MARKS BOOSTER

SERIES
CSAT
(Formula Book)

FOR UPSC CIVIL SERVICES PRELIMS EXAM 2025

 Table of Contents
NUMBER SYSTEM .......................................................................................................................................................... 1
LCM & HCF ..................................................................................................................................................................... 3
SURDS & INDICES ........................................................................................................................................................... 3
ELEMENTARY ALGEBRA ................................................................................................................................................. 4
AVERAGE........................................................................................................................................................................ 4
PERCENTAGE.................................................................................................................................................................. 5
RATIO, MIXTURE & PROPORTION .................................................................................................................................. 6
PROFIT, LOSS & DISCOUNT ............................................................................................................................................ 6
SIMPLE & COMPOUND INTEREST .................................................................................................................................. 7
TIME, SPEED & DISTANCE .............................................................................................................................................. 8
TIME & WORK ................................................................................................................................................................ 9
PERMUTATION & COMBINATION ................................................................................................................................ 10
PROBABILITY ................................................................................................................................................................ 11
SEQUENCE & SERIES .................................................................................................................................................... 12
STATISTICS ................................................................................................................................................................... 13
GEOMETRY................................................................................................................................................................... 13
DATA INTERPRETATION ............................................................................................................................................... 17
SERIES .......................................................................................................................................................................... 17
CODING & DECODING.................................................................................................................................................. 18
CLOCK .......................................................................................................................................................................... 19
CALENDAR ................................................................................................................................................................... 20
AGE BASED PROBLEMS ................................................................................................................................................ 20
DIRECTION & DISTANCE............................................................................................................................................... 21
SITTING ARRANGEMENT.............................................................................................................................................. 21
RANKING ...................................................................................................................................................................... 22
VENN DIAGRAM ........................................................................................................................................................... 22
INEQUALITIES............................................................................................................................................................... 24
SYLLOGISM................................................................................................................................................................... 25
CUBES & DICE .............................................................................................................................................................. 25
READING COMPREHENSION ........................................................................................................................................ 26
 1 NUMBER SYSTEM

FACTORS AND MULTIPLES Divisibility by ‘5’ : If the last digit of a number

is either 0 or 5.
1. Number of Factors (divisors): Divisibility by ‘6’ : If the number is divisible by
If ‘N’ is a composite number such that ‘N’ 2 and 3 both.
= 𝒂𝒙 × 𝒃𝒚 × 𝒄𝒛 × . .. Divisibility by ‘7’ : If the difference between
Here, ‘a’, ‘b’ and ‘c’ are prime numbers and twice the last digit and the number formed by the
‘x’, ‘y’ and ‘z’ are natural numbers. remaining digits is either zero or a multiple of 7.
• Number of factors of ‘N’ Divisibility by ‘8’ : If the last three digits of a
= (x + 1) × (y + 1) × (z + 1) × … number is divisible by 8 or the last three digits
• Sum of Factors of ‘N’ = (a0 + a1 + a2 … + ax) of a number is zero.
× (b0 + b1 + b2 …+ by) × (c0 + c1 + c2 …+ cz) Divisibility by ‘9’ : If the sum of the digits of a
×… number is divisible by 9.
𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒇𝒂𝒄𝒕𝒐𝒓𝒔
• Product of Factors of ‘N’ = 𝑵( 𝟐
) Divisibility by ‘11’ : If the difference between
• Number of prime factors of ‘N’ = a, b, c, … the sum of its digits at odd places and the sum
of its digits at even places is either 0 or a
2. Even and Odd factors of a number: multiple of 11.
If N = 2𝑥 × 𝑏 𝑦 × 𝑐 𝑧 × . ..
Divisibility by ‘12’ : If the number is divisible
○ If ‘N’ is even, then x > 0 but if ‘N’ is odd,
by 3 as well as 4.
then x = 0
• Total number of even factors Note:
= x × (y + 1) × (z + 1) × ... • A four digit number of the form ‘abab’
• Sum of even factors = (21 + 22 …. + 2x) × (b0 will always be divisible by ‘101’
+ b1 + b2 ….+ by) × (c0 + c1 + c2 ….+ cz) × … • A six digit number of the form ‘ababab’
• Total number of odd factors will always be divisible by ‘3’, ‘7’, ‘13’ and
= (y + 1) × (z + 1) × … ‘37’
• Sum of odd factors = (b0 + b1 + b2 ….+ by) × • A number of the form ‘aaa’ will always be
(c0 + c1 + c2 ….+ cz) ×… divisible by ‘3’ and ‘37’
Also, total number of factors = Number of a • A six digit number of the form ‘aaaaaa’
even factors + Number of odd factors will always be divisible by ‘3’, ‘7’, ‘13’ and
‘37’
• A six digit number of the form ‘abcabc’
DIVISIBILITY
will always be divisible by ‘3’, ‘7’ and ‘11’.
Divisibility by ‘2’ : If the unit's digit of a
number is 0, 2, 4, 6 or 8. CYCLICITY AND UNIT DIGIT
Divisibility by ‘3’ : If the sum of the digits of a
The unit’s digit of an expression can be
number is divisible by 3. calculated by getting the remainder when the
Divisibility by ‘4’ : If the last 2 digits of a expression is divided by 10. Thus, unit digit of
number is divisible by 4 or the last 2 digits of a (. . . . . . . . . . . . 𝑨𝑩𝑪𝑫)𝒙 will depend upon the unit
number is zero. digit of 𝑫𝒙 .

