1
TABLE OF CONTENTS
Divisibility Rules .................................................................................................................................. 3
Progression ........................................................................................................................................... 5
Important Algebraic Formulas ........................................................................................................... 7
Time, Speed and Distance ................................................................................................................... 8
Simple and Compound Interest ........................................................................................................ 10
Profit and Loss ................................................................................................................................... 11
Averages .............................................................................................................................................. 12
Percentage ........................................................................................................................................... 14
Alligation ............................................................................................................................................. 15
LCM and HCF ................................................................................................................................... 17
Permutation and Combination & Factorials ................................................................................... 18
Probability .......................................................................................................................................... 20
Venn Diagram .................................................................................................................................... 21
Roots of Quadratic Equation ............................................................................................................ 22
Important Series Sum ........................................................................................................................ 22
Ratios and Proportions ...................................................................................................................... 24
Surds and Indices ............................................................................................................................... 25
Squares and Cubes ............................................................................................................................. 26
Calendar.............................................................................................................................................. 27
Problems on Age ................................................................................................................................ 29
Clock……………................................................................................................................................ 29
Direction.............................................................................................................................................. 31
Numbers .............................................................................................................................................. 32
Unit Digit............................................................................................................................................. 33
Mean, Median and Mode .................................................................................................................. 34
Geometry ............................................................................................................................................ 35
Time and Work .................................................................................................................................. 38
Trigonometry...................................................................................................................................... 39
Syllogism Diagrams ........................................................................................................................... 41
Mensuration........................................................................................................................................ 43
Data Interpretation ............................................................................................................................ 45
Coding Decoding ................................................................................................................................ 46
ENGLISH
How to solve English comprehension in CSAT exam? ................................................................... 48
2
1
❖
Divisibility Rules
Important Tricks:
•
2 – Last digit is 0, 2, 4, 6 or 8
•
3 – Sum is divisible by 3
•
4 – Last two digits are divisible by 4
•
5 – Last digit is 0 or 5
•
6 – Number is divisible by 2 and 3
•
7 – Using rule of Triplet, if alternating sum is divisible by 7
•
8 - Last three digits are divisible by 8
•
9 – Sum is divisible by 9
•
10 – Unit digit is 0
•
11 – Difference of odd digits and even digits is 0 or divisible by 11
•
12 – Number divisible by 3 and 4
•
13 - Using rule of Triplet, if alternating sum is divisible by 13
RECENTLY ASKED QUESTIONS
1.
An Identity Card has the number ABCDEFG, not necessarily in that order, where each letter
represents a distinct digit (1, 2, 4, 5, 7, 8, 9 only). The number is divisible by 9. After deleting the first
digit from the right, the resulting number is divisible by 6. After deleting two digits from the right of
the original number, the resulting number is divisible by 5. After deleting three digits from the right
of the original number, the resulting number is divisible by 4. After deleting four digits from the
right of the original number, the resulting number is divisible by 3. After deleting five digits from
the right of the original number, the resulting number is divisible by 2. Which of the following is a
possible value for the sum of the middle three digits of the number? (2022)
(a) 8
(b) 9
(c) 11
(d) 12
Ans. (a)
Solution:
Identity card’s number has 7 digits denoted by: ABCDEFG. Each letter represents a distinct digit (1, 2, 4,
5, 7, 8, 9 only.
After deleting 1 digit from the right, the resulting number (ABCDEF) is divisible by 6.
(Divisibility Rule for 6 – Number is divisible by 2 and 3)
It means that, F will be an even number (i.e. 2, 4 or 8).
After removing G, the remaining number is divisible by 3. So the sum of the remaining numbers should be
divisible by 3. Thus the value of G can only be 9.
After deleting 3 digits from the right, the remaining number (ABCD) is divisible by 4 i.e. Last two digits
are divisible by 4 also. It means that, D will be an even number (i.e. 2, 4 or 8).
3
After deleting 5 digits from the right, the remaining number (AB) is divisible by 2. It means that, B will be
an even number (i.e. 2, 4 or 8).
So, F, D and B are even numbers (2, 4 or 8). And, A, C, E, and G are odd numbers (1, 5, 7 or 9).
After deleting 2 digits from the right, the resulting number (ABCDE) is divisible by 5.
It means that, E = 5. So, we just have to find C + D + E = C + D + 5, which must be an even number as C
is odd (1, or 7), and D is even (2, 4, or 8).
By looking at the options, we can see that C must be 1 and D must be 2. So, C + D + E = 1 + 2 + 5 = 8.
Hence, the correct answer is (a).
2.
How many 3-digit natural numbers (without repetition of digits) are there such that each digit is odd
and the number is divisible by 5? (2022)
(a) 8
(b) 12
(c) 16
(d) 24
Ans. (b)
Solution:
A three-digit number having all digits different, and is divisible by 5 hence the last digit is 0 or 5. It is an
odd number thus the last digit will be 5.
The number of ways we can fill the first two digits from amongst 4 distinct digits =first digit in 4 ways ×
second digit in 3 ways = Total 12 ways.
Hence, the correct answer is (b).
4
2
Progression
Arithmetic Progression
𝑆𝑢𝑚 𝑜𝑓 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛𝑠
Arithmetic Mean = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛𝑠
=
𝑥1+𝑥2+⋯…+𝑥𝑛
𝑛
Tn = a + (n-1)d
𝑛
Sn = 2 [2a + (n-1)d]
𝑛
2
= (a + l)
If numbers are consecutive numbers starting from 1 with difference = 1,
Then a = 1 and d = 1
𝑛
Sn = 2 [2 (1) + (n-1) (1)] =
𝑛 (𝑛 + 1)
2
Where a = First term
d = Common difference
Tn = nth term
Sn = Sum of n terms
While solving three unknown terms in an Arithmetic Progression whose sum or product is given should be
assumed as a-d, a, a+d.
While solving four terms in an Arithmetic Progression whose sum or product is given, it should be assumed
as a-3d, a-d, a+d, a+3d.
Geometric Progression
𝑛
Geometric Mean = √𝑥1 ∗ 𝑥2 ∗ 𝑥3 ∗ … . . 𝑥𝑛
Tn = arn-1
𝑎 (𝑟 𝑛 −1)
if r > 1
𝑟−1
𝑎 (1 − 𝑟 𝑛 )
Sn = 1 − 𝑟 if r < 1
𝑎
SInfinity = 1−𝑟
Sn =
Where a = First term
r = Common ratio
Tn = nth term
Sn = Sum of n terms
SInfinity = Sum of infinite terms with decreasing common ratio r
𝑎
While solving three unknown Term in an G.P whose sum or product is given should be assumed as (𝑟 ), a, ar.
Harmonic Progression
Harmonic Mean =
𝑛
1
1
1
1
+ + +⋯..+
𝑥1 𝑥2 𝑥3
𝑥𝑛
1
Tn = 𝑎+ (n-1)d
If A, G and H are respectively the arithmetic, geometric and harmonic means, then
5
A≥G≥H
A * H = G2, i.e. A, G, H are in GP.
RECENTLY ASKED QUESTIONS
1.
Consider the following addition problem : 3P+4P+PP+PP = RQ2; where P, Q and R are different
digits. What is the arithmetic mean of all such possible sums? (2021)
(a) 102
(b) 120
(c) 202
(d) 220
Ans. (c)
Solution:
3P + 4P + PP + PP = RQ2
Or (30 + P) + (40 + P) + (10 x P + P) + (10 x P + P) = (100 x R) + (10 x Q) + 2
Or 24P + 70 = 100R + 10 Q + 2
Or 20P + 70 + 4P = 100R + 10 Q + 2
The unit digit of the resultant is 2.
It will be obtained when 4 is multiplied by P.
So, P must be 3, or 8.
If P = 3, then: 24P + 70 = 24 × 3 + 70 = 72 + 70 = 142
If P = 8, then: 24P + 70 = 24 × 8 + 70 = 192 + 70 = 262
Arithmetic sum of 142 and 262 = (142 + 262)/2 = 202.
Hence, the correct answer is (c).
2.
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? (2021)
(a) P = 12 + 50n
(b) P = 50 + 12n
(c) P = 50 (2)12n
(d) P = 50 (2)n/12
Ans. (c)
Solution:
The population is doubling every 12 years.
So, this is a case of Geometric Progression. So, r = 2
Initial population = 50
Being a case of GP, the population should be in the form of arn-1 or 50(2)t
Thus, option A and B are eliminated since they are not represented in this particular manner.
From C and D (putting a value of n =12 for doubling of population), option C gives the correct answer.
Hence, the correct answer is (c).
❖❖❖
6
3
❖
Important Algebraic Formulas
Formulae:
➢ (a + b)2 = a2 + b2 + 2ab
➢ (a – b)2 = a2 + b2 – 2ab
➢ a2 – b2 = (a + b) (a – b)
➢ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
➢ (a + b)3 = a3 + 3a2 b + 3ab2 + b3
➢ (a−b) 3 = a 3− 3a2b + 3ab2 − b3
➢ a3 + b3 = (a + b) (a 2− ab + b2)
➢ a3−b3 = (a−b) (a2 + ab + b2)
RECENTLY ASKED QUESTION
1.
