CSAT – UPSC 2022

Paper Structure
 Total: 80 questions

 2.5 Marks per question = 200 Marks

 Negative Marking: 1/3rd

 Qualifying Marks required: 33%

 =66/200 Marks
Question Distribution

Comprehension Quant/Reasoning
Year
(Marks) (Marks)

30 50
2017
(75) (135)

26 54
2018
(65) (135)

30 50
2019
(75) (125)

25 55
2020
(62.5) (137.5)
Questions to be attempted
 Minimum Attempt: 40Q =100 Marks
 Buffer of 34 marks

Questions Right Wrong Marks

attempted
40 30 (75) 10 (8.3) 66.67
50 33 (82.5) 17 (14) 68

 For those weak in quant:

 Comprehension 25 Q = 18R + 7W
 Need 12 R in quant

 Optimum attempt: 50 Q
 What you need is:
 Speed
 Accuracy
 Practice
 Even targeted focus on half the paper will sail you through
Time Distribution
 80 Q in 2 hr = 1.5 min/Q
 50 Q in 2 hr = 2.4 min/Q

 Comprehension
 30-45 minutes

 Quant
 1:15 – 1:30 Hr

 Can be flexible
 Comprehension

Pros Cons
Bit easier Somewhat command on English
Comparatively less time consuming Subjectivity
Less accuracy
Only 25-30 questions
Quant/Reasoning

Pros Cons
Higher accuracy Time consuming
50-55 Questions Difficulty level higher
Objectivity
 Strategy
 Attempt easy questions

 Don’t waste time on difficult ones

 Not necessary to solve questions serially

 Focus on getting more correct answers than solving more questions

 Prioritise Comprehension/Quant based on your strengths

 Practice to identify the types of questions you find easy

 Even options can get you answers

 Quant
 Identify types of questions

 You should be familiar with types of questions

 Maximum 2-3 minutes per question. No more than that

 Play to your strength

 Don’t waste time in difficult questions

 At times, can even use options to get the answers.

 Can use numbers/values in unknowns in equation

 Drastic variation in number of question asked in each type eg. Data interpretation
 Types of questions
 Numbers: Prime, Natural ; Square, Cube
 Basic division, multiplication: Tests of divisibility
 Geometry: Triangle, Square, Rectangle: Area, Perimeter, Volume
 Average
 Time distance
 Percentage
 Profit & Loss
 Sale & purchase
 Age based
 Time & work
 Blood Relations
 Direction sense : Pythagoras
 Boat and streams
 Trains
 Syllogism
 Ratio and proportion
 Coding-Decoding
 Sequence series: Number/Letter/Image
 Data sufficiency
 Data interpretation
 Calendar
 Clock
Numbers
 Natural Numbers: 1,2,3,4,5,….
 Integers: ….-3,-2,-1,0,1,2,3,…
 Even Numbers: 2,4,6,8,10,…
 Odd Numbers: 1,3,5,7,9,…
 Prime Numbers: 2, 3, 5, 7, 11, 13…

 Squares of Numbers: till 15 at least

 Cubes of Numbers: till 5 at least

 These will be needed to solve sequence series

 Basic division and multiplication

 Divisibility Tests
 For 2: if its unit digit is 0,2,4,6,8
Eg. 25796, 685732 etc
 For 3: sum of digits should be divisible by 3
Eg. 654891: sum of digits= 33
 For 9: sum of digits should be divisible by 9
Eg. 845613: sum of digits = 27
 For 4: last two digits of a number should be divisible by 4
Eg. 6587928: 28 divisible by 4, hence divisible
 For 8: last 3 digits of a number should be divisible by 8
eg. 468352: 352 divisible by 8, hence divisible
 For 5: last digit should be 0 or 5
eg. 265, 689890
 For 10: last digit should be 0
Eg. 65640, 64460
 For 11: difference of sum of digits at odd places and that of even places
should be either 0 or divisible by 11
Eg. 4658793: (4+5+7+3)-(6+8+9) = (19)-(23) = -4 : not divisible
 957924: (9+7+2)-(5+9+4)= 0 : divisible by 11
 To speed up calculation:

 For 3: sum of digits should be divisible by 3

Eg. 654291:

 For 9: sum of digits should be divisible by 9

Eg. 845613
 Q. Two statements are given followed by two Conclusions:
Statements:
All numbers are divisible by 2.
All numbers are divisible by 3.
Conclusion-I:
All numbers are divisible by 6.
Conclusion-II:
All numbers are divisible by 4.
Which of the above Conclusions logically follows/follow from the two given
Statements?
a) Only Conclusion-I
b) Only Conclusion-II
c) Neither Conclusion-I nor Conclusion-II
d) Both Conclusion-I and Conclusion-II
 Conclusion-I is true, because all numbers which are divisible by 2 and 3
must also be divisible by 6.
Conclusion-ll is false, because all numbers which are divisible by 2 and 3
need not be divisible by 4.
Hence, only conclusion l logically follows.
 Q. Two Statements S1 and S2 are given below followed by a Question:
S1 : n is a prime number.
S2 : n leaves a remainder of 1 when divided by 4.
Question:
If n is a unique natural number between 10 and 20, then what is n?
Which one of the following is correct in respect of above Statements and
the Question?
a) S1 alone is sufficient to answer the Question.
b) S2 alone is sufficient to answer the Question.
c) S1 and S2 together are sufficient to answer the Question, but neither S1
alone nor S2 alone is sufficient to answer the Question.
d) S1 and S2 together are not sufficient to answer the Question.
 n is a unique natural number between 10 and 20.
According to S1, n is a prime number.
So, our number n may be: 11, 13, 17 or 19

 According to S2, on dividing n by 4 we get a remainder of 1.