1
 Different If the If the If the If the are co-prime to each other, then the
values of values values values of values of remainder of the given expression will
‘D’ of ‘x’ is of ‘x’ is ‘x’ is ‘x’ is always be ‘1’.
[4n + 1] [4n + 2] [4n + 3] [4n + 4] 𝑎𝑛
7. If an expression is given as , then the
0 0 0 0 0 𝑏
remainder of the following expression will
1 1 1 1 1 be (remainder obtained when ‘a’ prime is
divided by ‘b’)n.
5 5 5 5 5
8. The square of any odd number when
6 6 6 6 6 divided by 8 will leave 1 as the remainder.
4 4 6 4 6
NUMBER OF ZEROS
9 9 1 9 1
Number of zeros in a given expression depends
2 2 4 8 6 on the number of pairs of 2’s and 5’s available in
3 3 9 7 1 the expression. If in a given expression
• Number of 2’s < Number of 5’s, then
7 7 9 3 1
number of zeros in the expression =
8 8 4 2 6 Number of 2’s
• Number of 5’s < Number of 2’s, then
number of zeros in the expression =
REMAINDER Number of 5’s
𝑎±𝑏
1. Remainder of ( )
𝑛 Note:
𝑎 𝑏
= Remainder of ( ) ± Remainder of ( ) If the expression contains 10 (= 2 × 5) it means
𝑛 𝑛
If after addition, the result is more than ‘n’, the expression will have as many additional
then it will be further divided by ‘n’ and the zeros at the 10 as the number of 10’s in the
resulting remainder will be our answer. expression.
𝑎×𝑏
2. Remainder of ( )
𝑛 Process to find the number of zeros at the end
𝑎 𝑏
= Remainder of ( ) × Remainder of ( ) of ‘n!’:
𝑛 𝑛
If after multiplication, the result is more Since, n! = n × (n - 1) × (n - 2) × ...........× 1
than ‘n’, then it will be further divided by ‘n’ Number of zeros = power of ‘5’
and the resulting remainder will be our In the case of factorial, the power of ‘5’ is a
answer. limiting factor as ‘5’ is less likely to occur than
3. If an expression can be written in the form ‘2’.
(𝑎𝑥 + 1)𝑛
, then the remainder of the given Maximum power of ‘5’ in ‘n!’
𝑎 𝒏 𝒏 𝒏
expression will always be 1. = + 𝟐 + 𝟑 +. ..
𝟓 𝟓 𝟓
4. If an expression can be written in the form
(𝑎𝑥 − 1)𝑛 [Consider the integral part only]
, then the remainder of the given
𝑎
expression will be: Sum of ‘n’ Sum of squares Sum of cubes
consecutive
• ‘-1’ or (a – 1) if ‘n’ is odd. numbers
• ‘1’ if ‘n’ is even.
5. If an expression can be written in the Natural 𝑛(𝑛+1) 𝑛(𝑛 + 1)(2𝑛 + 1) 𝑛(𝑛 + 1) 2
(𝑎)𝑛 Numbers 2 6 [ ]
form of , then the remainder will be: 2
𝑎+1
• ‘a’ if ‘n’ is odd. Even 𝑛(𝑛 + 1) 2𝑛(𝑛 + 1)(2𝑛 + 1) 2{𝑛(𝑛 + 1)}2
• ‘1’ if ‘n’ is even. Numbers 3

