c/d; then a=bc/d(i) (a+b)/b = (c+d)/d(ii) (a-b)/b = (c-d)/d(iii) a/c = b/d(iv) b/a = d/c(v) (a+b)/(a-b) = (c+d)/(c-d)If a/b=c/d; then simplest possible value of a=c, b=d(vi) In the ratio a : b, if a > b, then (a + x / b + x) < a / b (x > 0)(vii) In the ratio a : b, if a < b, then (a + x / b + x) > a / b (x > 0)(viii) In the ratio a : b, if a = b, then (a + x / b + x) = a / b (x > 0)Proportion & Variation:If a is directly proportional to b; then a=kb

If a is inversely proportional to b; then a=k/bIf a is directly proportional to b and inversely proportional to c, then a is directlyproportional to b/c=> a= kb/cExercise Questions1. Ram, Sham and Suresh start business investing in the ratio 1/2 : 1/3:1/6. The time for whicheach of them invested their money was in the ratio 8:6:12 respectively. If

they get profit of Rs.18000from the business, then how much share of profit will Ram get?a. Rs.4000

b. Rs.6000

c. Rs.8000d. Rs. 100002. The ratio of the number of boys and girls in a college is 7 : 8. If the

percentage increase in thenumber of boys and girls be 20% and 10% respectively, what will be thenew ratio?a. 8 : 9b. 17 : 18c. 21 : 22d. Cannot be determined

3. p,q and r are three positive numbers and Q=(p+q+r)/2; If (Q-p):(Qq):(Q-r) = 2:5:7, then findthe ratio of p,q and r ?a. 4:3:7

b. 12:9:7c.9:7:4d. 4:3:2

4.A and B together have Rs. 1210. If 4/15 of A's amount is equal to 2/5of B's amount, how muchamount does B have?a. Rs. 460b. Rs. 484c. Rs. 550d. Rs. 6645. Two numbers are respectively 20% and 50% more than a third

number.The ratio of the twonumbers is

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a. 2 : 5b. 3 : 5

c. 4 : 5d. 6 : 76. The ratio of the cost prices of two articles A and B is 4:5.The articles are sold at aprofit with their

selling prices being in the ratio 5:6.If the profit on article A is half of its cost price,find the ratio ofthe profits on the articles A and B?a. 7:10b. 9:11c. 5: 9d. 10:117. A sum of money is to be distributed among A, B, C, D in the proportionof 5 : 2 : 4 : 3. If C getsRs. 1000 more than D, what is the total amount?a. Rs. 14000

b. Rs. 15000c. Rs. 20000

d. None of these8. If Rs. 782 be divided into three parts, proportional to :2/3:3/4, then

the first part is:a. Rs. 182b. Rs. 190c. Rs. 196d. Rs. 2049.A bag contains 50 paisa, 20 paisa and 10 paisa coins in the ratio5:3:1.If the total amount in thebag is 640 Rs,find the difference in the amounts contributed by 50 paisaand 20 paisa coins.a. Rs.300

b. Rs.400c. Rs.380d. None of these10. The speed of an engine is proportional to the square root of the

number of wagons attached toit. Without any wagons attached to it the speed of the engine is 60km/hr.With 16 wagons attachedto it the speed of the engine is 40km/hr; find the maximum number of

wagons that can be attachedso that the train moves.a. 144

b. 145c. 142d. 143Answer Key:

1.c; 2.c; 3.b; 4.b; 5.c; 6.d; 7.a; 8.d; 9.c; 10.d

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AVERAGEAverage= (x1+x2+x3+......xn)/n where x1,x2,x3, are quantities.Weighted Average= (n1x1+n2x2+n3x3+.....nkxk) / n1+n2+n3+......nkwhere x1,x2,x3.... xk are the quality factorsn1,n2,n3,........nk are the quantity factorsEg: If the average height of boys= 172cms and that of girls=154 cms, then find average

height of the class with 18 boys and 12 girls?Here n1 and n2 are no. of boys and girls (Quantity factor)x1 and x2 are average heights (Quality factor)MixturesFor mixtures, Average, x'= (n1x1+n2x2)/ (n1+n2)Alligation:=> n1/n2 = (x2-x')/x'-x1Exercise questions1. The average wages of a worker during a fortnight comprising 15 consecutiveworking days was Rs.90 per day. During the first 7 days, his average wages wasRs.87/day and the average wagesduring the last 7 days was Rs.92 /day. What was his wage on the 8th day?

a)83b) 92c)90d)972. The average of 5 quantities is 6. The average of 3 of them is 8. What is theaverage of theremaining two numbers?a) 6.5 b) 4 c) 3 d) 3.53. The average temperature on Wednesday, Thursday and Friday was 250. Theaveragetemperature on Thursday, Friday and Saturday was 240. If the temperature onSaturday was 270,

what was the temperature on Wednesday?a) 240b)210c) 270d) 3004. The average age of a group of 12 students is 20years. If 4 more students join thegroup, theaverage age increases by 1 year. The average age of the new students isa) 24b)26c)23d) 225. When a student weighing 45 kgs left a class, the average weight of the remaining