How many pairs of natural numbers are there such that the difference of whose squares is 63? (2020)
(a) 3
(b) 4
(c) 5
(d) 2
Ans. (a)
Solution:
Let the required natural numbers be a and b As per statement, a2 – b2 = 63
Using formula a2 – b2 = (a + b) (a – b)
(a + b) (a – b) = 63
63 = 9*7
Or 63 = 21*3
Or 63 = 63*1
There will be a total of three possible cases in which product of two numbers is 63
Case 1:
(a + b) = 9 and (a – b) = 7 Then a = 8 and b = 1
Case 2:
(a + b) = 21 and (a – b) = 3
Then a = 12 and b = 9
Case 3:
(a + b) = 63 and (a –b) = 1
Then a = 32 and b = 31
Total three such pair will be there
Hence, the correct answer is (a).
❖❖❖
7
4
Time, Speed and Distance
Formulae:
Speed =
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑇𝑖𝑚𝑒
Average Speed =
𝑇𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑
𝑇𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
Relative Speed:
𝑇𝑖𝑚𝑒 =
Time =
𝑆𝑢𝑚 𝑜𝑓 𝑙𝑒𝑛𝑔𝑡ℎ𝑠
𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑆𝑝𝑒𝑒𝑑
𝐿1 + 𝐿2
𝑆1±𝑆2
Note: Speeds to be added if two objects are travelling towards each other and subtracted if going away
from each other.
Problems on Trains:
Time =
𝑺𝒖𝒎 𝒐𝒇 𝒍𝒆𝒏𝒈𝒕𝒉𝒔
𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝑺𝒑𝒆𝒆𝒅
=
𝑳𝟏 + 𝑳𝟐
𝑺𝟏 ± 𝑺𝟐
Problems on Boats:
Speed Downstream = Speed of boat in still water + Speed of Stream
-> Direction of boat
- > Direction of Stream
As both are in same direction, speeds will be added
Speed Upstream = Speed of boat in still water - Speed of Stream
<- Direction of boat
- > Direction of Stream
As both are in opposite direction, speeds will be subtracted
1
2
Speed of the boat in still water = (Downstream speed + Upstream speed)
Conversion from Km/h to m/s -> Multiply by 5/18
An object covers equal distance at speed S1 and other equal distance at speed S2 then his average speed
for the distance is
𝟐(𝑺𝟏)(𝑺𝟐)
𝑺𝟏+𝑺𝟐
8
RECENTLY ASKED QUESTIONS
1.
X and Y run a 3 km race along a circular course of length 300m. Their speeds are in the ratio 3:2. If
they start together in the same direction, how many times would the first one pass the other (the
start-off is not counted as passing)? (2022)
(a) 2
(b) 3
(c) 4
(d) 5
Ans. (b)
Solution:
Formula: Speed =
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑇𝑖𝑚𝑒
As per the question, Speed of X and Y are in ratio 3:2, the faster runner i.e. X will cross the slower one i.e.
Y after covering extra 300 m.
Assume that their speeds are 3 m/sec and 2 m/sec.
So, their relative speed = 3 – 2 = 1 m/sec.
So, the time taken by the X to cross Y = Distance/Relative Speed = 300/1 = 300 seconds. Thus X will cross
Y every 300 seconds.
Now, the time taken for the X to complete the race = Total Distance/Speed = 3000/3 = 1000 seconds.
Therefore, the Xr will cross Y three times during the entire race i.e. after 300 seconds, 600 seconds, and
900 seconds.
Hence, the correct answer is (b).
2.
On one side of a 1.01 km long road, 101 plants are planted at equal distance from each other. What
is the total distance between 5 consecutive plants? (2022)
(a) 40 m
(b) 40.4 m
(c) 50m
(d) 50.5 m
Ans. (b)
Solution:
Length of the road = 1.01 km = 1010 m. 101 plants are planted at equal distance from each other.
So, there will be 100 gaps between those plants and each gap = 1010/100 = 10.1 m in length.
Now, there must be 4 gaps between 5 consecutive plants.
So, required distance = 4 × 10.1 = 40.4 m
Hence, the correct answer is (b).
❖❖❖
9
5
❖
Simple and Compound Interest
Formulae:
➢ Amount = Principal + Interest
➢
Simple Interest =
𝑃𝑟𝑖𝑛𝑐𝑖𝑝𝑎𝑙∗𝑅𝑎𝑡𝑒∗𝑇𝑖𝑚𝑒
100
➢ Compound Interest = P * (1+
✓
Where P = Principal
✓
R = Rate
✓
T = Time
𝑹
𝟏𝟎𝟎
)𝑻 − 𝑷
➢ Doubling of money – Rule of 72 i.e. Rate * Time = 72 (approx.)
➢ Tripling of money – Rule of 114 i.e. Rate * Time = 114 (approx.)
➢ Quadrupling of money – Rule of 144 i.e. Rate * Time = 144 (approx.)
RECENTLY ASKED QUESTION
1.
A person bought a refrigerator worth Rs. 22800 with 12.5% interest compounded yearly. At the end
of first year he paid Rs. 8650 and at the end of second year Rs. 9125. How much will he have to pay
at the end of third year to clear the debt? (2018)
(a) Rs. 9990
(b) Rs. 10000
(c) Rs. 10590
(d) Rs. 11250
Ans. (d)
Solution:
Using the Formula Simple Interest =
1st year’s Interest = 22800 x
12.5
=
100
𝑷𝒓𝒊𝒏𝒄𝒊𝒑𝒂𝒍∗𝑹𝒂𝒕𝒆∗𝑻𝒊𝒎𝒆
𝟏𝟎𝟎
2850
Amount at the end of I Year = 22800 + 2850 = 25650.
Payment made for first year = 25650 - 8650 = 17000
Interest on 17,000: 17000 x
12.5
100
= 2125
Total Amount after Second year: 17000 + 2125 = 19125
Payment made at the end of second year: 9125.
Amount Remaining: 19125-9125= 10000.
Interest at the end of third year: 10000 + 10000 x
12.5
=
100
1250
Total amount to be paid: 10000 + 1250 = 11250.
Hence, the correct answer is (d).
❖❖❖
10
6
❖
Profit and Loss
Formulae:
➢ Profit = Selling Price – Cost Price
➢ Loss = Cost Price – Selling Price
➢ Profit % =
➢
Loss % =
𝑃𝑟𝑜𝑓𝑖𝑡
𝐶𝑜𝑠𝑡 𝑃𝑟𝑖𝑐𝑒
𝐿𝑜𝑠𝑠
𝐶𝑜𝑠𝑡 𝑃𝑟𝑖𝑐𝑒
* 100
* 100
➢ Discount = Marked Price – Selling Price
➢ Discount %
=
𝐷𝑖𝑠𝑐𝑜𝑢𝑛𝑡
* 100
𝑀𝑎𝑟𝑘𝑒𝑑 𝑃𝑟𝑖𝑐𝑒
➢ Successive Discount formula = (x + y -
𝑥𝑦
100
)% Where x and y refer to successive
discounts offered
RECENTLY ASKED QUESTIONS
1.
A person bought a car and sold it for Rs. 3,00,000. If he incurred a loss of 20%, then how much did
he spend to buy the car? (2020)
(a) Rs. 3,60,000
(b) Rs. 3,65,000
(c) Rs. 3,70,000
(d) Rs. 3,75,000
Ans. (d)
Solution:
𝐿𝑜𝑠𝑠
Loss % = 𝐶𝑜𝑠𝑡 𝑃𝑟𝑖𝑐𝑒 x 100
Loss % =
𝐶𝑜𝑠𝑡 𝑃𝑟𝑖𝑐𝑒 − 𝑆𝑒𝑙𝑙𝑖𝑛𝑔 𝑃𝑟𝑖𝑐𝑒
𝐶𝑜𝑠𝑡 𝑃𝑟𝑖𝑐𝑒
x 100
Substituting values as given in question: 20% = 1 -
3,00,000
𝐶𝑃
x 100
Solving the equation, We get CP = 3,75,000.
Hence, the correct answer is (d).
2.
Rakesh had money to buy 8 mobile handsets of a specific company. But the retailer offered a very
good discount on that particular handset. Rakesh could buy 10 mobile handsets with the amount he
had. What was the discount the retailer offered? (2019)
(a) 15 %
(b) 20 %
(c) 25 %
(d) 30 %
Ans. (b)
Solution:
Discount = Marked Price – Selling Price
Marked Price of 10 mobile = 100
Marked Price of 80 mobile = 80
Marked Price of 1 mobile = 10
11
Rakesh bought 10 mobiles for 80.
SP = 80
Discount = 20 %
Hence, the correct answer is (b).
❖❖❖
7
❖
Averages
Formulae:
➢ Average =
𝑆𝑢𝑚 𝑜𝑓 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛𝑠
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛𝑠
➢ Weighted Average =
𝑤1𝑥1+𝑤2𝑥2+⋯……..𝑤𝑛𝑥𝑛
𝑥1+𝑥2+⋯…..+𝑥𝑛
➢ If the value of each unit in a set is increased or decreased by some value x, then
the average of the set also increases or decreases respectively by x.
RECENTLY ASKED QUESTIONS
1.
The average weight of A, B, C is 40 kg, the average weight of B, D, E is 42 kg and the weight of F is
equal to that of B. What is the average weight of A, B, C, D, E and F? (2022)
(a) 40.5 kg
(b) 40.8 kg
(c) 41 kg
(d) Cannot be determined as data is inadequate
Ans. (c)
Solution:
As per the question,
(A + B + C)/3 = 40 Or (A + B + C) = 120
(B + D + E)/3 = 42 Or (B + D + E) = 126
F=B
From the above information:
A + B + C + B + D + E = 120 + 126
Or A + B + C + D + E + B = 246
Or A + B + C + D + E + F = 246 (as F = B)
So, average weight of A + B + C + D + E + F = 246/6 = 41.
Hence, the correct answer is (c).