6. Fermat’s Theorem:
Odd 𝑛2 𝑛(2𝑛 + 1)(2𝑛 − 1) 𝑛2 (2𝑛2 − 1)
𝑎 (𝑃 − 1) Numbers
If an expression can be expressed as 3
𝑃
such that ‘P’ is a prime number and ‘a’ & ‘P’

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 2 LCM & HCF
L.C.M. H.C.F.
1. LCM of Fractions: 1. H.C.F. of Fractions:
Express the given fraction in their lowest terms.
𝑳.𝑪.𝑴.𝒐𝒇 𝒏𝒖𝒎𝒆𝒓𝒂𝒕𝒐𝒓𝒔
Express the given fractionin their lowest terms.
L.C.M of given fractions = 𝑯.𝑪.𝑭.𝒐𝒇 𝒏𝒖𝒎𝒆𝒓𝒂𝒕𝒐𝒓𝒔
𝑯.𝑪.𝑭 𝒐𝒇 𝒅𝒆𝒏𝒐𝒎𝒊𝒏𝒂𝒕𝒐𝒓𝒔
H.C.F of fractions =
𝑳.𝑪.𝑴.𝒐𝒇 𝒅𝒆𝒏𝒐𝒎𝒊𝒏𝒂𝒕𝒐𝒓𝒔
2. Applications of LCM:
• The smallest number which is exactly
2. Applications of HCF:
divisible by ‘x’, ‘y’ and ‘z’ = L.C.M of x, y, z. • The H.C.F of two or more numbers is
• The L.C.M of two or more numbers is greater smaller than or equal to the smallest number
than or equal to the greatest number of given of given numbers.
numbers.
• Let ‘K’ be the largest number which when
• The number which when divided by ‘x’, ‘y’
and ‘z’ leaves a remainder ‘R’ in each case. divide a,b,c gives the same remainder 'r' &
Then, the required number quotients are x, y, z respectively then,
= (L.C.M of x, y, z) × K + R, K = HCF of (a – b, b – c, c – a).
where K is a constant • The greatest number which divides x, y and
• The smallest number which when divided
z to leave the remainder R is
by ‘x’, ‘y’ and ‘z’ leaves the remainder of ‘a’,
‘b’, ‘c’ such that common difference(d) H.C.F of (x – R), (y – R) and (z – R).
= x - a = y – b = z – c. • The greatest number which divides x, y, z to
Then the required number leave remainders a, b, c is
= (L.C.M of x, y and z) × K – d,
where K is a constant H.C.F of (x – a), (y – b) and (z – c).
• If some bells ring after an interval of ‘a’
seconds, ‘b’ seconds and ‘c’ seconds, PRODUCT RULE
respectively, then together they will ring
after LCM of (a, b and c) seconds. If ‘p’ and ‘q’ are two numbers then,
Number of times they will ring together in Product of numbers (p × q)
𝑇 = (H.C.F. of ‘p’ and ‘q’) × (L.C.M. of ‘p’ and ‘q’)
‘T’ seconds= + 1 (if they
𝐿𝐶𝑀 𝑜𝑓 (𝑎,𝑏 𝑎𝑛𝑑 𝑐)
start ringing together initially)

3 SURDS & INDICES

LAWS OF SURDS LAWS OF INDICES
𝟏
𝒏
1. √𝒂 = 𝒂 𝒏 1. na × nb = n(a + b)
𝟏 𝟏 𝟏
𝒎 𝒎 𝒎
2. √𝒂𝒃 = √𝒂 × √𝒃 = 𝒂𝒎 × 𝒃𝒎 = (𝒂𝒃)𝒎 2.
𝒏𝒂
= 𝒏(𝒂 − 𝒃)
𝟏
𝒎 𝒏𝒃
𝒎 𝒂 √𝒂 𝒂 𝒎
3. √𝒃 = 𝒎 = ( ) 3. (na)b = nab
√𝒃 𝒃
𝒎
4.
𝒏
( √𝒂 ) 𝒎 = 𝒂 𝒏
𝒏
= √𝒂 𝒎 4. (𝒂𝒃)𝒎 = 𝒂𝒎 × 𝒃𝒎
𝒎 𝟏 𝟏 5. n0 = 1
√ √𝒂 = √𝒂𝒏 = 𝒂𝒎𝒏
𝒎 𝒏
5.