59 studentsincreased by 200g. What is the average weight of the remaining 59 students?a) 57 kgsb) 56.8 kgsc)58.2 kgsd)52.2 kgs6. The average of 5 quantities is 10 and the average of 3 of them is 9. What is theaverage of theremaining 2?a) 11b) 12c) 11.5

d) 12.57. The average age of a family of 5 members is 20 years. If the age of the youngest

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member be 10years then what was the average age of the family at the time of the birth of theyoungest member?a) 13.5b) 14c)15

d) 12.58. Average cost of 5 apples and 4 mangoes is Rs.36. The average cost of 7 applesand 8 mangoes isRs.48. Find the total cost of 24 apples and 24 mangoes.a) Rs.1044b) RS.2088c)Rs.720d) Rs.3249. Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety in theratio 1 : 1 : 2.If the mixture is worth Rs. 153 per kg, the price of the third variety per kg will be:a) Rs. 169.50

b) Rs. 170c) Rs. 175.50d) Rs. 1810. A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. Howmuch of themixture must be drawn off and replaced with water so that the mixture may be halfwater and half syrup?a) 1/3b) 1/4c) 1/5d) 1/7Answer Key

1.d; 2.c; 3.d; 4.a; 5.a; 6.c; 7.d; 8.b; 9.c; 10.cMail to us contact@upscjob.com (For Any Suggestion,Help,Doubt,Advice.Problems Regarding Jobs)

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TIME SPEED AND DISTANCE

Important Formula and Equations1. Speed, Time and Distance:Speed = Distance / timeTime = distance /speedDistance =speed*time

2. km/hr to m/sec conversion:x km/hr =[x*5/18] m/sec3. m/sec to km/hr conversion:x m/sec =[x* 18/5] km/h4. If the ratio of the speeds of A and B is a : b, then the ratio of thetimes taken by them to cover thesame distance is 1/a:1/b or b : a5. Suppose a man covers a certain distance at x km/hr and an equaldistance at y km/hr. Then, theaverage speed during the whole journey is [xy/(x+y)] km/hrKey Note:Caution average speed should not be calculated as average of different

speeds, i.e., Ave. speed(Sum of speed / No. of different Speed)There are two different cases when average speed is required.Case IWhen time remains constant and speed varies :If a man travels at the rate of x km/h for t hours and again at the rateof y km/h for another t hours,then for the whole journey, his average speed is given byAverage speed= Total distance/ Total time taken = (xt+yt)/(t+t)= (x+y)/2 kmphCase IIWhen the distance covered remains same and the speeds vary :When a man covers a certain distance with a speed of x km/h andanother equal distance at the rateof y km/h. then for the whole journey, the average speed is given byAverage speed =2xy/(x+y) km/h.Velocity :The speed of a moving body is called as its velocity. If thedirection of motion is also takeninto consideration Velocity =(Net displacement of the body)/(Time taken)

Relative speed:a) Bodies moving in same direction

When two bodies move in the same direction, then the difference of their speeds is called the relative speedof one with respect to the other.

When two bodies move in the same direction, the distance between them increases (or decreases) at the

rate of difference of their speeds.

b) Bodies moving in opposite direction The distance between two bodies moving towards each other will get

reduced at the rate of their relative speed (i.e., sum of their speeds). Relative speed of one body with

respect to other body is sum of their speeds. Increase or decrease in distance between them is the product

of their relative speed and time.

Key notes to solve problems

When a moving body covers a certain distance at x km/h and another same distance at the speed of y km/h,

then average speed of moving body during its entire journey will be [2xy/(x+y)]km/h

A man covers a certain distance at x km/h by car and the same

distance at y km/h by bicycle. If the time taken by him for the whole journey by t hours, then Totaldistance covered by him is equal to 2txy/(x+y) km.

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A boy walks from his house at x km/h and reaches the school ' t1 ' minutes late. If he walks at y

km/h he reaches ' t2 ' minutes earlier. Then, distance between the school and the house

= ((xy)/(y-x))* (t1+t2)/60 km

If a man walks with (x/y) of his usual speed he takes t hours more to cover a certain distance, then

the time to cover the same distance when he walks with his usual speed, (xt)/(y-x) hours.

If two persons A and B start at the same time in opposite directions from the points and after passing each

other they complete the journeys in 'x ' and ' y ' hrs. respectively, then A's speed: B'sspeed=speed= ROOT OVER OF y: ROOT OVER OF xIf the speed is (a/b) of the original speed, then the change in time taken to cover the same distance is given

by Change in time = ((b/a)-1)*original time Key notes to solve problems on Trains The time taken by a train

in passing a signal post or a telegraph pole or a man standing near a railway line =

(Length of the train)/ (speed of the train)

The time taken by a train passing a railway bridge or a platform or a tunnel or a train at rest=

(x+y)/Speed where, x = length of the train, y = length of the bridge or platform or standing train or tunnel

Time taken by faster train to pass the slower train in the same

direction= (x+y)/(u-v); where, x = length of the first train ; y = length of the second train ; u = speed of the

first train ; v = speed of

the second train and u > v

Time taken by the trains in passing each other while moving inopposite direction =(x+y)/(u+v)

Time taken by the train to cross a man = x/(u-v) where, both

are moving in the same direction and x= length of the train; u= speed of the train and v= speed of

the man.