So, n can be: 13, 17

 Either of the two statements alone is not sufficient to answer the question.
 Two Statements S1 and S2 are given below with regard to two numbers
followed by a Question:
S1 : Their product is 21.
S2 : Their sum is 10.
Question:
What are the two numbers?
Which one of the following is correct in respect of the above Statements
and the Question?
a) S1 alone is sufficient to answer the Question.
b) S2 alone is sufficient to answer the Question.
c) S1 and S2 together are sufficient to answer the Question, but neither S1
alone nor S2 alone is sufficient to answer the Question.
d) S1 and S2 together are not sufficient to answer the Question.
 As per S1 : Product is 21
 21 = 1*21 or 3*7

 As per S2: Sum is 10

 10= 1+9 or 2+8 or 3+7 or 4+6 or 5+5

 Hence both S1 and S2 are needed to identify the numbers which are
3 and 7
 Q.1. Let XYZ be a three-digit number, where (X + Y + Z) is not a multiple of 3.
What is XYZ + YZX + ZXY not divisible by?
a) 3
b) 9
c) 37
d) (X+Y+Z)
 Value of 567=(100*5+10*6+7)
 Similarly:
 Value of XYZ=(100X+10Y+Z)
 YZX=(100Y+10Z+X)
 ZXY=(100Z+10X+Y)

 XYZ+YZX+ZXY=111(X+Y+Z)
 Or just put values in XYZ
 Let X=1, Y=2, Z=3
 XYZ=123 , and X+Y+Z=6
 but (X+Y+Z) is not multiple of 3
 Then Let XYZ=124

 Solving:
 XYZ+YZX+XZY= 777
 Q.2. Number 136 is added to 5B7 and the sum obtained is 7A3, where A
and B are integers. It is given that 7A 3 is exactly divisible by 3. What is the
only possible value of B?
a) 2
b) 5
c) 7
d) 8
 Since 7A3 is divisible by 3, the possible options for A are 2, 5 and 8

136
+ 5B7
7A3

11
136
+ 5B7
7A3

4+B = A
Since there is a carry for next digit, B must be greater than 5
 Q.3. The number of times the digit 5 will appear while writing the integers
from 1000 is
a) 269
b) 271
c) 300
d) 302
 At unit’s place:
5-95: 10 times in 0-100
At unit place 5 will come in each subsequent 100s, i.e 100-200, 200-300,….
No of 5 at unit place : 10*10 = 100
 At ten’s place:
50-59 : 10 times
As above: 10*10= 100
 At hundred’s place:
500 to 599 : 100 times

 Adding each: 100+100+100 = 300

 Q. 4. A printer numbers the pages of a book starting with 1 and uses 3089
digits in all. How many pages does the book have?
a) 1040
b) 1048
c) 1049
d) 1050
Single digit pages (from 1 to 9) = 9 digits
Double digit pages (from 10 to 99) = 2 × 90 = 180 digits
Three digit pages (from 100 to 999) = 3 × 900 = 2700 digits
Total digits used = 9 + 180 + 2700 = 2889
Digits left = 3089 - 2889 = 200
Number of pages with four digits = 200/4 = 50
Required number of total pages = 999 + 50 = 1049
 Q. 5. Consider all 3-digit numbers (without repetition of digits) obtained
using three non-zero digits
which are multiples of 3. Let S be their sum.
Which of the following is/are correct?
1. S is always divisible by 74.
2. S is always divisible by 9.
Select the correct answer using the code given below:

a) 1 only
b) 2 only
c) Both 1 and 2
d) Neither 1 nor 2
 Three non-zero digits which are multiples of 3 are: 3, 6, and 9.

 Using these 3 digits, we can make 6 three-digit numbers.