2.
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. (2021)
Consider the following statements:
1. The average score of Class-B will definitely decrease.
2. The average score of Class-A will definitely increase.
12
Which of the above statements is/are correct?
(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2
Ans. (a)
Solution:
Average = Sum of observations / Number of observations
As 4 students were moved from A to B, the average of class B will definitely decrease because the range
of marks for class A is less than the range of marks for class B.
That is, the new students must be having lesser marks than the previous minimum marks of Class B.
Statement 1 is correct.
However, we cannot be sure regarding the marks of the 4 students that have been moved from class A.
They may have marks close to 21, which will lead to a decrease in the average marks of Class A.
If they have marks close to 17, it will lead to an increase in the average marks of Class A.
Therefore, we cannot determine whether the average of class A will increase or decrease.
Statement 2 is incorrect.
Hence, the correct answer is (a).
❖❖❖
13
8
❖
Percentage
Fraction
Percentage
1/12
8.33%
½
50%
1/13
7.69%
1/3
33.33%
1/14
7.14%
¼
25%
1/15
6.66%
1/5
20%
1/20
5%
1/6
16.66%
1/24
4.16%
1/7
14.28%
1/25
4%
1/8
12.5%
1/30
3.33%
1/9
11.11%
1/40
2.5%
1/10
10%
1/50
2%
1/11
9.09%
1/100
1%
Formulae:
𝐹𝑖𝑛𝑎𝑙 𝑉𝑎𝑙𝑢𝑒−𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑉𝑎𝑙𝑢𝑒
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑉𝑎𝑙𝑢𝑒
➢ Percentage Change =
➢ Successive change of x% and y%
➢ Total change in percentage = ( x + y +
* 100
𝑥𝑦
)%
100
RECENTLY ASKED QUESTIONS
1.
The increase in the price of a certain item was 25%. Then the price decreased by 20% and then again
increased by 10%. What is the resultant increase in the price? (2022)
(a) 5%
(b) 10%
(c) 12.5%
(d) 15%
Ans. (b)
Solution:
Let the initial price be Rs. 100.
After 25% rise, the new price = 100 + 25% of 100 = Rs. 125
After 20% fall, the new price = 125 – 20% of 125 = Rs. 100
After 10% rise, the new price = 100 + 10% of 100 = Rs. 110
So, resultant percentage increase in price = 10%.
Hence, the correct answer is (b).
2.
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? (2021)
(a) 45%
(b) 40%
(c) 30%
(d) 20%
Ans. (d)
Solution:
14
Let the total number of students in the class be 100.
Indian students = 60% of 100 = 60
So, foreign students = 100 – 60 = 40 students
Total number of girls students = 50% of 100 = 50
According to the question, Total number of Indian girl students = 30% of 60 = 18 students
So, foreign girl students = 50 – 18 = 32 As total foreign students = 40 So, foreign boy students = 40 – 32
=8
So, percentage of boys among foreign students = (8/40) × 100 = 20%
Hence, the correct answer is (d).
❖❖❖
9
❖
Alligation
Formulae:
➢ Rule of Allegation:
➢
𝒘𝟏
𝒘𝟐
=
𝒙𝟐−𝒙
𝒙−𝒙𝟏
➢ Cross method:
RECENTLY ASKED QUESTION
1.
There are two containers X and X contains 100 ml of milk and Y contains 100 ml of water. 20 ml of
milk from X is transferred to Y. After mixing well, 20 ml of the mixture in Y is transferred back to
X. If m denotes the proportion of milk in X and n denotes the proportion of water in Y, then which
one of the following is correct? (2022)
(a) m=n
(b) m> n
(c) m<n
(d) Cannot be determined due to insufficient data
Ans. (a)
Solution:
Container X contains 100 ml of milk and container Y contains 100 ml of water.
If 20 ml of milk is transferred from container X to container Y, then:
Amount of milk left in container X = 100 – 20 = 80 ml
Amount of solution in container Y becomes = 100 ml water + 20 ml milk = 120 ml
Ratio of milk and water in container Y = 20 : 100 = 1 : 5
Amount of milk in 20 ml solution of container Y = (1/6) × 20 = 3.33 ml
15
Amount of water in 20 ml solution = 20 – 3.33 = 16.67 ml
If 20 ml of this solution is transferred from container Y to container X, then:
Amount of solution in container X becomes = (80 + 3.33), i.e. 83.33 ml milk + 16.67 ml water
Amount of solution in container Y becomes = (100 – 16.67), i.e. 83.33 ml water + (20 – 3.33), i.e. 16.67
ml milk
As per the question, m denotes the proportion of milk in X, and n denotes the proportion of water in Y. So,
m = 83.33 ml and n = 83.33 ml Thus, m = n
Hence, the correct answer is (a).
1.
There is a milk sample with 50% water in it. If 1/3rd of this milk is added to equal amount of pure
milk, then water in the new mixture will fall down to :
(2017)
(a) 25%
(b) 30%
(c) 35%
(d) 40%
Ans. (a)
Solution:
1
Original 18 liters (9 milk + 9 water), i.e., 18 x 3 = 6 (3 𝑚𝑖𝑙𝑘 + 3 𝑤𝑎𝑡𝑒𝑟)
Pure milk 18 liters (18 milk + 0 water) + 3 milk +3 water
New mixture = 24 liter (21 milk + 3 water) = 25%.
Hence, the correct answer is (a).
❖❖❖
16
10
❖
LCM and HCF
Formulae:
➢ HCF of fraction =
➢ LCM of fraction =
𝐻𝐶𝐹 𝑜𝑓 𝑁𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟
𝐿𝐶𝑀 𝑜𝑓 𝐷𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟
𝐿𝐶𝑀 𝑜𝑓 𝑁𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟
𝐻𝐶𝐹 𝑜𝑓 𝐷𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟
➢ LCM * HCF = Product of two numbers
➢ The least number which when divided by a, b and c leaves a
remainder R in each case. Required number = (LCM of a, b, c) +
R
➢ The greatest number which divides a, b and c to leave the
remainder R is HCF of (a – R), (b – R) and (c – R)
➢ The greatest number which divide x, y, z to leave remainders a, b,
c is HCF of (x – a), (y – b) and (z – c)
RECENTLY ASKED QUESTIONS
1.
𝟏
𝟑
What is the greatest length x such that 3𝟐 m and 8𝟒 m are integral multiples of x? (2020)
1
1
(a)
12m
(b) 13m
(c)
1 m
1
4
(d) 1 m
3
4
Ans. (d)
Solution:
1
7
3
32 = 2and 84=
35
4
x = H.C.F. of (7/2) and (35/4) =
𝐻.𝐶.𝐹.𝑜𝑓 7 𝑎𝑛𝑑 35
𝐿.𝐶.𝑀.𝑜𝑓 2 𝑎𝑛𝑑 4
7
=4
3
Hence, = 7/4 m or 14 m
Hence, the correct answer is option (d).
❖❖❖
17
Permutation and Combination &
Factorials
11
❖
Formulae:
➢ n! = 1 * 2 * 3 * 4 * …….. n
➢ n! = n * (n - 1)!
𝑛!
➢ Permutation nPr = (𝑛−𝑟)!
➢ Combination nCr =
➢
➢
➢
➢
n
𝑛!
𝑟!(𝑛−𝑟)!
n
Cr = Cn-r
n
C0 + nC1 + nC2 + ….. + nCn = 2n
Total number of Handshakes possible among total n people = nC2
Total number of Triangles that can be formed by joining sides of
polygon of n sides = nC3
➢ Total number of diagonals of a polygon of n sides =
➢
𝑛∗(𝑛−3)
2
Total number of circular permutations if clockwise and anti
clockwise are taken as different= (n-1)!
Number
Factorial
0
1
1
1
2
2
3
6
4
24
5
120
6
720
7
5040
8
40320
9
362880
10
3628800
RECENTLY ASKED QUESTIONS
1.
The digits 1 to 9 are arranged in three rows in such a way that each row contains three digits, and
the number formed in the second row is twice the number formed in the first row; and the number
formed in the third row is thrice the number formed in the first row. Repetition of digits is not
allowed. If only three of the four digits 2, 3, 7 and 9 are allowed to be used in the first row, how many
such combinations are possible to be arranged in the three rows? (2022)
18
(a)
(c)
4
2
(b) 3
(d) 1
Ans. (c)
Solution:
We can only use three of the four digits – 2, 3, 7, and 9, in the first row.
The first digit in the first row cannot be 7 or 9, as otherwise thrice the number will not be a three-digit
number.
So, the first digit in the first row can either be 2, or 3.
The possible cases are: 237, 273, 239, 293, 279, 297, 327, 372, 329, 392, 379, or 397.
On eliminating the numbers whose 3x is not a three-digit number, we are left with: 237, 273, 239, 293,
279, 297, 327, and 329.
On Checking the following numbers:
237 × 2 =474 (digit repetition, and so eliminated)
273 × 2 = 546; 273 × 3 = 819
239 × 2 = 478; 239 × 3 = 717 (digit repetition, and so eliminated) 293 × 2 = 586; 293 × 3 = 879 (digit
repetition, and so eliminated) 279 × 2 = 558 (digit repetition, and so eliminated)
297 × 2 = 594 (digit repetition, and so eliminated)
327 × 2 = 654; 327 × 3 = 981
329× 2 = 658; 329× 3 = 987 (digit repetition, and so eliminated) So, only two cases are possible.
Hence, the correct answer is (c).
2.