Time taken by the train to across a man running in the

opposite direction= x/ (u+v) If two trains start at the same time from two points A and B towards each other

and after crossing, take a and b hours in reaching B and A respectively. Then, A's speed: B's speed= b: a

A train starts from a place at u km/h and another fast train

starts from the same place after t hours at v km/h in the same direction. Find at what distance from the

starting place both the trains will meet and also find the time of their meeting.

Distance= uvt/(v-u) km

Time=ut/(v-u) hoursThe distance between two places A and B is x km. A train starts from A to B at u km/h. One another train

after t hours starts from B to A at v km/h. At what distance from A will both the train meet and also find the

time of their meeting

Time=(x-ut)/ (u+v) + t hours

Distance from A = u(((x-ut)/(u+v))+t) km

Two trains starts simultaneously from the stations A and B towards each other at the rates of u and

v km/h respectively. When they meet it is found that the train had traveled x km more than

the first. Then the distance between the two stations (i.e., between A and B) is x(u+v)/ (v-u) km.

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Time & Work(M1D1HI)/ W1 = (M2D2H2)/W2Where M1 & M2 represents no of labourers; D1 & D2 represents no of days; H1 &H2 represents no of hours; W1&W2 represents work done.If there are 2 persons A & B such that A can do work in a days and B can do workin bdays. Such that a is a multiple of b, then, time taken by them to complete the work together

= Bigger no/Sum of ratiosEg: A can do work in 9 days, B can do work in 18 days. In how many days they willcomplete the work together.Bigger no=18, Ratio=9:18=1:2No of days = 18/(1 + 2) = 6 daysIf a is not a multiple of b, then time taken by A&B to complete the work together= (LCM)/(Sum of ratios)Eg: A can do a piece of work 30 days. B can do work in 45 days. In how many daysthey will complete the work together?LCM = 90, Ratio= 30:45=2:3No of days= 90/(2 + 3) = 90/5 = 18If there are 3 persons A, B & C whose time taken a,b,c days respectively, to

complete a certain work. Time taken by them to complete the work =LCM of (a, b, c)/(LCM/a + LCM/b + LCM/c)Note: For 3 persons the common format isStep1: Find the LCMStep2: Find the individual share of work i.e. LCM/a, LCM/b, LCM/c.Step3: Rest methods depend on the question i.e. follow the question patterns.Eg: A, B and C can do a work in 15,20,45 days respectively. In how many days theycan completethe work together.LCM=180No of days= [180/ (180/15 + 180/20 + 180/45)= [180/ (12+9+4)]

= [180/25]= 7.2 daysPipes & CisternsIf there are 2 pipes A & B such that A (inlet pipe) & B (outlet pipe). Such that A canfill tank in a minutes and B can empty the tank in b minutes , then time taken to fill the tank ifboth are operated together = Bigger no/Difference of ratiosEg: A can fill tank in 9 minutes, B can empty the tank in 18 minutes.. In what timethe tank be filled,if both pipes work simultaneously?Bigger no=18, Ratio=9:18=1:2Time taken to fill the tank = 18/(2 - 1)= 18 minutesIf a is not a multiple of b, then time taken by A&B to fill the tank.

= (LCM)/(Difference of ratios)Eg.: An inlet pipe can fill the tank in 30 minutes. B an outlet pipe can empty the tankin 45 minutes.In what time the tank be filled if both pipes work simultaneously?Time taken to fill the tank= LCM = 90= Ratio= 30:45=2:3= 90/(3 - 2) = 90/1 = 90 minutesIf there are 3 pipes A, B & C, in which A, B are inlet pipes which takes a,b,minutesrespectively to fill the tank and C an outlet pipe which takes c minutes to empty the tankTime taken by them to fill the tank, if all of them are operated together.= LCM of abc/ (LCM/a + LCM/b - LCM/c)Eg: A, B two inlet pipes takes 15,18 minutes to fill the tank and C an oulet pipe takes

45 minutes to empty the tank respectively. In what time the tank be filled if all of them areoperated together?

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LCM=90No of days= [90/(90/15 + 90/18 - 90/45)= [90/(6+5-2)]= [90/9]= 10 minutesNote: In case of division of money with respect to share of each persons work then

share of A = bc/ab+bc+acIn case of division of money with respect to share of each persons work then shareof B = ac/ab+bc+acIn case of division of money with respect to share of each persons work then shareof C =ab/ab+bc+acSame as Share of A:(LCM/a)/ (LCM/a + LCM/b + LCM/c)Share of B:(LCM/b)/ (LCM/a + LCM/b + LCM/c)Share of C:(LCM/c)/ (LCM/a + LCM/b + LCM/c)Eg: A,B,C can do a work in 15,20,45 days respectively. They get Rs 500 for their work. What is theshare of A?LCM = 180

Share of A = (LCM/a x Total amount)/LCM/a + LCM/b + LCM/c= (180/15)/(180/15 +180/20 + 180/45)= (12/25) * 500= Rs.240Exercise questions1. Two workers A and B manufactured a batch of identical parts. A worked for 2hours and B worked for 5 hours and they did half the job. Then they worked together for another 3hours and they had to do (1/20)th of the job. How much time does B take to complete the job, if heworked alone?