 369, 396
 639, 693
 963, 936

 So their sum, S = 369 + 396 + 639 + 693 + 936 + 963 = 3996

 Which is divisible by both 74 and 9

Q. In an objective type test of 90 questions, 5 marks are allotted for every
correct answer and 2
marks are deducted for every wrong answer. After attempting all the 90
questions, a student
got a total of 387 marks. What is the number of incorrect responses ?

a) 9
b) 13
c) 27
d) 43
 Total 90 questions
 Let X be no of correct answers
 Hence wrong answers are= (90-X)

 Marks for correct answers = 5*X

 Marks deducted for wrong answers = 2*(90-X)

 5X-2(90-X) =387
 5X-180-2X =387
 7X=567
 X= 81 = correct answers
 Wrong answers= 90-81=9
 Q. How many zeroes are there at the end of the following product?
1 x 5 x 10 x 15 x 20 x 25 x 30 X 35 x 40 x 45 x 50 x 55 x 60

a) 10
b) 12
c) 14
d) 15
 Zeros at the end means multiple of 10, i.e. (2 × 5).
So, the number of 2s or the number of 5s (whichever is lesser) will decide
the number of zeros at the end of the expression.
 Now, 1 × 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50 × 55 × 60
Now arrange the numbers in their prime number format
= 1 × 5 × (5 × 2) × (5 × 3) × (5 × 22) × (5 × 5) × (5 × 3 × 2) × (5 × 7) × (5 × 23) × (5 ×
9) × (5 × 5 × 2) × (5 × 11) × (5× 22 × 3)
Here, number of 2s = 10
And number of 5s = 14
The lesser of the two will determine the number of zeros.
Hence, there will be 10 zeros at the end in the given expression.
 Alternatively;
 1 × 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50 × 55 × 60

 Divide the numbers which are directly divisible by 10

 1 × 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50 × 55 × 60

 1 × 5 × 1 × 15 × 2 × 25 × 3 × 35 × 4 × 45 × 5 × 55 × 6 × (10 6 )

 1 × 5 × 1 × 15 × 2 × 25 × 3 × 35 × 4 × 45 × 5 × 55 × 6 × (10 6 )

 1 × 10× 1 × 15 × 100 × 3 × 35 × 45 × 30 × 55 × (10 6 )

 1 × 1 × 15 × 3 × 35 × 45 × 3 × 55 × (10 10 )
 Q. How many five-digit prime numbers can be obtained by using all the
digits 1, 2, 3, 4 and 5 without repetition of digits?
a) Zero
b) One
c) Nine
d) Ten
 We have to form prime number.

 Hence there cannot be 2, 4 and 5 at the units place

 As then it would be divisible by 2/4 and 5… by applying test of divisibility

 So only 1 or 3 can be at units place.

 So we can for numbers such as 24513, 24531, 45213, 45231, and so on


 Now we have to check for divisibility by 3, 7, 11 etc

 As the sum of 1, 2, 3, 4, 5 = 15 which is divisible by 3

 No Prime number can be formed
 LCM and HCF
 LCM: Least number which is exactly divisible by each number
 Eg. For 2 and 3 LCM is 6
 HCF: Highest common factor

 How to derive LCM:

 Divide the numbers into its factors of prime numbers.

 For 15 and 20
 15= 3*5
 20= 4*5

 HCF = 5
 LCM=5(Common factor)*3*4(Remaining Factors)= 60
 For 60 and 72

 60= 2*2*3*5
 72=2*2*2*3*3

 HCF = 2*2*3=12
 LCM= 2*2*3 (Common factors in both: taken only once) * 5*2*3 (Remaining
Factors)
= 360
 Q.1. In a school every student is assigned a unique identification number. A
student is a football player if and only if the identification number is divisible
by 4, whereas a student is a cricketer if and only if the identification number
is divisible by 6. If every number from 1 to 100 is assigned to a student, then
how many of them play cricket as well as football?
a) 4
b) 8
c) 10
d) 12
If a student plays cricket as well as football, then identification number must
be divisible by both 4 and 6.

LCM of (4, 6) = 12

There are 8 multiples of 12 between 1 and 100.

Basic Geometry
 Pythagoras Theorem
Find X and Y
Average

𝑆𝑢𝑚 𝑜𝑓 𝑎𝑙𝑙 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑋1+𝑋2+𝑋3+...+𝑋𝑛

 Average = =
𝑇𝑜𝑡𝑎𝑙 𝑁𝑢𝑚𝑏𝑒𝑟𝑠 𝑛

21+22+23+24+25 115
 Eg. Avg of 21,22,23,24,25 = 5
= 5
= 23
 Q.1. There are two sections A and B of a class consisting
of 36 and 44 students respectively. If the average weight of
section A is 40 kg and that of section B is 35 kg, find the average weight of
the whole class?

 Total weight of 80(36+44) students

=36×40+44×35kg
=1440+1540=2980 kg.

∴ Avg weight of the total class=2980/80kg =37.25 kg

 Q.2. The average of 25 results is 18. The average of first twelve of them is 14
and of last twelve is 17. Find the thirteenth result.