In a tournament of Chess having 150 entrants, a player is eliminated whenever he loses a match. It
is given that no match results in a tie/draw. How many matches are played in the entire tournament?
(a) 151
(b) 150
(c) 149
(d) 148
Ans. (c)
Solution:
The tournament starts with 150 players.
After the first round (in which 75 matches are held): 75 players are eliminated, and 75 remain.
After the second round (in which 37 matches are held): 37 players are eliminated, and 38 remain.
After the third round (in which 19 matches are held): 19 players are eliminated, and 19 remain.
After the fourth round (in which 9 matches are held): 9 players are eliminated, and 10 remain.
After the fifth round (in which 5 matches are held): 5 players are eliminated, and 5 remain.
After the sixth round (in which 2 matches are held): 2 players are eliminated, and 3 remain.
After the seventh round (in which 1 match is held): 1 player is eliminated, and 2 remain.
After the eighth round (in which 1 match is held): 1 player is eliminated, and 1 remains.
So, total number of matches = 75 + 37 + 19 + 9 + 5 + 2 + 1 + 1 = 149.
Hence, the correct answer is (c).
❖❖❖
19
12
❖
Probability
Formulae:
➢ Random Experiment – An experiment whose result cannot be
predicted e.g. Dice, coin etc
➢ Probability of an event always lies between 0 and 1
➢ P (Not A) = 1 – P(A)
➢ Probability of an event =
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑎𝑣𝑜𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
𝑆𝑎𝑚𝑝𝑙𝑒 𝑠𝑝𝑎𝑐𝑒 𝑜𝑟 𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
➢ Odds in favour of event =
➢
Odds against an event =
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑎𝑣𝑜𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑢𝑛𝑓𝑎𝑣𝑜𝑢𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑢𝑛𝑓𝑎𝑣𝑜𝑢𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑎𝑣𝑜𝑢𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
RECENTLY ASKED QUESTIONS
1.
A bag contains 15 red balls and 20 black balls. Each ball is numbered either 1 or 2 or 3. 20% of the
red balls are numbered 1 and 40% of them are numbered 3. Similarly, among the black balls, 45%
are numbered 2 and 30% are numbered 3. A boy picks a ball at random. He wins if the ball is red
and numbered 3 or if it is black and numbered 1 or 2. What are the chances of his winning? (2018)
(a) 1/ 2
(b) 4/7
(c) 5/9
(d) 12/13
Ans. (b)
Solution:
Total red balls = 15
Total black balls = 20
Red Ball
Probability of picking a random ball is Red = 15/ 35
Probability of picking Red and Number 3 = 15/35*40/100 -Eq 1
(As 40% red balls are Number 3)
Black Ball
Probability of picking a random ball is Black= 20/35
Black and Number 1 = 20/35*25/100 -Eq 2
(As Black Number 1 balls = 100% - Black
Number 2 balls – Black Number 3 balls = 100% - 45% - 30% = 25%)
Black and Number 2 = 20/35*45/100 - Eq 3
(As 45% black balls are Number 2)
Total probability = Eq 1 + Eq 2 + Eq 3 = 4/7
Hence, the correct answer is (b).
❖❖❖
20
13
❖
Venn Diagram
Formulae:
➢ n(A ⋃ B) = n(a) + n (B) – n(A ⋂ B)
➢ n(A ⋃ B ⋃ C) = n(a) + n(B) + n(c) – n(A ⋂ B) – n(B ⋂ C) – n(A ⋂ C) + n(A ⋂ B
⋂ C)
RECENTLY ASKED QUESTION
1.
In a group of 120 persons, 80 are Indians and the rest are foreigners. Furthermore, 70 persons in the
group can speak English. The number of Indians who can speak English is:
(2021)
(a) 20
(b) 30
(c) 30 or less
(d) 30 or more
Ans. (d)
Solution:
Out of 120 persons, 80 are Indians and 40 are foreigners.
Out of 120 persons, 70 can speak English, and the rest cannot.
The maximum possible number of Indians who can speak English is 70, i.e. if all the English-speaking
people are Indians.
The minimum possible number of Indians who can speak English is 30, i.e. if all the foreigners speak
English.
So, English-speaking Indians will fall in the range of 30 to 70.
Hence, the correct answer is (d).
❖❖❖
21
14
❖
Roots of Quadratic Equation
The roots of the quadratic equation ax2 + bx + c = 0 :
−𝒃±√𝒃𝟐 −𝟒𝒂𝒄
𝟐𝒂
PRACTICE QUESTION
The roots of the equation 6x2 – 17x + 12 =0 are:
(a) 5/2 and 3/2
(b) 3/2 and 4/3
(c) 2/3 and 3/2
(d) 5/2 and 3/4
Ans.(b)
Solution:
1.
−𝑏±√𝑏2 −4𝑎𝑐
2𝑎
−(17)±√(−17)2 −4(6)(12)
2(6)
(17)±1
12
Using Formula,
Roots =
Roots =
Roots = 3/2 and 4/3
Hence, the correct answer is (b).
❖❖❖
15
❖
Important Series Sum
Formulae
➢ Sum of first n Natural numbers =
𝑛(𝑛+1)
2
➢ Sum of first n Odd numbers = n2
➢ Sum of first n Even numbers = n (n+1)
➢ Sum of squares of first n Natural numbers =
➢
Sum of cubes of first n Natural numbers = [
22
𝑛(𝑛+1)(2𝑛+1)
6
𝑛(𝑛+1) 2
2
]
RECENTLY ASKED QUESTIONS
1.
What is the value of X in the sequence 20, 10, 10, 15, 30, 75, X? (2022)
(a) 105
(b) 120
(c) 150
(d) 225
Ans. (d)
Solution:
The given series is:
20, 10, 10, 15, 30, 75, X?
The terms are decreasing in the initial half, and then they start increasing. The speed at which they increase
at the latter half suggests that multiplication may be involved.
The pattern is as follows:
20 × 0.5 = 10
10 × 1 = 10
10 × 1.5 = 15
15 × 2 = 30
30 × 2.5 = 75
75 × 3 = 225
Hence, the correct answer is (d).
2.
Replace the incorrect term by the correct term in the given sequence 3, 2, 7, 4, 13, 10, 21, 18, 31, 28,
43, 40 where odd terms and even terms follow the same pattern. (2021)
(a) 0
(b) 1
(c) 3
(d) 6
Ans. (a)
Solution:
On separating the Odd and the Even Series, we observe the following pattern:
ODD SERIES EVEN SERIES
32
3+4=70+4=4
7 + 6 = 13 4 + 6 = 10
13 + 8 = 21 10 + 8 = 18
21 + 10 = 31 18 + 10 = 28
31 + 12 = 43 28 + 12 = 40
On observing the given two series, we observe that the first term in EVEN SERIES should be replaced by
0.
Hence, the correct answer is (a).
❖❖❖
23
16
❖
Ratios and Proportions
Formulae:
𝒂
𝒄
𝒃
𝒅
➢ If = then
𝑏
𝑑
𝑎
𝑎
𝑐
𝑏
𝑐
𝑑
𝑎+𝑏
▪ As per Invertendo law, =
▪ As per Alternendo law, =
▪ As per Componendo law,
▪ As per Dividendo law,
𝑎−𝑏
𝑏
𝑏
=
=
𝑐+𝑑
𝑑
𝑐−𝑑
𝑑
▪ As per Componendo and Dividendo,
➢ For a proper fraction
then
▪
𝑎+𝑐
𝑏+𝑐
𝑎
𝑎−𝑐
𝑏
𝑏−𝑐
> and
𝒂
𝒃
𝑎+𝑏
𝑎−𝑏
=
𝑐+𝑑
𝑐−𝑑
i.e. a <b, then for a positive quantity c,
<
𝒂
𝑎
𝑏
➢ For improper fraction i.e. a >b, then for a positive quantity c,
𝒃
then
▪
𝑎+𝑐
𝑏+𝑐
𝑎
𝑎−𝑐
𝑏
𝑏−𝑐
< and
>
𝑎
𝑏
RECENTLY ASKED QUESTIONS
1.
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 papers
did he score less than 60% of the maximum marks? (2021)
(a) 2
(b) 3
(c) 4
(d) 5
Ans. (b)
Solution:
Let total marks in each subject be 100.
Therefore, total marks for all 6 subjects = 600
Overall marks scored = 60% of 600 = 360
According to the question, 5x + 6x + 7x + 8x + 9x + 10x = 360 or 45x = 360 or x = 8
So, marks in the given 6 subjects must be: 5 × 8 = 40 6 × 8 = 48 7 × 8 = 56 8 × 8 = 64 9 × 8 = 72 10 × 8 =
80
Therefore, in 3 subjects the student has scored less than 60% marks.
Hence, the correct answer is (b).
2.
An amount of money was distributed among A, B and C in the ratio p : q : r.
Consider the following statements :
1. A gets the maximum share if p is greater than (q+r).
24
2.
C gets the minimum share if r is less than (p+q).
Which of the above statements is/are correct ? (2021)
(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2
Ans. (a)
Solution:
Ratio of distribution of money among A, B and C is p : q : r.
Statement 1 is correct: If p > (q + r), then p is definitely the largest number.
So, A must have got the maximum share.
Now
Statement 2 is incorrect: If r < (p + q), then r may or may not be the smallest number. For example, 5 <
(2 + 4) So, C may or may not have got the minimum share.
Hence, the correct answer is (a).