A) 24 hoursB) 12 hoursC) 15 hours

D) 30 hours2. Pipe A can fill a tank in 'a' hours. On account of a leak at the bottom of the tank it takes thrice aslong to fill the tank. How long will the leak at the bottom of the tank take to empty afull tank, when pipe A is kept closed?

A) (3/2)a hoursB) (2/3)aC) (4/3)aD) (3/4)a3. A and B working together can finish a job in T days. If A works alone and completes the job, hewill take T + 5 days. If B works alone and completes the same job, he will take T + 45 days. What isT?

A) 25B) 60

C) 15D) None of these4. A man can do a piece of work in 60 hours. If he takes his son with him and both work togetherthen the work is finished in 40 hours. How long will the son take to do the same job, if he workedalone on the job?

A) 0 hoursB) 120 hoursC) 24 hoursD) None of these5. A, B and C can do a work in 5 days, 10 days and 15 days respectively. They started together to dothe work but after 2 days A and B left. C did the remaining work (in days)

A) 1

B) 3C) 5

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D) 46. X alone can do a piece of work in 15 days and Y alone can do it in 10 days. X and Y undertook todo it for Rs.720. With the help of Z they finished it in 5 days. How much is paid to Z?

A) Rs.360B) Rs.120C) Rs.240

D) Rs.3007. Ram starts working on a job and works on it for 12 days and completes 40% of the work. To helphim complete the work, he employs Ravi and together they work for another 12 days and the workgets completed. How much more efficient is Ram than Ravi?

A)50%B) 200%C) 60%D)100%

8. A red light flashes 3 times per minute and a green light flashes 5 times in two minutes at regular

intervals. If both lights start flashing at the same time, how many times do they flash together in

each hour?

A) 30

B) 24C) 20D) 609. A and B can do a piece of work in 21 and 24 days respectively. They started the work togetherand after some days A leaves the work and B completes the remaining work in 9 days. After howmany days did A leave?

A) 5B) 7C) 8D) 610. Ram, who is half as efficient as Krish, will take 24 days to complete a work if he worked alone. IfRam and Krish worked together, how long will they take to complete the work?

A) 16 daysB) 12 daysC) 8 daysD) 18 days

Answer Key1.C; 2.A; 3.C; 4.B;5.D; 6.B; 7.D; 8.A; 9.B; 10.CMail to us contact@upscjob.com (For Any Suggestion,Help,Doubt,Advice.Problems Regarding Jobs)

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PercentagesBy a certain percent, we mean that many hundredths. Thus, x percent means x

hundredths, writtenas x%.To express x% as a fraction, we have x%=x/100.Thus, 20%= 20/100= 1/5

To express a/b as a percent, we have, a/b= (a/b)*100%.Thus, 1/4= (1/4)*100%= 25%.1. If A is R% more than B, then B is less than A by R/ (100+R)*1002. If A is R% less than B, then B is more than A by R/(100-R)*100

3. If the price of a commodity increases by R%, then reduction in consumption,not to increase the expenditure is: R/(100+R)*1004. If the price of a commodity decreases by R%, then the increase in consumption, not

to decrease the expenditure is: R/(100-R)*100Results on population:Let the population of a town be P now and suppose it increases at the rate of R%per annum, then;

1. Population after n years= p (1+(R/100))n

2. Population n years ago= P/(1+(R/100))n3. If a number is increased by x% and thereafter reduced by x%, then the number willbe reduced by x2/100 percent4. If a number is reduced by x% and there after increased by x% then the number willbe reduced by x2/100 percent5. If in an examination, in which the minimum pass percentage is x%, a candidate

secures y marks and falls by z marks, then the total number of marks in thisexamination will be 100*(y+z)/x6. If in an examination x% and y% candidates respectively fail in two different subjectswhile z% candidates fail in both the subjects, then the percentage of candidates who

pass in both the subjects will be [100-(x+y+z)]%Tips:

1. If an object's price is increased or decreased by x% and the other factor is decreased by y% thenthe net effect is given by: Net Effect= [x+y+xy/100]%2. If the net effect is nil, ie, there is no loss or no gain, then the above formula becomes:y=100x/100+x3. If the price of an article is successively increased by x%,y% and z% then single equivalentincrease in the price will be [x+y+z+{xy+yz+zx}/(100)+xyz/1002]%4. If after spending p1% first, then p2% from the remaining and so on, B is the balance amount,then the total (original) amount is given by:Total amount= B*100*100...../ (100-p2).....Population formula: 1)If the population increases by x% during the first year, by y% during thesecond year, by z% during the third year, the population after three years will be:P(1+x/100)(1+y/100)(1+z/100)

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Profit and LossImportant formula and Equations

Gain= SP-CPLoss= CP-SPGain Percentage= (Gain*100)/ CPLoss Percentage= (Loss*100)/ CP

Selling Price=((100+Gain %)/100)*CP or ((100-Loss%)/100)*CPCost Price= (100*SP)/(100+Gain%) or (100*SP)/(100-Loss%)When a person sells two similar items, one at a gain of say x%, and the other at a lossof x%, then the seller always incurs a loss given by:

Loss %= (Common loss ans gain %)2/10 = (x/10)2If a trader professes to sell his goods at cost price, but uses false weights, thenGain%=((Error)/(True value)-(Error))*100%

Key NotesWhen an article is sold at a profit of x%. If it would be sold for Rs.n less, there would bea loss of y%, then the cost price of the article CP=(n*100)/(x+y)A man sells an article at a gain of x%. If it would have been sold for Rs.n more, there

would have a profit of y%, then CP= (n*100)/(y-x)

A person brought two articles for Rs.n. On selling one article at x% profit and other aty% profit, he get the same selling price of each, thenCP of first article= Rs. (100+y)n/200+x+yCP of second article= Rs. (100+x)n/ 200+x+yWhen m articles are brought for Rs.n and n articles are sold for Rs.m and m>n, thenprofit%= ((m2-n2)*100)/n2

If A sells an article to B at a profit of r1 %, B sells it to C at a profit of r2 % and C sells itto D at a profit of r3 %, then, cost price of D= Cost Price of A(1+r1/100)(1+r2/100)(1+r3/100)If A sells an article to B at a loss of r1 %, B sells it to C at a loss of r2 % and C sells it to

D at a loss of r3 %, then, cost price of D= Cost Price of A(1-r1/100)(1-r2/100)(1-r3/100)

A dealer purchases a certain number of articles at x articles for a rupee and the samenumber at y articles for a rupee. He mixes them together and sells at z articles for arupee.Then his gain or loss %=([2xy- 1]/z(x+y))*100; according to positive or negative sign.If P1 is rate gain w.r.t. selling price S1 and P2 is rate gain w.r.t. selling price S2Then CP=(100/P1-P2)* difference between selling pricesIf P1 is rate gain w.r.t. selling price S1 and P2 is rate loss w.r.t. selling price S2Then CP= (100/ P1+P2) * difference between selling prices

When a man sells two things at the same price each and in this process his loss on firstthing is x% and gain on second thing is x%, then in such a type question, there isalways a loss andLoss= 2*SP/((100/x)2 -1)

When a man buys two things on equal price each and in those things one is sold on theprofit of x% and another is sold on the loss of x%, then there is no loss or no gainpercent.A sells an article at a profit of r1 % to B and B again sells it to C at a profit of r2 %. If C

pays Rs. P to B, then CP of the article forA= Rs. 100*100*P/(100+r1)(100+r2)When a shopkeeper on selling an article for Rs.n, gains as much percent as the costprice of it,then =Rs[-50+/-10 square roots of(25+r)]CP of the articleIf there is loss in place of profit,=Rs[-50+/-10 square roots of(25-r)]

then CP of the article=If two articles are sold at the same price (i.e., the selling prices are equal) and the

magnitude ofpercentage of profit x on one article is the same as the magnitude of percentage of loss

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x on thesecond article, then there is an overall loss and the percentage of loss is x2/100.

If a shopkeeper claims to sell the goods at cost price and gives x units less than theactual weight,then the profit percentage made by the shopkeeper is [x / actual weight x] x 100.In the above case, the error percentage = [x / actual weight] x 100

If two articles are bought for the same price (i.e., the cost prices are equal) and one issold at a profit of p1% and the second is sold at a profit of p2%, then the overallpercentage of profit is ((p1+ p2 )/2) x 100

If the selling price of m articles is equal to cost price of n articles, where m > n, thenprofitpercentage is ((m n )/m)x 100.

If m < n, then loss percentage = ((n m)/m) x 100.DiscountDiscount% = Discount / Marked price * 100%An article sold at selling price(SP1) at a loss of x% is to be sold at selling price(SP2) to

gain y%,

then SP2 = SP1(100 + y)/ (100-x)If selling an object for Rs.x a person loses a certain sum and selling for Rs.y he gains thesameamount, CP is given by CP = (x+y)/2.When the price of an article is reduced by p% a man can buy x quantity of the article forRs.y then

reduced price = 1/x ( y * p / 100) per unit.original price = reduced price * 100 / (100 - p).If the MP (marked price) of an article above CP is M% and after allowing a discount ofd%, the gain is g%,

Then M% = d+g * 100% / 100 - d, and if there is a loss of l%, then M% = d-l * 100% /100-d.

A person sells goods at a profit of x%. Had he sold it for Rs. X more, y% would havebeen gained.Then CP is given by Rs. X *100 / y-x.A person sells goods at a loss of x%. Had he sold it for Rs. X more, he would havegained y% . ThenCP is given by Rs. X * 100/ y+x.When there are two successive profits of x% and y% the net gain% is given by: Net gain= [ x + y +{ xy / 100 }]%.

When there are two successive losses of x% and y% the net loss% is given by: Net loss= [ - x - y +{ xy / 100 }]%.10)When there is a gain of x% and a loss of y% the net effect is given by: Net effect =[ x - y - { xy /100}]%.

l. If d1, d2, d3.. are percentages of successive discounts on a marked price MP, thenthe sellingprice SP = MP (1 d1/100) (1 d2/100) (1 d3/100)2. If d1, d2, d3. are the percentages of successive discounts offered, then the effective

discount isd% = 100[1- (1 d1/100) (1 d2/100) (1 d3/100)]3. If x and y are two successive discount percentages, then it is equivalent to a singlediscount percentage of x + y xy/100.