Average of 25 = 18
Sum of 25 numbers = 25×18=450
Average of 1st 12 no.= 14
Sum of 12 numbers =14×12=168
Average of last 12 no.= 17
Sum of last twelve numbers =17×12=204

 Sum of 25 numbers = Sum of 1st 12 no. + 13th no. + Sum of last 12 no.
 450=168 + 13th no. + 204
 13th no. = 450-168-204 = 78
 Q.3.The average weight of A, B and C is 45 kg. If the average weight of A
and B be 40 kg and that of B and C be 43 kg, then the weight of B is :

 Total weight of A, B, C = 45×3=135 kg

 Total weight of A and B = 40×2=80 kg
 Total weight of B and C = 43×2=86 kg

 For Weight of B
B=(A+B)+(B+C)-(A+B+C)
= 80+86−135=31 kg
 Q. The average of six observations is 12. The average decreases by 1 when
a new observation is included. What is the seventh observation?

a) 1
b) 3
c) 5
d) 6
 Avg of Six obs = 12
 Sum = 12* 6 = 72
 After adding 7th obs Avg decreases by 1

 Hence, new sum

 Sum = 11*7 = 77

 Hence 7th obs = 77-72 = 5

Q. The average salary of the entire staff in an office is ? 200 per day. The
average salary of officers is ? 550 and that of non-officers is ? 120. If the
number of officers is 16, then find the numbers of non-officers in the office.

a) 60
b) 70
c) 80
d) 90
 Let number of non-officers = x
 Then, 120x + 550 x 16 = 200 (16 + x)
 12x + 55 x 16 = 20 (16 + x)
 3x + 55 x 4 = 5 (16 + x)
 3x + 220 = 80 + 5x
 140 = 2x
 X = 70
 Q.4. The average marks of 100 students are given to be 40. It was found
later that marks of one student were 53 which were misread as 83. The
corrected mean marks are
a) 39
b) 39.7
c) 40
d) 40.3
 Total marks = Average × Number of students
 Total marks = 40 × 100 = 4000
 Corrected total marks = 4000 - 83 + 53 = 3970
 New average = 3970/100 = 39.7
 Q. A club level soccer goalkeeper in his first 246 matches has average
goals saved score of 4. After 266th match his average goals saved score
increases by 3. What was his average goals saved score in his last 20
matches?
a) 43.9
b) 46
c) 46.5
d) 48.2
 Average goals saved score after 246 matches = 4
 Total goals saved score after 246 matches = 246 × 4 = 984
 Average goals saved score after 266 matches = 7
 Total goals saved score after 266 matches = 266 × 7 = 1862
 So, total goals saved in the last 20 matches = 1862 – 984 = 878
 Average goals saved in the last 20 matches = 878/20 = 43.9
 Q. 5. A family has two children along with their parents. The average of the
weights of the children and their mother is 50 kg. The average of the
weights of the children and their father is 52 kg. If the weight of the father is
60 kg, then what is the weight of the mother?
a) 48 kg
b) 50 kg
c) 52 kg
d) 54 kg
 Consider combined weight of children be W kg and weight of mother be M
kg.
 Average weight of children and father = 52
 (W + 60)/3 = 52
 W = 156 - 60 = 96
 Average weight of children and mother = 50
 (W + M)/3 = 50
 M = 150 - 96 = 54.
 Q.6. The average monthly income of P and Q is Rs.5050. The average
monthly income of Q and R is Rs.6250 and the average monthly income of
P and R is Rs.5200. The monthly income of P is
a) 3500
b) 4000
c) 4050
d) 5000
 P+Q=(5050×2)=10100....(i)
Q+R=(6250×2)=12500.....(ii)
P+R=(5200×2)=10400....(iii)
 Add (i) and (iii) and subtract (ii)
(P+Q)+ (P+R) –(Q+R) = P+Q+P+R-Q-R= 2P
10100+10400-12500= 2P
8000= 2P
P= 4000

 Alternatively
Adding (i), (iii) and (iii), we get: 2(P+Q+R)=33000 or P+Q+R=16500...(iv)
Subtracting (ii) from (iv), we get P=4000.
∴ P's monthly income =Rs.4000.
 Q.7. There are two Classes A and B having 25 and 30 students respectively.
In Class-A the highest
score is 21 and lowest score is 17. In Class-B the highest score is 30 and
lowest score is 22.
Four students are shifted from Class-A to Class-B.
Consider the following statements:
1. The average score of Class-B will definitely decrease.
2. The average score of Class-A will definitely increase.
Which of the above statements is/are correct?

a) 1 only
b) 2 only
c) Both 1 and 2
d) Neither 1 nor 2
 A B

25 30

H: 21 H: 30

L: 17 L: 22

Since 4 students are moved from A to B, therefore average of class B will

definitely decrease as range
of marks for class A is less than range of marks for class B. That is, the new
students must be having
less marks than the previous minimum marks of Class B.
 Q. A student appeared in 6 papers. The maximum marks are the same for
each paper. His marks in these papers are in the proportion 5 : 6 : 7 : 8 : 9 :
10. Overall he scored 60%. In how many number of papers did he score less
than 60% of the maximum marks?

a) 2
b) 3
c) 4
d) 5
 Average = 60%
5x+6x+7x+8x+9x+10x
 Average of 6 papers = = 60
6
 45x=360

 X=8

 Actual Marks are:

 45; 48; 56; 64; 72; 80
 In 3 papers marks are below 60
Time and Distance
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
 Speed = 𝑇𝑖𝑚𝑒

i.e. v=d/t
i.e. d=v*t
i.e. t=d/v

5
 X km/hr = X ∗ m/sec
18

18
 X m/sec = X* 𝑘𝑚/ℎ𝑟
5
 Q.1 A cyclist covers a distance of 750 m in 2 min 30 sec. What is the speed
in km/hr the cyclist ?