17
Surds and Indices
❖ Formulae:
x0 = 1
xa * xb = xa+b
𝑥𝑎
𝑥
𝑎
a-b
𝑏 = x
x-a =
1
𝑎
𝑎
𝑏
1
𝑥𝑎
𝑏
(𝑥) = √𝑥 𝑎
xaya = (xy)a
√𝑥 = (𝑥)
(xa)b = xab
√𝑎 * √𝑏 = √𝑎𝑏
RECENTLY ASKED QUESTION
Q. What is the largest number among the following? (2020)
(a) ( 1/2 ) -6
(b) ( 1/4 ) −3
(c) ( 1/3 ) −4
(d) ( 1/6 ) −2
Ans: (c)
Solution:
Given numbers can be simplified as follows
Option (a): (½)-6 = 26 = 64
Option (b): (¼)-3 = 43 = 64
Option (c): (⅓)-4 = 34 = 81
Option (d): (⅙)-2 = 62 = 36
We can see that 81 is the largest number among the given four numbers.
Hence, the correct answer is (c).
25
18
Squares and Cubes
Number
Square
1
1
2
4
3
9
4
16
5
25
6
36
7
49
8
64
9
81
10
100
11
121
12
144
13
169
14
196
15
225
16
256
17
289
18
324
19
361
20
400
21
441
22
484
23
529
24
576
25
625
26
676
27
729
28
784
29
841
30
900
26
Number
Cube
1
1
2
8
3
27
4
64
5
125
6
216
7
343
8
512
9
729
10
1000
11
1331
12
1728
13
2197
14
2744
15
3375
RECENTLY ASKED QUESTIONS
1.
If 32019 is divided by 10, then what is the remainder? (2021)
(a) 1
(b) 3
(c) 7
(d) 9
Ans. (c)
Solution:
It’s given that: 32019 is divided by 10.
Now,
31 = 3
32 = 9
33 = 27
34 = 81
35 = 243
36 = 729
Since, the unit place of the power of 3 repeats after every 4 steps (i.e. it has a cyclicity of 4).
Now, on dividing 2019 by 4 we get a remainder of 3.
Hence, 32019 will have the same last digit as that of 33 , i.e. 7. (33 )/10 = 27/10
Hence, the correct answer is (c).
❖❖❖
19
❖
Calendar
Important Points:
➢ Normal Year – 365 days or 52 weeks and 1 day
➢ Leap Year – 366 days or 52 weeks and 2 days
➢ Century Leap Year – If century year is divisible by 400
➢ e.g. 2000 is leap year , but 1900 is not a leap year
➢ Leap Years in 400 year time period - 97
➢ Leap Years in 100 year time period – 24 or 25 depending whether 100 year end is in
century leap year or not
RECENTLY ASKED QUESTIONS
1.
Which date of June 2099 among the following is Sunday? (2022)
(a) 4
(b) 5
(c) 6
(d) 7
Ans. (d)
Solution:
Method I:
It can be inferred that 1st January, 2001 was a Monday.
(Just like 1st January, 1601, or 1st January, 1201, i.e. every 400 years).
In 100 years from 1st January, 2001 to 31st December, 2100, there will be 24 leap years (as 2100 is not a
leap year).
27
So, the number of odd days from 1st January, 2001 to 31st December, 2100 = (24 × 2) + 76 = 48 + 76 =
124 = 5 odd days. (every leap year has 2 odd days, and every non-leap year has 1 odd day)
So, the day on 1st January, 2101 must be Monday + 5 = Saturday.
So, the day on 1st January, 2100 must be Saturday – 1 = Friday (there is 1 odd day in any non-leap year)
So, the day on 1st January, 2099 must be Friday – 1 = Thursday (there is 1 odd day in any non-leap year)
The number of odd days in the year 2099 are: January – 3;
February – 0;
March – 3;
April – 2;
May – 3;
So, the number of odd days from 1st January, 2099 to 31st May, 2099 = 3 + 0 + 3 + 2 + 3 = 11 = 4 So, the
day on 1st June, 2099 must be Thursday + 4 = Monday
So, the first Sunday in the month of June, 2099 will fall on 7 .
Method II:
Considering 5th June 2022 as the reference date (date on which UPSC CSE prelims 2022 was conducted),
which was a Sunday.
Difference between both the years = 2099 – 2022 = 77 years
Number of leap years in between = 77 /4 = 19 (we will ignore the remainder 1) Hence, till 2099 we will
have 58 normal years and 19 leap years.
Number of odd days = (58 × 1) + (19 × 2) = 58 + 38 = 96
96/7 gives 5 as the remainder.
So, 5th June 2099 will be Sunday + 5 = Friday
And, so 7th June 2099 will be a Sunday.
Hence, the correct answer is (d).
2.
How many seconds in total are there in x weeks, x days, x hours. x minutes and X seconds? (2022)
(a) 11580x
(b) 11581x
(c) 694860x
(d) 694861x
Ans. (d)
Solution:
The required number of seconds are calculated as under:
x weeks = 7x days = (7 × 24) x hours = 168x hours = (168 × 60) x minutes = 10080x minutes = (10080 ×
60)x seconds = 604800x seconds
x days = 24x hours = 1440x minutes = 86400x seconds
x hours = 60x minutes = 3600x seconds
x minutes = 60x seconds
So, the total seconds in x weeks, x days, x hours. x minutes and X seconds = 604800x + 86400x + 3600x
+ 60x + x = 694861x seconds.
Hence, the correct answer is (d).
❖❖❖
28
20
Problems on Age
RECENTLY ASKED QUESTIONS
1.
X said to Y, "At the time of your birth I was twice as old as you are at present." If the present age of
X is 42 years, then consider the following statements:
(2021)
1. 8 years ago, the age of X was five times the age of Y.
2. After 14 years, the age of X would be two times the age of Y.
Which of the above statements is/are correct?
(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2
Ans. (b)
Solution:
Present age of X is 42 years.
Let the present age of Y be y years.
As per the question, 42 – y = 2y Or 3y = 42 Or y = 14 years.
So, at present the ages of X and Y are 42 and 14 respectively.
Statement 1 is incorrect: Eight years ago, the ages of X and Y must have been 34 and 6 respectively.
We can see that the age of X was not 5 times the age of Y.
Statement 2 is correct: After fourteen years, the ages of X and Y will be 56 and 28 respectively.
We can see that the age of X would indeed be two times the age of Y.
Hence, the correct answer is (b).
❖❖❖
21
❖
Clock
Important Points:
➢ Degrees covered by Minute hand in 1 min = 6°
➢ Degrees covered by Second hand in 1 second = 6°
1
➢ Degrees covered by Hour hand in 1 min = 2°
➢ Angle between Hour and Minute Hand at a particular
60∗𝐻𝑜𝑢𝑟−11∗𝑀𝑖𝑛𝑢𝑡𝑒
time =
2
29
RECENTLY ASKED QUESTIONS
1.
A man started from home at 14:30 hours and drove to the village, arriving there when the village
clock indicated 15:15 hours. After staying for 25 minutes, he drove back by a different route of length
1.25 times the first route at a rate twice as fast, reaching home at 16:00 hours. As compared to the
clock at home, the village clock is
(2022)
(a) 10 minutes slow
(b) 5 minutes slow
(c) 10 minutes fast
(d) 5 minutes fast
Ans. (d)
Solution:
Total time taken by the man to come back home = 16 – 14.5 = 1.5 hours = 90 minutes
Out of which he stayed in the village for 25 minutes.
So, his total travelling time = 90 – 25 = 65 minutes
The return route was 1.25 times the initial route. So, time taken must have increased by 25% too. So, if the
initial time was 100 units, now it must be 125 units.
But it is also given that while returning he drove twice as fast. So, time taken must have been halved. So,
time taken while returning back = 125/2 = 62.5 units
So, 100 + 62.5 = 65 minutes
Or 162.5 units = 65 minutes
So, 100 units = (65/162.5) × 100 = 40 minutes
So, the man took 40 minutes to reach the village.
So, the actual time at that moment = 14:30 + 40 minutes = 15:10 hoursIt’s pretty evident that the village
clock is 15:15 – 15:10 = 5 minutes fast.
Hence, the correct answer is (d).
2.
Consider the following statements :
1. Between 3:16 p.m. and 3:17 p.m., both hour hand and minute hand coincide.
2. Between 4:58 p.m. and 4:59 p.m.. both minute hand and second hand coincide.
Which of the above statements is/are correct?
(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2
(2022)
Ans. (c)
Solution:
From statement 1:
At 3 o’clock, the minute hand is 15 minutes apart from the hour hand.
To be coincident, it must gain 15 minute spaces.
We know that 55 minutes are gained in 60 minutes.
So, 15 minutes are gained in (60/55) × 15 = 180/11 minutes = 16.36 minutes
Thus, hour hand and minute hand will coincide at 3:16:36, which is between 3:16 pm and 3:17 p.m.
Hence, statement 1 is correct.
From statement 2:
At 4:58 p.m. the second hand is at 12. In the next minute, the second will definitely cross the minute hand.
Thus, between 4:58 p.m. and 4:59 p.m. the minute hand and second hand will definitely coincide.
Hence, statement 2 is correct.
Hence, the correct answer is (c).
❖❖❖
30
22
Direction
RECENTLY ASKED QUESTIONS
1.
Two friends X and Y start running and they run together for 50 m in the same direction and reach
a point. X turns right and runs 60 m, while Y turns left and runs 40m. Then X turns left and runs
50m and stops, while Y turns right and runs 50 m and then stops. How far are the two friends from
each other now?
(2022)
(a) 100 m
(b) 90 m
(c) 60 m
(d) 50 m
Ans. (a)
Solution:
The path taken by them has been represented below:
It’s pretty clear that they are 40 + 60 = 100 m apart at the end of their run. Hence, the correct answer is (a).