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SIMPLE AND COMPOUND INTEREST

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PARTNERSHIP

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PERMUTATION & COMBINATIONnPr = n!/ (n-r)!nCr = n!/r!(n-r)!nCr = nC(n-r)nCr = nPr/r!0!=1

1!=1nPn=n!nP1=nnC1=nnCn=1Exercise questions1. How many words can be formed by re-arranging the letters of the word ASCENTsuch that A andT occupy the first and last position respectively?

A)5!B)4!C)6! - 2!

D)6! / 2!2. There are 2 brothers among a group of 20 persons. In how many ways can thegroup bearranged around a circle so that there is exactly one person between the twobrothers?

A) 2 * 19!B)18! * 18C) 19! * 18D)2 * 18!3. There are 12 yes or no questions. How many ways can these be answered?

A) 1024B) 2048

C) 4096D)1444. How many ways can 4 prizes be given away to 3 boys, if each boy is eligible forall the prizes?

A) 256B) 12C) 81D) None of these5. A team of 8 students goes on an excursion, in two cars, of which one can seat 5and the otheronly 4. In how many ways can they travel?

A) 9B)26

C)126D) 39206. How many numbers are there between 100 and 1000 such that at least one oftheir digits is 6?

A) 648B) 258C) 654D)2527. How many ways can 10 letters be posted in 5 post boxes, if each of the postboxes can take morethan 10 letters?

A) 510

B) 105C) 10P5

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D) 10C58. In how many ways can the letters of the word EDUCATION be rearranged so thatthe relativeposition of the vowels and consonants remain the same as in the word EDUCATION?

A) 9!/4B) 9!/(4!*5!)

C) 4!*5!D) None of these9. In how many ways can 15 people be seated around two round tables with seatingcapacities of 7and 8 people?

A) 15!/(8!)B) 7!*8!C) (15C8)*6!*7!D)2*(15C7)*6!*7!10. If the letters of the word CHASM are rearranged to form 5 letter words such thatnone of theword repeat and the results arranged in ascending order as in a dictionary what is

the rank of theword CHASM?

A) 24B)31C) 32D)3011. How many words of 4 consonants and 3 vowels can be made from 12consonants and 4 vowels,if all the letters are different?

A) 16C7 * 7!B) 12C4 * 4C3 * 7!C) 12C3 * 4C4

D) 12C4 * 4C312. In how many ways can 5 letters be posted in 3 post boxes, if any number ofletters can beposted in all of the three post boxes?

A) 5C3B) 5P3C) 53D)3513. How many number of times will the digit '7' be written when listing the integersfrom 1 to 1000?

A) 271B) 300C) 252

D)30414. There are 6 boxes numbered 1, 2,...6. Each box is to be filled up either with ared or a greenball in such a way that at least 1 box contains a green ball and the boxes containinggreen balls areconsecutively numbered. The total number of ways in which this can be done is

A) 5B) 21C) 33D) 6015. What is the value of 1*1! + 2*2! + 3!*3! + ............ n*n!, where n! means nfactorial or n(n-

A(n-2)...1A) n(n-A(n-A!))

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B) (n+A!)/(n(n-A))C) (n+A! - n!) D) (n + A! - 1!)

Answers1.B; 2.D; 3.C; 4.C; 5.C; 6.D; 7.A; 8.C; 9.C; 10.C; 11.B; 12.D; 13.B; 14.B; 15.DMail to us contact@upscjob.com (For Any Suggestion,Help,Doubt,Advice.Problems Regarding Jobs)

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PROBABILITYProbability is a measure of how likely a particular outcome to an event is to happen.It ranges from 0to 1. A probability of 0 means that the outcome cannot happen. A probability of 1 means that theoutcome will definitely happen. And in between 0 and 1 means that the outcome may happen.Example with a coin.When a coin is tossed the outcome (or event) can be heads or tails. What is the probability it is tails?

Since each outcome, heads or tails, is equally likely we can say that the probability of each is 0.5.p(coin toss is tails)= 1/2Basic rule of probabilityMore generally we can say that where there are n equally likely outcomes then the probability ofeach of these possibilities will be 1/n. So we can say that p(outcome)= (number of ways it canhappen)/(total number of possible outcomes)Example with dieWhat is the probability of rolling a 6 when you throw a 6 sided die.Each number from 1 to 6 is equallylikely to be thrown and only one of those outcomes is a 6 so using the general rule we can say thatp(throw a six)= 1/6Example with cardsIf you pick a card at random from a deck of cards what is the probability that it is an ace?There are 52 cards in a pack and those there are 4 aces so

p (an ace)= 4/52 - 1/13The probability two outcomes for independent events both occur can be found by multiplying theirprobabilities.p(A and B)- p(A)*p(B)Example with coinsWhat is the probability of throwing two heads in a row when tossing a coin?This is the same as asking what the probability that the first coin tossed will be head AND the secondcoin tossed will be a head.So the probability that of tossing two heads in a row is Example with a jarA jar contains 2 red balls and 4 green balls. What is the probability that two balls selected at randomfrom the jar are both green?Each ball is equally likely to be selected from the jar so we can work out the probability of the first