a) 5
b) 10
c) 15
d) 18
Distance covered by cyclist =750 m

Time taken by the cyclist =2m30sec=150sec

∴ Speed = time/distance
= 750/150 m/s
=5 m/s
=5*18/5 km/hr
=18 km/hr
 Q.2. A boy rides his bicycle 10 km at an average speed of 12 km/ hr and again
travels 12 km at an average speed of 10 km/hr. His average speed for the entire
trip is approximately:

A) 10.4 km/hr

B) 10.8 km/hr

C) 11.0 km/hr

D) 12.2 km/hr
 𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
 Avg speed = 𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒

 Total Distance= 10+12 =22 km

10 12 10∗10+12∗12 244
 Total Time=12 + 10 = =120
12∗10

22
 Avg Speed= 244/120 = 10.8 km/hr
 Q. 3. A car travels the first one third of a certain distance with a speed of 10
km/hr, the next one third with a speed of 20 km/hr and the last one third
distance with a speed of 60 km/hr The average speed of the car for the
whole journey is

a) 18 km/hr
b) 24 km/hr
c) 30 km/hr
d) 36 km/hr
 Let the total distance travelled be x km
 Then total time taken =
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑥/3 𝑥/3 𝑥/3
Time= = + +
𝑆𝑝𝑒𝑒𝑑 10 20 60
𝑥 𝑥 𝑥
= 30 + 60 + 180
6𝑥+3𝑥+𝑥
= 180
10𝑥
= 180
𝑥
= 18

𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑥
Speed = = 𝑥/18 = 18 𝑘𝑚/ℎ𝑟
𝑇𝑖𝑚𝑒
 Alternatively:
 Max speed is 60km/hr
 Let us consider it travels 1/3rd distance to be of 60km:
 Hence,
for 60km/hr = 1 hr time
For 20km/hr = 3 hr time
For 10km/hr = 6 hr time

As 1/3rd distance is 60km, total is 60*3=180km

𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 180
Speed = = = 18 𝑘𝑚/ℎ𝑟
𝑇𝑖𝑚𝑒 10
 Q.5. If a train runs at 40 km/hr, it reaches its destination late by 11 minutes,
but if it runs at 50 km/hr, it is late by 5 minutes only. The correct time for the
train to complete the journey is

a) 13 min
b) 15 min
c) 19 min
d) 21 min
 Let the correct time be ‘t’ min.
 Let ‘d’ be the distance

 Distance = Speed* time

 d= 40*(t+11)
 d= 50*(t+5)
 Since Distance for both is same

 40*(t+11)= 50*(t+5)
 4t+44=5t+25
 T=44-25= 19 min
 Q. A person X from a place A and another person Y from a place B set out
at the same time to walk towards each other. The places are separated by
a distance of 15 km. X walks with a uniform speed of 1.5 km/hr and Y walks
with a uniform speed of 1 km/hr in the first hour, with a uniform speed of
1.25 km/hr in the second hour and with a uniform speed of 1.5 km/hr in the
third hour and so on.

Which of the following is/are correct?

1. They take 5 hours to meet.
2. They meet midway between A and B.

Select the correct answer using the code given below:

a) 1 only
b) 2 only
c) Both
d) Neither
 Distance between places A and B = 15 km
Speed of X = 1.5 km/hr
So, Distance covered by X in 5 hours = 7.5 km
Speed of Y in 1st hour = 1 km/hr
So, Distance covered by Y in 1st hour = 1 km
Similarly, Distance covered by Y in 2nd hour = 1.25 km
Distance covered by Y in 3rd hour = 1.5 km
Distance covered by Y in 4th hour = 1.75 km
Distance covered by Y in 5th hour = 2 km
So, the total distance covered by Y in 5 hours = 1 + 1.25 + 1.5 + 1.75 + 2 = 7.5
km
Both the given statements are true
Percentage

 Representation:
𝑋
 X%= 100

 Representation of fraction in percentage

𝑎 𝑎
 = 𝑏 ∗ 100
𝑏
3 3
 eg. 4 = 4 ∗ 100= 75%
5 5 1 1
= ∗ 100= 25% ; = ∗ 100= 16.66%
20 20 6 6
 Decimal and percentage conversion
 A) Percentage to decimal or fraction
15 3 17
Eg. 15% = 0.15 = = ; 85% = 0.85 =
100 20 20

B) 20 % of X = 0.2*X
Eg. 20 % of 200 = 0.2*200 = 400
4% of 200 = 0.04*200 = 8

B) 30% increase in X:: Final value = (X+0.3X) = 1.3 * X

Eg. 30% increase in 50 :: Final Value= 1.3 * 50 = 65

C) 15% decrease in X:: Final value = 0.85 *X

Eg 15% decrease in 100:: Final value = 0.85 *100 = 85
 Q. 1. When a runner was crossing the 12 km mark, she was informed that
she had completed only 80% of the race. How many kilometers was the
runner supposed to run in this event?
a) 14
b) 15
c) 16
d) 16.5
 Let the total length of event be x km.
 80% of x = 12
 80*X/100 = 12
 80X = 1200
 X = 15
 Q.2. Raman's salary was decreased by 50% and subsequently increased
by 50%. How much percent does he lost?

a) 20
b) 35
c) 25
d) 15
 Consider Original salary = 100 Rs.
 After 50% decrease, salary = 100-50 =50 Rs.