❖❖❖
31
23
❖
Numbers
Important Points:
➢ Representation of a Number
✓ ab = 10a + b
✓ abc = 100a + 10b + c
✓ abcd = 1000a + 100b + 10c + d
➢ Natural Number – Counting numbers containing set of positive integers from 1 to
infinity
➢ Whole Number – Natural numbers with 0 adjoined
➢ Prime Number – Natural number greater than 1 having only 1 and itself as factors
e.g. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37 etc
➢ Odd Number – Numbers not divisible by 2 or having ending in 1, 3, 5, 7 or 9
➢ Even Number – Numbers divisible by 2 or having ending in 0, 2, 4, 6 or 8
𝑝
➢ Rational Number – Can be expressed in term of 𝑞
1
✓
Terminating rational e.g. 4 = 0.25
✓
Non Terminating rational e.g.
13
6
= 2.16666
𝑝
➢ Irrational Number – Cannot be expressed in term of 𝑞
✓
✓
✓
➢ Integer – Set of all whole numbers with set of negative natural numbers
Positive Integer - 1, 2, 3, 4….. is the set of positive integers
Negative Integers: −1, −2, −3, -4….. is the set of negative integers
Non-Positive and Non-Negative Integers: 0 is neither positive nor negative
𝑝
➢ Proper Fraction – Numbers in form of 𝑞 where p < q
𝑝
➢ Improper Fraction – Numbers in form of 𝑞 where p > q
➢ Co Prime Numbers – Numbers with HCF 1
➢ Composite Numbers – Numbers having more than 2 factors
➢ Twin Prime – Pair of Prime numbers where difference is of two e.g. (5, 7), (11, 13)
etc
RECENTLY ASKED QUESTIONS
1.
A bill for 1,840 is paid in the denominations of 50, 20 and 10 notes. 50 notes in all are used.
Consider the following statements:
1. 25 notes of 50 are used and the remaining are in the denominations of 20 and 10.
2. 35 notes of 20 are used and the remaining are in the denominations of 50 and 10.
3. 20 notes of 10 are used and the remaining are in the denominations of 50 and 20.
Which of the above statements are not correct?
(a) 1 and 2 only
(b) 2 and 3 only
(c) 1 and 3 only
(d) 1, 2 and 3
(2022)
Ans. (d)
Solution:
32
The given total amount is Rs. 1840, which is made up of Rs. 50, Rs. 20, and Rs. 10 notes in unknown
quantities denoted as a, b, and c, respectively. The statements given about the number of notes used are
then analyzed.
Statement 1, which claims that 25 notes of Rs. 50 were used, is proven to be incorrect since the remaining
amount of Rs. 590 cannot be obtained even if all the remaining notes are of Rs. 20 denomination.
Similarly, Statement 2, which claims that 35 notes of Rs. 20 were used, is also proven to be incorrect since
the remaining amount of Rs. 1140 cannot be obtained even if all the remaining notes are of Rs. 50
denomination.
Finally, Statement 3, which claims that 20 notes of Rs. 10 were used, is also proven to be incorrect since
the remaining amount of Rs. 1640 cannot be obtained even if all the remaining notes are of Rs. 50
denomination.
Hence, the correct answer is (d).
2.
Which number amongst 𝟐𝟒𝟎 ,𝟑𝟐𝟏 , 𝟒𝟏𝟖 and 𝟖𝟏𝟐 is the smallest?
(a) 240
(b) 321
(c) 418
(d) 812
(2022)
Ans. (b)
Solution:
The given numbers are: 𝟐𝟒𝟎 ,𝟑𝟐𝟏 , 𝟒𝟏𝟖 and 𝟖𝟏𝟐
We can also write them as: 𝟐𝟒𝟎 ,𝟑𝟐𝟏 , 𝟐𝟑𝟔 , and𝟐𝟑𝟔 . (as 4 = 22, and 8 = 23)
So, we basically need to find the smallest one among 𝟑𝟐𝟏 , 𝟐𝟑𝟔 . As we cannot have two correct answers, it
must be 𝟑𝟐𝟏 .
Hence, the correct answer is (b).
❖❖❖
24
❖
Unit Digit
Important Points:
➢ Unit digit when a number is raised to some power
✓ 0 – Always 0
✓ 1 – Always 1
✓ 2 – 2,4,8,6
✓ 3 – 3,9,7,1
✓ 4 – 4, 6
✓ 5 – Always 5
✓ 6 – Always 6
✓ 7 – 7,9,3,1
✓ 8 – 8,4,2,6
✓ 9 – 9, 1
33
25
❖
Mean, Median and Mode
Important Points:
𝑆𝑢𝑚 𝑜𝑓 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛𝑠
➢ Mean = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛𝑠
➢
➢
➢
➢
➢
❖
Median = Middlemost observation when data is arranged in ascending order
Mode = Most frequently occurring value in data set
2 Mean + Mode = 3 Median
Range = Maximum value – Minimum value
Symmetric Data - Data sets whose values are evenly spread around centre
Skewed Data – Data sets that are not symmetric
RECENTLY ASKED QUESTIONS
1.
If for a sample data Mean < Median < Mode then the distribution is
(a) Symmetric
(b) Skewed to the right
(c) Neither symmetric nor skewed
(d) Skewed to the left
(2017)
Ans. (d)
Solution:
When Mean < Median < Mode as discussed above, then the distribution is skewed to the left. Hence, the
correct answer is (d).
❖❖❖
34
26
Geometry
❖
Triangle and its properties:
❖
❖
❖
❖
❖
❖
❖
❖
Sum of three angles of triangle = 180°
Three types of triangle based on angle–
○ Acute – All angles less than 90°
○ Right – One angle equal to 90° and P2 + B2 = H2
○ Obtuse – One angle greater than 90°
Three types of triangle based on sides
○ Scalene triangle – no sides are equal
○ Isosceles triangle - two sides are equal
○ Equilateral – all sides are equal
Sum of two sides is always greater than third side
Difference of two sides is always less than third side
Sum of interior angles = 180°
Exterior angle is always equal to sum of two opposite angles
Sum of exterior angles = 360°
Perimeter of triangle = Sum of sides
❖
Semi - perimeter, s = 𝟐 of perimeter
❖
Area of triangle using Heron’s Formula = √(𝒔)(𝒔 − 𝒂)(𝒔 − 𝒃)(𝒔 − 𝒄)
❖
Inradius of triangle = 𝑆𝑒𝑚𝑖−𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟
❖
Circumradius = 4 ∗ 𝐴𝑟𝑒𝑎 , where a, b and c are three sides of triangle
❖
In a right angle triangle, Cirumradius is equal to half of Hypocentre
❖
Area of Triangle= 𝟐 * base * height
❖
Area of Triangle = 𝟐ab sinQ where Q is angle between a and b
❖
𝟏
𝐴𝑟𝑒𝑎
𝑎∗𝑏∗𝑐
𝟏
𝟏
35
𝟑𝒂𝟐
,
𝟒
❖
Are of equilateral triangle = √
❖
Important Pythagorean Triplets:
➢ Numbers that follow P2 + B2 = H2
➢ 3,4,5
➢ 6, 8, 10
➢ 7, 24, 25
➢ 5, 12, 13
➢ 9, 40, 41
➢ 20, 21, 29
❖
Quadrilateral
➢
➢
➢
Trapezium
➢
I.
where a is side of triangle
Total number of sides =4
Sum of interior angles = 360°
5 major types of Quadrilateral
Two sides are parallel and two sides are non
parallel
➢ Area = ½* sum of parallel sides * distance
between parallel sides =½ * (a+b) * h
➢ Perimeter = a + b + c + d
II. Parallelogram
➢ Opposite sides are equal and parallel
➢ Opposite angles are equal
➢ Perimeter= 2(a+b)
➢ Area = ½ * product of diagonals* sinQ
➢ (D1)2 + (D2)2 = 2 (a2+ b2)
✓ Where D1 = Diagonal 1 and D2 = Diagonal 2
✓ a = length of one side
✓ b = length of other unequal side
III. Rectangle
➢ Opposite sides are equal, parallel and all angles are
of 90°
➢ Perimeter = 2(l+b)
➢ Area = l*b
✓
IV. Square
➢ Diagonal = √𝒍𝟐 + 𝒃𝟐
➢ Diagonals are equal
Where l= Length and b = Breadth of rectangle
➢
➢
➢
➢
➢
All sides are equal
All angles are of 90°
Perimeter = 4a
Diagonal = √𝟐 a
Area = a2
36
V.
Rhombus
➢
➢
➢
➢
➢
All sides are equal
Opposite angles are equal
Perimeter = 4a
Area = ½ * product of diagonals
(As angles intersect at 90°, so sin Q =1)
RECENTLY ASKED QUESTIONS
1.
There are eight equidistant points on a circle. How many right-angled triangles can be drawn using
these points as vertices and taking the diameter as one side of the triangle? (2022)
(a) 24
(b) 16
(c) 12
(d) 8
Ans. (a)
Solution:
In the following figure, we have drawn eight equidistant points on a circle - A, B, C, D, E, F, G, and H.
When we consider AE as the diameter and one side of the right-angled triangle, we can draw 6 right- angled
triangles.
Similarly, if we consider BF as the diameter and one side of the right-angled triangle, we can draw 6 rightangled triangles. The same can be done when we consider CG and DH as the diameter and one side of the
right-angled triangle. Hence, the total number of right-angled triangles that can be drawn = 6 + 6 + 6 + 6 =
24. Hence, the correct answer is (a).
2.