ball selected being green.Here is where we need to be careful, once we have taken 1 green ball out of the jar, the jarcontains only 3 green balls ans 2 red balls so Now we can say that,So the probability that of pickingout two green balls is 2/5.Total Probability Formula.P(A) = P(AjH1)P(H1) + + P(AjHn)P(Hn):Total probabilityEvents H1;H2; : : : ;Hn form a partition of the sample space S if(i) They are mutually exclusive (Hi Hj = ;; i 6= j) and(ii) Their union is the sample space S; i=1 Hi=S;The events H1; : : : ;Hn are usually called hypotheses and from their definition follows that P(H1) + + P(Hn) = 1 (= P(S)):Let the event of interest A happens under any of the hypotheses Hi with a known (conditional)

probability P(AjHi): Assume, in addition, that the probabilities of hypotheses H1; : : : ;Hn are known.Then P(A) can be calculated using the total probability formula.Total Probability Formula:P(A) = P(AjH1)P(H1) + + P(AjHn)P(Hn):The probability of A is the weighted average of the conditional probabilities P(AjHi) with weightsP(Hi):Bayes Formula:Let the event of interest A happens under any of hypothesesHi with a known (conditional) probabilityP(AjHi): Assume, in addition, that the probabilities of hypotheses H1; : : : ;Hn are known (priorprobabilities). Then the conditional (posterior) probability of the hypothesis Hi; i = 1; 2; : : : ; n,given that event A happened, isP(HijA) = P(AjHi)P(Hi) P(A) ; where P(A) = P(AjH1)P(H1) + + P(AjHn)P(Hn):Assume that out of N coins in a box, one has heads at both sides. Such two-headed coin can bepurchased in Spencer stores. Assume that a coin is selected at random from the box, and without

inspecting it, flipped k times. All k times the coin landed up heads. What is the probability that twoheaded coin was selected?

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Denote with Ak the event that randomly selected coin lands heads up k times. The hypothesesareH1-the coin is two headed, and H2 the coin is fair. It is easy to see that P(H1) = 1=N and P(H2) =(N 1)=N.The conditional probabilities are P(AkjH1) = 1 for any k, and P(AkjH2) = 1=2k:By total probability formula,P(Ak) = (2k + N- 1)/2kN and

P(H1jAk) = 2k/(2k + N- 1)Conditional ProbabilityProblem: A math teacher gave her class two tests. 25% of the class passed both tests and 42% of theclass passed the first test. What percent of those who passed the first test also passed the secondtest:Analysis: This problem describes a conditional probability since it asks us to find the probability thatthe second test was passed given that the first test was passed. In the last lesson, the notation forconditional probability was used in the statement of Multiplication Rule 2.Multiplication Rule 2:When two events, A and B, are dependent, the probability of both occurring is P(A and B)=P(A).P(B|A)The formula for the Conditional probability of an event can be derived from Multiplicationon Rule 2 as follows.Step1: P(A and B)=P(A).P(B|A) start with multiplication rule 2.

Step2: P(A and B)/P(A) = (P(A).P(B|A))/P(A) Divide both sides of the equation by P(A)Step3: P(A and B)/P(A) = P(B|A) Cancel P(A)s on right-hand side of the equationStep4: P(A and B)/P(A) = We have derived the formula for conditional probabilityProblem: A math teacher gave her class two tests. 25% of the class passed both tests and 42% ofthe class passed the first test. What percent of those who passed the first test also passed thesecond test?Solution: P(Second|First) = P(First and Second)/P(First)= = 0.25/0.42 = 0.60 = 60%

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BOATS AND STREAMSThe water of a stream, usually, keeps flowing at a certain speed, in aparticular direction. Thisspeed is called the current of the stream. A boat develops speedbecause of its engine power. Thespeed with which it travels when there is no current is called speed of

boat in still water. When theboat moves in the direction of the current is said to be with the stream/current or downstream.When the boat moves in the direction opposite to that of the current, itis said to be against thestream is called upstream.Eg:-If the speed of a boat in still water is ukm/hr and the speed of thestream is vkm/hr then:* Speed downstream=(u+v)km/hr* Speed upstream = (u-v)km/hrIf the speed downstream is u km/hr and the speed upstream is v km/hr,then:

* Speed of boat in still water = (u+v)km/hr* Speed (Rate) of stream = (u-v)km/hrExamplesa) A man can row a boat 12 km/h with the stream and 8km/h againstthe stream. Find his speed in still water.a) 2km/hrb) 4km/hrc) 8km/hrd) 10km/hrSolution: Speed of boat in still water = (u+v) km/hr = (12+8)=10km/hr

b) A man can row a boat 27km/h with the stream and 11km/h againstthe stream. Find speed of streama) 2km/hrb)4km/hrc)8km/hrd)10km/hrSolution: Speed (Rate) of stream = (u-v) km/hr = (27-11)=8km/hr

c) A boat running downstream covers a distance of 16km in 2 hourswhile for covering the samedistance upstream, it takes 4 hours. What is the speed of the boat in stil lwater?a)4km/hrb)6km/hrc)8km/hr

d) None of theseRate of downstream=(16/2 ) kmph=8kmphRate of upstream =(16/4) kmph=4kmphTherefore Speed in still water=1/2(8+4) kmph=6kmphNote: If ratio of downstream and upstream speeds of a boat is a:b.Then ratio of time taken= b:aSpeed of stream=a-b/a+b *Speed in still waterSpeed in still water =a+b/a-b *Speed of streamExercise Questions1. A man rows downstream 32 km and 14km upstream. If he takes 6hours to cover each distance,then the velocity (in kmph) of the current is:

a)1/2b)1c)1and

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d)2Solution: Rate downstream=(32/6)kmph; Rate upstream=(14/6)kmphVelocity of current=1/2(32/6-14/6)kmph=3/2kmph=1.5kmph