50
 Now 50% increase in salary = 50+ 50*(100) = 75 Rs.

𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙−𝑁𝑒𝑤 100−75
 % Decrease in salary = = 100 = 25%
𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙
 Q. If income of Ravi is 20% more than that of Ram, then income of Ram is
how much percent less than that of Ravi?
 Let Income of Ram = 100 Rs
 Hence income of Ravi = 120 Rs.

 Income of Ravi compared to Ram is:

120−100
 = * 100
120
20
 = * 100
120
 = 16.66%
 Q. A student has to score 30% marks to get through. If he gets 30 marks and
fails by 30 marks, then find the maximum marks set for the examination.

a) 100
b) 150
c) 200
d) 250
 Let maximum marks be x
 Marks req for passing = 30% of x

 Students scores 30 marks and fails by 30 marks

 Hence total passing marks = 30+30 = 60

 60 = 30% x
30
 60 = *x
100

 X= 200
 Q.3. Raju has Rs. 9000 with him and he wants to buy a mobile handset; but
he finds that he has only 75% of the amount required to buy the handset.
Therefore, he borrows Rs. 2000 from a friend. Then
a) Raju still does not have enough amount to buy the handset.
b) Raju has exactly the same amount as required to buy the handset.
c) Raju has enough amount to buy the handset and he will have Rs. 500 with
him after buying the handset.
d) Raju has enough amount to buy the handset and he will have Rs. 1000 with
him after buying the handset.
 Let the price of handset be Rs. x
 75% of x = 9000
75
 ( )*X = 9000
100
3
 (4)*X=9000
4
 X= 9000*(3)

 X = Rs. 12000
 After borrowing Rs.2000 from a friend,
 Raju has Rs.9000 + Rs.2000 = Rs.11000
 But, he needs Rs. 12000 to buy mobile handset.
 Q.4. In an examination, A has scored 20 marks more than B. If B has scored
5% less marks than A, how much has B scored?
a) 360
b) 380
c) 400
d) 420
 Marks of A=B+20

 B scored 5% less than A

𝐴−𝐵 5
𝐴
= 100= 0.05

 Now inserting value of A in equation,

𝐴−(𝐴−20)
= 0.05
𝐴

20
= 0.05
𝐴

A= 20/0.05 = 400

B=A-20 = 380
 Q. In a class, 60% of students are from India and 50% of the students are
girls. If 30% of the Indian students are girls, then what percentage of foreign
students are boys?

a) 45%
b) 40%
c) 30%
d) 20%
 Consider total 100 students.
 Indian: 60; Hence, Foreign: 40
 Girls: 50; Hence Boys:50

 30% of Indian students are Girls = 60*30/100 = 18

 Remaining Indian Students/Boys = 42

 Total Boys = 50
 Foreign Boys = Total – Indian Boys
= 50-42 = 8

 % of Foreign students who are boys = 8/40 = 20%

 Q. If the price of an article is decreased by 20% and then the new price is
increased by 25%, then what is the net change in the price?

a) 0%
b) 5% increase
c) 5% decrease
d) Cannot be determined due to insufficient data
 Consider Original Price to be 100 Rs

 Hence: 20% decrease= 80 Rs

25
 Then increased by 25% = 80* = 20
100

 Hence Final Price = 20+80= 100 = original price

 Hence 0% increase.
 Q. 1. If the numerator of a fraction be increased by 15% and its
denominator be diminished by 8% , the value of the fraction is 15/16. Find
the original fraction?
 Let the original numerator be x and denominator be y

 New Numerator = 15% increase = 115% of x = 1.15*x = 15

 New Denominator = 8% decrease= 92% of y= 0.92y =16

 Now x= 15/1.15
 And y= 16/0.92

𝑥 15/1.15
 Then 𝑦= 16/0.92= 3/4
 Q.2. At an election involving two candidates, 68 votes were declared
invalid. The winning candidate secures 52% and wins by 98 votes. The total
number of votes polled are :
a) 2382
b) 2450
c) 2518
d) None
 Winning candidate: 52% vote
 Losing candidate: 48% vote
 Difference= 4%
 He wins by 98 votes

 Then, 4%=98
 Hence, total valid votes= 100% votes=
 4%*25=98*25
 100%=2450

 Total vote cast= valid + invalid

 =2450+68
 2518
 Q.3. One type of liquid contains 30% of water and the second type of liquid
contains 45% of water. A glass is filled with 10 parts of first liquid and 5 parts
of second liquid. The percentage of water in the new mixture in the glass is:

a) 15
b) 20
c) 35
d) 40
 1st type of liquid: 30% water
 2nd type of liquid: 45% water