Consider the Question and two Statements given below in respect of three cities P, Q and R in a
State:
Question: How far is city P from city Q ?
Statement-1: City Q is 18 km from city R.
Statement-2: City P is 43 km from city R.
Which one of the following is correct in respect of the Question and the Statements? (2022)
(a) Statement-1 alone is sufficient to answer the Question
(b) Statement-2 alone is sufficient to answer the Question
(c) Both Statement-1 and Statement-2 are sufficient to answer the Question
(d) Both Statement-1 and Statement-2 are not sufficient to answer the Question
Ans. (d)
Solution:
Since, we do not know the respective positions of P and Q, therefore we cannot find the distance between
them, despite the information in both the statements.
Hence, the correct answer is (d).
❖❖❖
37
27
❖
Time and Work
Formulae
➢ Days
required
to
complete
work=
1
𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑖𝑛 1 𝑑𝑎𝑦
➢ Efficiency α
➢
1
𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
If M1 persons can do W1 work in D1 days working T1 hours each
day with E1 efficiency and M2 persons can do W2 work in
D2 days working T2 hours each day with E2 efficiency, then
𝑀1𝐷1𝑇1𝐸1
𝑊1
=
𝑀2𝐷2𝑇2𝐸2
𝑊2
RECENTLY ASKED QUESTIONS
1.
24 men and 12 women can do a piece of work in 30 days. In how many days can 12 men and 24
women do the same piece of work? (2022)
(a) 30 days
(b) more than 30 days
(c) Less than 30 days or more than 30 days
(d) Data is inadequate to draw any conclusion
Ans. (d)
Solution:
Since the comparative efficiencies of men and women are not known, we cannot determine the time taken
by 12 men and 24 women to complete the given work.
Hence, the data is inadequate to draw any conclusion.
Hence, the correct answer is (d).
2.
A man completes 7/8 of a job in 21 days. How many more days will it take him to finish the job if the
quantum of work is further increased by 50% .
(2021)
(a) 24
(b) 21
(c) 18
(d) 15
Ans. (d)
Solution:
Time taken to complete 7/8 of a job = 21 days. Quantum of work further increased by 50%
Time required to complete the job =
𝟐𝟏
𝟕
𝟖
= 24 days.
So time taken by the man to complete 50% work, t1 = 24/2 = 12 days
Also, remaining work after 21 days = 1 - 7/8 = 1/8 and time required to complete 1/8 of the job, t2, is
t2 = Time required to complete whole work - time taken to complete 7/8 of the job i.e. = 24 - 21 = 3 days.
38
Now the number of days required = t1 + t2 = 12 + 3 = 15 days.
Hence, the correct answer is (d).
❖❖❖
28
❖
Trigonometry
Formulae:
𝑃
➢ Sin θ = 𝐻
➢ Cos θ =
➢ Tan θ =
𝐵
𝐻
𝑃
𝐵
➢ Cosec θ =
𝐻
𝑃
𝐻
➢ Sec θ = 𝐵
𝐵
➢ Cot θ = 𝑃
✓
Where B = Base
▪ P = Perpendicular
▪ H = Hypotenuse
RECENTLY ASKED QUESTIONS
1.
If you have two straight sticks of length 7.5 feet and 3.25 feet, what is the minimum length you can
measure?
(2020)
(a) 0.05 foot
(b) 0.25 foot 35
(c) 1 foot
(d) 3.25 feet
Ans. (c)
Solution:
It’s given that:
Length of stick S1 = 7.5 feet and Length of stick S2 = 3.25 feet
To get the minimum length which we can measure, we will use stick S2 and measure the length of stick
S1. Therefore, we get:
Total length of S1 = 3.25 + 3.25 + remaining length of S1
Or 6.5 + remaining length of S1 = 7.5
Or the remaining length of S1 = 7.5 – 6.5 = 1 foot.
Hence, the correct answer is (c).
2.
AB is a vertical trunk of a huge tree with A being the point where the base of the trunk touches the
ground. Due to a cyclone the trunk has been broken at C which is at a height of 12 metres, broken
part is partially attached to the vertical portion of the trunk at C. If the end of the broken part B
touches the ground at D which is at a distance of 5 metres from A, then the original height of the
trunk is
(2016)
(a) 20 m
(b) 25 m
(c) 30 m
(d) 35 m
39
Ans. (b)
Solution:
Pythagoras theorem CD2 =AC2 +AD2
=122 + 52 ⇒144 + 25 =169
CD2 =169⇒CD= 169=13
Decimal height = AC + CB
=12+13 = 25 meters.
Hence, the correct answer is (b).
❖❖❖
40
29
Syllogism Diagrams
❖
All A are B
❖
No A are B
❖
Some A are B
❖
Some A are not B
RECENTLY ASKED QUESTIONS
1.
Two Statements followed by four Conclusions are given below. You have to take the Statements to
be true even if they seem to be at variance from the commonly known facts. Read all the Conclusions
and then decide which of the given Conclusions logically follows follow from the Statements,
disregarding the commonly known facts : (2022)
Statement-1: All pens are books.
Statement-2: No chair is a pen.
Conclusion-I: All chairs are books.
Conclusion-II: Some chairs are pens.
Conclusion-III: All books are chairs.
Conclusion-IV: No chair is a book.
Which one of the following is correct?
(a) Only Conclusion-1
(b) Only Conclusion-11
(c) Both Conclusion-III and Conclusion-IV
(d) None of the Conclusion follows
Ans. (d)
Solution:
We can draw the following possible Venn diagrams based on the given two statements:
41
Hence, the correct answer is (d).
2.
Three Statements followed by three Conclusions are given below. You have to take the Statements
to be true even if they seem to be at variance from the commonly known facts. Read all the
Conclusions and then decide which of the given Conclusions logically follows/ follow from the
Statements, disregarding the commonly known facts: (2022)
Statement-1 : Some doctors are teachers
Statement-2 : All teachers are engineers.
Statement-3 : All engineers are scientists.
Conclusion-I : Some scientists are doctors.
Conclusion-II : All engineers are doctors.
Conclusion-III: Some engineers are doctors.
Which one of the following is correct?
(a) Only Conclusion-I
(b) Only Conclusion-II
(c) Both Conclusion-I and Conclusion-III
(d) Both Conclusion-I and Conclusion-II
Ans. (c)
Solution:
We can draw the following possible Venn diagrams based on the given two statements:
We can see that conclusions I and III follow. Hence, the correct answer is (c).
❖❖❖
42
30
❖
Mensuration
Circle
➢ Radius = r=
𝑫𝒊𝒂𝒎𝒆𝒕𝒆𝒓
𝟐
➢ Circumference = 2 * π * r
➢ Area = π * r2
𝜃
➢ Area of Arc= 𝟑𝟔𝟎 * π * r2
❖
Cube
➢
➢
➢
➢
➢
All sides are equal with length a
Base of Cube is a square
Volume of cube = a3
Length of Diagonal = √𝟑 a
Lateral surface area = Perimeter of base * Height
= 4a2
➢ Total surface area = Lateral Surface Area + 2
Base Area = 6a2
➢ Open area = Lateral Surface Area + Base Area =
5a2
❖
Cuboid
➢ Base of Cuboid is a rectangle
➢ Volume = Length * Breadth * Height
➢ Let Length =l, Breadth = b, Height =h, then
❖
➢
➢
➢
➢
Length of Diagonal = √𝒍𝟐 + 𝒃𝟐 + 𝒉𝟐
Lateral Surface Area = 2(l+b) *h
Open area = 2(l+b)*h + l*b
Total Surface Area = 2(l + b) *h + l*b =
2(lb + bh + lh).
➢
➢
➢
➢
Base of cylinder is a circle with radius r and having height h
Volume = π * r2 * h
Curved Surface Area = 2 * π * r * h
Total Surface Area = Curved Surface Area + 2 Base area = 2
* π * r * h + 2 * π * r2 = 2 * π * r (h + r)
Cylinder
RECENTLY ASKED QUESTION
1.
Consider the following statements in respect of a rectangular sheet of length 20 cm and breadth 8
cm:
1. It is possible to cut the sheet exactly into 4 square sheets.
43
2.
It is possible to cut the sheet into 10 triangular sheets of equal area.
Which of the above statements is/are correct?
(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2
(2022)
Ans. (c)
Solution:
The rectangle is of dimensions 20 cm × 8 cm. Statement I: You may think that as the area of a rectangle is
not a perfect square, it is not possible to cut it into exactly 4 square sheets. But there’s a catch. The statement
never says that the 4 squares have to be equal in area. We can do so as follows:
Statement II: You may think that we can cut the rectangle into 8 triangles of equal area (as shown below),
but not in 10.
But again, this can be done. It has been shown below:
Hence, the correct answer is (c).
❖❖❖
44
31
Data Interpretation
RECENTLY ASKED QUESTION
1.
Three persons A, B and C are standing in a queue not necessarily in the same order. . There are 4
persons between A and B, and 7 persons between B and C. If there are 11 persons ahead of C and 13
behind A, what could be the minimum number of persons in the queue? (2022)
(a) 22
(b) 28
(c) 32
(d) 38
Ans. (a)
Solution:
According to the question, there are 4 persons between A and B, and 7 persons between B and C. If there
are 11 persons ahead of C and 13 behind A. Assume, each person is denoted by a number, and by the
alphabet A, B, C. The following series captures the statements mentioned in the question:
1, 2, 3, B, 5, 6, 7, 8, A, 10, 11, C, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22.