2. In one hour, a boat goes 11km along the stream and 5km against thestream.The speed of the boat in still water (in km/hr)is:

a)3b)5c)8d)9Solution: Speed in still water=1/2(11+5)kmph=8kmph3. Speed of a boat in still water is 16km/h. If it can travel 20kmdownstream in the same time as itcan travel 12 km upstream, the rate of stream is.a)1km/hb)2km/hc)4km/hd)5km/h

Solution: Speed downstream: Speed upstream=20:12=5:3Speed of current=5-3/5+3*16=4km/h

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SOME IMPORTANT CHAPTERS FOR SOLVING SIMPLE CALCULATIONS ASKED IN VARIOUSEXAMS..

SIMPLIFICATION

I. 'BODMAS' Rule: This rule depicts the correct sequence in which the operations

are to beexecuted, so as to find out the value of a given expression. Here, 'B' stands for

'Bracket', 'O' for 'of','D' for Division', 'M' for 'multiplication', 'A' for 'Addition' and 'S' for 'Subtraction'.

Thus, in simplifyingan expression, first of all the brackets must be removed, strictly in the order(),{} and []. Afterremoving the brackets, we must use the following operations strictly in theorder:(i) of (ii) Division (iii) Multiplication (iv) Addition (v) Subtraction.Eg: (-5)(4)(2)(-1/2)(3/4)=?-5*4*2*-1/2*3/4= 15II. Modulus of a Real Number: Modulus of a real number a is defined asa|= a, if a>0 or

|a|= -a, if a 120x = 40x=1/3Number of floors covered by David in (1/3) min. = (1/3)*57= 19So, their paths cross at (11 +19) i.e., 30th floor.3. A man has some hens and cows. If the number of heads be 48 and the number offeet equals140, then the number of hens will be:a. 22b. 23c. 24

d. 26Explanation: Let the number of hens be x and the number of cows be y.

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Then, x + y = 48 .... (i)and 2x + 4y = 140 => x + 2y = 70 .... (ii)Solving (i) and (ii) we get: x = 26, y = 22.Therefore, The required answer = 26.4. In a regular week, there are 5 working days and for each day, the working hoursare 8. A man

gets Rs. 2.40 per hour for regular work and Rs. 3.20 per hours for overtime. If heearns Rs. 432 in 4weeks, then how many hours does he work for ?a. 160b. 175c. 180d. 195

Answer: Option BAnswer: Option BExplanation: Suppose the man works overtime for x hours.Now, working hours in 4 weeks = (5 * 8 * 4) = 160.Therefore, 160 * 2.40 + x * 3.20 = 432

=> 3.20x = 432 - 384 = 48=> x = 15.Hence, total hours of work = (160 + 15) = 175.Public Provident Fund. If he has Rs. 1,50,000 as total savings, how much has hesaved in PublicProvident Fund ?a. Rs. 30,000b. Rs. 50,000c. Rs. 60,000d. Rs. 90,000

Answer: Option CExplanation: Let savings in N.S.C and P.P.F. be Rs. x and Rs. (150000 - x)

respectively. Then,(1/3)x= (1/2)(150000-x)x/3+x/2=750005x/6=75000x=75000*6/5= 90000Therefore, Savings in Public Provident Fund = Rs. (150000 - 90000) = Rs. 600006. A sum of Rs. 1360 has been divided among A, B and C such that A gets 2/3 ofwhat B gets and Bgets 1/4 of what C gets. B's share is:a. Rs.120b. Rs. 160c. Rs.240d. Rs.300

Answer: Option CExplanation: Let C's share = Rs. xThen, B's share = Rs. x/4; A's share = Rs.2/3 * x/4= Rs. x/6Therefore x/6 + x/4 + x = 136017x/12 = 1360x= 1360*12/17= Rs.960Hence, B's share= Rs.960/4= Rs.2407. If a-b= 3 and a2+b2=29, find the value of aba. 10b. 12c. 15d. 18

Answer: Option AExplanation: 2ab = (a2 + b2) - (a - b)2

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= 29 - 9 = 20=> ab = 10.8. If 45-[28-{37-(15-*)}]= 58, then * is equal to:a. -29b. -19c. 19

d. 29Answer: Option cExplanation: 45-[28-{37-(15-*)}]= 58 => 45-[28-{37-15+*}]=5845-[28-37+15-*]=58 => 45[43-37-*]=5845-[6-*]=58 =>45-6+*=5839+*=58 => *=58-39= 19

SUDES AND INTEGERS

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LOGARITHMS

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FRACTIONS

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