 10 parts of 1st liquid and 5 parts of 2nd liquid= total 15 parts

10 5
 % of water= *30 + ∗45 = 20+15=35%
15 15
 #speed + percentage
 Q. If the speed is increased by 20% then by what percent time should be
reduced to cover the same distance?

a) 25%
b) 20%
c) 33.33%
d) 16.66%
 Speed = Distance/time
 Distance is constant

 Distant = Speed * time

 d = v*t…………… original
 d = 1.2v* T………..new
 1.2v*T = v*t
 T = t/1.2
 T= 0.8333t
 Reduction in time = t-0.8333t = 0.16667 = 16.67%
 #Geometry + percentage

 Q. If one of the sides of a rectangle is increased by 20% and the other is

increased by 5%, find the percent value by which the area changes?

a. 25 % increase
b. 26 % increase
c. 20 % increase
d. None of these
 Area of rectangle = L * B
 Length increased by 20% i.e. new length = 1.2 L
 Breadth increased by 5% i.e. new breadth = 1.05 L

 New area = 1.2L * 1.05B = 1.26 L*B

 Change in area = 1.26 – 1 = 0.26 = 26%

Ratio and Proportion

 Q.1. The ratio of a two-digit natural number to a number formed by

reversing its digits is 4:7. The number of such pairs is
a) 5
b) 4
c) 3
d) 2
 Let the digit at unit's place be a.
 Let the digit at ten's place be b.
 Original number = 10b + a
 Number formed on reversing the digits = 10a + b
 According to the question,
(10b + a) 4
 =
(10a + b) 7
 70b + 7a = 40a + 4b
 2b = a
 When b = 1, a = 2; numbers are 12 and 21.
 When b = 2, a = 4; numbers are 24 and 42.
 When b = 3, a = 6; numbers are 36 and 63.
 When b = 4, a = 8; numbers are 48 and 84.
 When b = 5, number become three-digit number.
 Alternate method:

 Ratio of double digit number is 4/7

 So multiply the ratio by a natural number:

Multiplied by
1: 4/7
2: 8/14
3: 12/21
4: 16/28
5: 20/35
6: 24/42
and so on..
 Q. Salaries of Akash, Bablu and Chintu are in the ratio of 2 : 3 : 5. If their
salaries were increased by 15%, 10% and 20% respectively, then what will
be the new ratio of their salaries?

a) 3 : 3 : 10
b) 23 : 33 : 60
c) 20 : 22 : 40
d) None of these
 Ratio of salaries :
 2: 3: 5
 Increase in salaries: 15%, 10% and 20%

 New salaries:
 15% increase for 2 = 1.15*2 = 2.3
 10% increase for 3 = 1.1*3 = 3.3
 20% increase for 5 = 1.2*5 = 6

 New ratio = 2.3: 3.3: 6 = 23: 33 : 60

 Q. A biology class at high school predicted that a local population of
animals will double in size every 12 years. The population at the beginning
of the year 2021 was estimated to be 50 animals. If P represents the
population after n years, then which one of the following equations
represents the model of the class for the population?

a) P = 12 + 50n
b) P = 50 + 12n
c) P = 50 (2)12n
d) P = 50 (2)n/12
 Year No. of years elapsed (n) Population (P)
2021 0 50
2033 12 100
2045 24 200
2057 36 400

 Let us put n= 12
 For this
 1) P = 12+50(12) ≠ 100
 2) P= 50+12(12) ) ≠ 100
 3) P= 50 (2)12n ≠ 100
 4)P = 50 (2)n/12 = 50*2=100
 Q. A pie diagram shows the percentage distribution of proteins, water and
other dry elements in the human body. Given that proteins correspond to
16% and water corresponds to 70%. If both proteins and the other dry
elements correspond to p%, then what is the central angle of the
sector representing p on the pie diagram?

a) 54°
b) 96°
c) 108°
d) 120°
 Protein: 16%; Water: 70%
 Hence Dry elements = 100-(16+70) = 14%
 Protein + Dry elements = 14+16 = 30%

 Since represented in Pie chart , i.e. in 360°

30
 Corresponding Angle = 360* = 108°
100
 Q. If the numerator and denominator of a proper fraction are increased by
the same positive quantity which is greater than zero, the resulting fraction
is
a) always less than the original fraction
b) always greater than the original fraction
c) always equal to the original fraction
d) such that nothing can be claimed definitely
 Let us take any simple fraction:

1
 = 0.5
2

 Add any number to both N and D

1+1 2
 = 3 = 0.66
2+1

1+2 3
 = = 0.75
2+2 4
 Q.3. A vessel full of water weighs 40 kg. If it is one third filled, its weight
becomes 20kg. What is the weight of the empty vessel?
a) 10 kg
b) 15 kg
c) 20 kg
d) 25 kg
 Let x = wt of vessel
 And y = wt of completely filled water