Therefore, there can be a total of at least 22 persons in the queue.
Hence, the correct answer is (a).
2.
P, Q, R, S, T and U are six members of a family. R is the spouse of Q, U is the mother of T and S is
the daughter of U. P's daughter is T and Rs son is P. There are two couples in the family. Which one
of the following is correct?
(a) is the grandfather of T
(b) is the grandmother of T
(c) R is the mother of P.
(d) T is the granddaughter of Q
Ans. (d)
Solution:
From the above question, the family tree can be drawn as:
R~Q
↓
P (male) ~ U (female)
↓↓
T (female) S (female)
Where, ~ sign denotes married couple.
Hence, the correct answer is (d).
❖❖❖
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32
Coding Decoding
Alphabet
Code
Q
17
A
1
R
18
B
2
S
19
C
3
T
20
D
4
U
21
E
5
V
22
F
6
W
23
G
7
X
24
H
8
Y
25
I
9
Z
26
J
10
K
11
L
12
M
13
N
14
O
15
P
16
❖ Trick to remember codes – EJOTY word
where
➢ E=5
➢ J = 10
➢ O = 15
➢ T = 20
➢ Y = 25
RECENTLY ASKED QUESTIONS
1.
If the order of the letters in the English alphabet is reversed and each letter represents the letter
whose position it occupies, then which one of the following represents 'LUCKNOW'? (2022)
(a) OGXPMLD
(b) OGXQMLE
(c) OFXPMLE
(d) OFXPMLD
Ans. (d)
Solution:
We need to find the opposite letter of the letters in the given word.
Opposite Letter Position = 27 – Letter Position So, in case of LUCKNOW, we get:
27 – 12 = 15 = O
27 – 21 = 6 = F
27 – 11 = 16 = P
27 – 15 = 12 = L
27 – 23 = 4 = D
The word that we get is OFXPMLD.
Hence, the correct answer is (d).
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2.
In the series AABABCABCDABCDE.., which letter appears at the 100th place? (2022)
(a) G
(b) H
(c) I
(d) J
Ans. (c)
Solution:
The given series is: AABABCABCDABCDE…
The pattern being followed here is: A, AB, ABC, ABCD, ABCDE, …
It’s kind of an arithmetic series wherein the first term has 1 letter, second term has 2 letters, and so on. The
length of the letter-series is to be estimated near the 100th letter.
In case of an arithmetic series, 1, 2, 3 …, then: Sum of first n terms = (n/2) (a + l), where ‘n’ is the number
of terms in the series, ‘a’ the first term, and ‘l’ the last term.
So, the sum of the first 14 terms = (14/2) (1 + 14) = 7 × 15 = 105 So, the term having 14 letters will have
the 100th letter of the series.
It will start at the 92nd letter and end at the 105th letter. Hence, the 100th letter must be I.
Hence, the correct answer is (c).
❖❖❖
47
ENGLISH
How to solve English comprehension in CSAT exam?
Solving English comprehension passages in the UPSC CSE CSAT exam can be challenging, but there are some
strategies that you can use to improve your performance. Here are some tips to help you:
❖ Improve your reading speed and comprehension skills: Reading is a skill that you can improve with
practice. Try to read as much as possible and pay attention to the structure of the passages. Focus on
understanding the main idea and the supporting details.
❖ Read the questions before you read the passage: By doing this, you will have a better idea of what to
look for while reading the passage. This will also help you to identify the relevant information more
quickly.
❖ Highlight or underline important information: When you come across important information, such as
dates, names, and key concepts, highlight or underline them. This will help you to find them quickly when
you need to answer the questions.
❖ Identify the type of question being asked: There are different types of questions that can be asked, such
as factual questions, inference questions, and vocabulary questions. Understanding the type of question
being asked will help you to approach the question correctly.
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Eliminate answer choices: If you are unsure about an answer, try to eliminate the choices that are clearly
incorrect. This will increase your chances of selecting the correct answer.
Practise previous year question papers: Practising previous year question papers will give you an idea
of the type of questions that can be asked in the exam. This will also help you to develop your time
management skills.
Look for context clues: Sometimes, unfamiliar words or phrases may appear in the passage. If you come
across such words, look for context clues such as synonyms or antonyms, to help you understand the
meaning of the word or phrase.
Focus on the structure of the passage: The structure of the passage can give you important clues about
the author's purpose and main ideas. Look for transition words, such as "however" and "therefore," to
understand the relationships between different ideas.
Make educated guesses: If you are unsure about an answer, try to make an educated guess by eliminating
obviously incorrect answers and making an educated guess based on the information you have.
Practise active reading: When reading a passage, try to actively engage with the text by asking yourself
questions and making connections to your own experiences and knowledge. This will help you to
understand the passage better and retain the information more effectively.
Manage your time wisely: Time management is crucial in the UPSC CSE CSAT exam. Make sure to
allocate your time wisely, and don't spend too much time on any one question. If you get stuck on a
question, move on and come back to it later if you have time.
Remember, the key to success in English comprehension passages is to practise regularly and develop your
skills over time.
RECENTLY ASKED QUESTIONS
2022
Passage–1
The main threat to maintaining progress in human development comes from the increasingly evident
unsustainability of production and consumption patterns. Current production models rely heavily on fossil fuels.
We now know that this is unsustainable because the resources are finite. The close link between economic
48
growth and greenhouse gas emissions needs to be served for human development to become truly sustainable.
Some developed countries have begun to alleviate the worst effects by expanding recycling and investing in
public transport and infrastructure. But most developing countries are hampered by the high costs and low
availability of clean energy sources. Developed countries need to support developing countries' transition to
sustainable human development.
Q1. Unsustainability in production patterns is due to which of the following?
1. Heavy dependence on fossil fuels
2. Limited availability of resources
3. Expansion of recycling
Select the correct answer using the code given below:
(a) 1 and 2 only
(b) 2 only
(c) 1 and 3 only
(d) 1, 2 and 3
Ans. (a)
Solution:
❖ Statement 1 is correct: In the passage, it is directly mentioned that current production models rely
heavily on fossil fuels. Hence, this statement regarding the heavy dependence of fossil fuel is correct.
❖ Statement 2 is correct: In the passage, it is directly mentioned that the current production model is
unsustainable as the resources are finite. Therefore, this makes statement 2 also correct.
❖ Statement 3 is incorrect: This statement is opposite of what the author is trying to indicate. According
to the passage, “Some developed countries have begun to alleviate the worst effects by expanding
recycling”. Therefore, expansion of recycling would rather help in rectifying the unsustainable
production model and not aggravate it. Hence, this statement is incorrect.
Q2. Consider the following statements:
Developed countries can support developing countries' transition to sustainable human development
by
1. making clean energy sources available at low cost
2. providing loans for improving their public transport at nominal interest rates
3. encouraging them to change their production and consumption patterns.
Which of the statements given above is/are correct?
(a) 1 only
(b) 1 and 2 only
(c) 2 and 3 only
(d) 1, 2 and 3
Ans. (d)
Solution:
❖ Statement 1 is correct: This statement can be ascertained directly from the passage as the author
mentions that the developed countries should directly help the developing countries in their transition
to sustainable human development. Hence, the above statement is correct.
❖ Statement 2 is correct: This statement can be inferred from the passage. Even though not explicit, it
is implied by the author where it is stated that “Developed countries need to support developing
countries' transition to sustainable human development.” Providing loans for improving their public
transport at nominal interest rates would eventually result in ensuring this.
❖ Statement 3 is correct: Similarly, developing countries should be encouraged by developed countries
to change their production and consumption patterns. There should be efforts to ensure that sustainable
consumption and production practices are adopted everywhere.
Passage–2
Unless the forces and tendencies which are responsible for destroying the country's environment are checked in
the near future and afforestation of denuded areas is taken up on a massive scale, the harshness of the climatic
conditions and soil erosion by wind and water will increase to such an extent that agriculture, which is the
mainstay of our people, will gradually become impossible. The desert countries of the world and our own desert
areas in Rajasthan are a grim reminder of the consequences of large-scale deforestation. Pockets of desert-like
49
landscape are now appearing in other parts of the country including the Sutlej-Ganga Plains and Deccan Plateau.
Where only a few decades back there used to be lush green forests with perennial streams and springs, there is
only brown earth, bare of vegetation, without any water in the streams and springs except in the rainy season.
Q3. According to the passage given above, deforestation and denudation will ultimately lead to which of
the following?
1. Depletion of soil resource
2. Shortage of land for the common man
3. Lack of water for cultivation
Select the correct answer using the code given below:
(a) 1 and 2 only
(b) 2 and 3 only
(c) 1 and 3 only
(d) 1, 2 and 3
Ans. (c)
Solution:
❖ Statement 1 is correct: According to the passage, a few decades back, there would be lush green
forests with perennial streams and springs, while now there is only brown earth, bare of vegetation,
without any water in the streams and springs except in the rainy season. The phrase ‘brown earth’
denotes the absence of vegetation, which can be due to depletion of soil resources as one of the reasons.
Hence, this statement is correct.
❖ Statement 2 is incorrect: There is no mention of the shortage of land for the common man anywhere
in the passage. Hence, this statement goes beyond the context of the passage.
❖ Statement 3 is correct: “Where only a few decades back there used to be lush green forests with
perennial streams and springs, there is only brown earth, bare of vegetation, without any water in the
streams and springs except in the rainy season”. The above lines from the passage directly support
that there is a lack of water for cultivation as the author mentions about the presence of water in
streams and springs only during the rainy season.
❖❖❖
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51
52
53
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