 Now,
 X+y = 40 …..(1)
𝑦
 X+ 3 = 20 …..(2)

 Subtract 2nd from 1st

2𝑦
 3
= 20

 Y= 30
 X= 10kg
 Q. A man completes 7/8 of a job in 21 days. How many more days will it
take him to finish the job if quantum of work is further increased by 50%

a) 24
b) 21
c) 18
d) 15
 7
 For work = 21 days
8

8
 For Complete work = work = 1 work
8
7
= 21
8
7 8 8
* = 21*
8 7 7
1 work= 24 days
For complete work 24 days are required
1 1
For remaining work= *24 = 3 days req
8 8
50% work is increased ; Hence 0.5 * 24 = 12 days more
1
Total days = Remaining work + 50% additional work
8
= 3 + 12 = 15 days
 A bottle contains 20 litres of liquid A. 4 litres of liquid A is taken out of its and
replace by same quantity of liquid B. Again 4 litres of the mixture is taken
out and replaced by same quantity of liquid B. What is the ratio of quantity
of liquid A to that of liquid B in the final mixture?
a) 4 : 1
b) 5 : 1
c) 16 : 9
d) 17 : 8
 1st stage
 Liquid A= 20 lts

 2nd stage
 Liquid A= 16 lts + Liquid B =4 lts

 Now 4 lts of liquid is taken out of the bottle

 Removed liquid =
4
 Liquid A = 16* = 16/5
20
4
 Liquid B = 4* = 4/5
20

 Remaining liquid A= 16 -16/5 = 64/5

 Remaining liquid B= 4 -4/5 = 16/5
 3rd stage

 4 lts liquid B is added

 Now total Liquid B = 4 + 16/5 = 36/5

 Liquid A = 64/5
 Liquid A/B = 64/36 = 16/9
 Q. A vessel A contains two chemicals p and q mixed in the ratio 2:3.
Another vessel B contains these two chemicals in the ratio of 4:7. Both A
and B are filled to the brim and then emptied into a third vessel C. What is
the ratio of chemicals p and q in vessel C?

a) 11:17
b) 17:11
c) 9:13
d) Cannot be determined as data is insufficient
 Insufficient data as no information is given about Volume of either vessels

 Depending on the volume of vessels, the final ratio may change

 Q. The ratio of incomes of Raman and Gagan is 4 : 3 and ratio of their
expenditures is 3 : 2. If each person saves 2500, then find their incomes and
expenditures
 Income = Expenditure + Savings

 Let their expenditure be 3x and 2x,

 Then their income would be
 Income = 3x +2500 and 2x + 2500

4 3x +2500
 =
3 2x + 2500
 8x + 10000 = 9x + 7500
 X = 2500

 Their expenditure = 5000 and 7500

 And income = 7500 and 10000
 Q. A bag contains One rupee, 50p and 25p coins in the ratio 5 : 3 : 2. If the
total amount is Rs.70, find the total number of coins.
a) 70
b) 80
c) 90
d) 100
 Let us suppose that number of coins are
 5x, 3x and 2x, respectively.
 Then total money will be
 5x * 1 + 3x * 0.5 + 2x * 0.25= 70
 5x + 1.5x + 0.5 x = 70

 7x = 70

 x = 10

 So, the total number of coins will be

 = (5x+ 3x + 2x) = 10x = 10 × 10 = 100
 Q. In a town, 80% of the population are adults of which the men and
women are in the ratio of 9 : 7, respectively. If the number of adult women
is 4.2 lakh, what is the total population of the town?

a) 12 lakh
b) 9.6 lakh
c) 9.8 lakh
d) 11.6 lakh
 Men : Women ratio:: 9:7

 Adult women population = 4.2 lakh

 Let Adult men population = x

x 9
 =
4.2 7

9
 X= * 4.2 = 9*0.6 = 5.4 lakh
7

 Adult Men and women population = 4.2+5.4 = 9.6 lakh

 Adult population = 80% of Total population = 0.8* Total population

 Total population = Adult population/0.8 = 9.6/0.8 = 12 lakh

 Q. Mr. Shrimant inherits 2505 gold coins and divides them among his three
sons; Bharat, Parat and Marat; in a certain ratio. Out of the total coins
received by each of them, Bharat sells 30 coins, Parat donates his 30 coins
and Marat losses 25 coins. Now, the ratio of gold coins with them is 46:41
:34, respectively. How many coins did Parat receive from his father?
a) 705
b) 950
c) 800
d) 850
 Total No of coins = 2505
 Divides in certain ratio
 Bharat sells 30 coins, Parat donates his 30 coins and Marat losses 25 coins
 After this ratio of remaining coins is 46:41:34
 Consider the proportion to be 46x, 41x and 34y
 Coins with Bharath before sell = 46x+30
 Coins with Parat before donation = 41x+30
 Coins with Marat before losing= 34x+25

 Total no. of coins given = 46x+30 + 41x+30 + 34x+25

 2505 = 121x+ 85
 2420 = 121x
 X = 20
 Coins with Parat = 41x+30 = 41*20+30 = 850