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 4 ELEMENTARY ALGEBRA

SOME IMPORTANT IDENTITIES QUADRATIC EQUATIONS

1. (a + b)2 = a2 + 2ab + b2 Sum of roots and Product of roots:

2. (a – b)2 = a2 – 2ab + b2
Let 𝛼, 𝛽 are real roots of the quadratic equation
3. a2 – b2 = (a + b)(a – b)
ax² + bx + c = 0
4. (a + b)3 = a3 + b3 + 3ab (a + b)
Since, ax² + bx + c = 0
5. (a – b)3 = a3 – b3 - 3ab (a – b)
Dividing each term br ‘a’, we get;
6. (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
𝑏 𝑐
𝑥 2 + ( ) 𝑥 + ( ) = 0 ...eq(1)
𝑎 𝑎
1 1 1
𝐼𝑓 𝑥 + = 𝐴 𝑥2 + = 𝐴2 − 2 𝑥3 + = 𝐴3 − 3𝐴
𝑥 𝑥3 𝒃
𝑥2 So, sum of roots = 𝜶 + 𝜷 = −
𝒂
1 1 1 𝒄
𝐼𝑓 𝑥 −
𝑥
= 𝐴 𝑥2 + = 𝐴2 + 2 𝑥3 −
𝑥3
= 𝐴3 + 3𝐴 And, product of roots = 𝜶 × 𝜷 =
𝑥2 𝒂

5 AVERAGE
𝑺𝒖𝒎 𝒐𝒇 𝒂𝒍𝒍 𝒐𝒃𝒔𝒆𝒓𝒗𝒂𝒕𝒊𝒐𝒏𝒔
Average or Mean =
𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒐𝒃𝒔𝒆𝒓𝒗𝒂𝒕𝒊𝒐𝒏𝒔

Facts about Average

• When each of the observations is increased/decreased by ‘p’, then the overall average will also
increase/decrease by ‘p’.
• When each observation is multiplied/ divided by ‘p’, then the overall average will also get
multiplied/divided by ‘p’.
If the average of ‘a’ numbers is ‘x’ and out of these ‘a’ numbers average of ‘b’ numbers is ‘y’, then
average of remaining numbers will be:
𝑎𝑥 − 𝑏𝑦
𝑎 − 𝑏
Application of average:
When a person leaves a group and is replaced by another person, then
Case-I: Increase Age/weight/height of new person = Age/weight/height of person who left
in average the group + (Number of persons in the group including new person × Increase
in average age/weight/height of the group)
Case-II: Age/weight/height of new person = Age/weight/height of person who left
Decrease in the group – (Number of persons in the group including new person × Decrease
average in average age/weight/height of the group)

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When a person leaves the group but nobody joins that group, then
Case I: Increase Age/Weight/Height etc of person left = Previous average (Number of persons
in average present × Increase in average)
Case II: Decrease Age/Weight/Height etc of person left = Previous average + (Number of
in average persons present x Decrease in average)
• If the average of ‘M’ number of observations is ‘N’ but some observations are misread as ‘a’, ‘b’
𝑀 × 𝑁 − (𝑎 + 𝑏 + 𝑐) + (𝑥 + 𝑦 + 𝑧)
and ‘c’ in place of ‘x’, ‘y’ and ‘z’, respectively, then the correct average =
𝑀

6 PERCENTAGE

PERCENTAGE INCREASE / • Then the population of town after 3 years

𝑥 𝑦 𝑧
DECREASE OF A QUANTITY = 𝑃 (1 ± ) (1 ± ) (1 ± )
100 100 100
‘+’ sign indicates increase in population and
• If an amount is increased by x% and then it ‘–’ sign indicates decrease in population.
is decreased by x% again, then there is
always an overall percentage decrease EXAMINATION MARKS
which is equal to: RELATED PROBLEMS
𝒙𝟐
Percentage Decrease = %
𝟏𝟎𝟎
In a certain examination, 'x' boys and 'y' girls
• If an amount is increased by x% and then it
participated, out of which ‘a%’ of boys and ‘b%’
is increased by y% again, then the resultant
of girls passed the examination, then,
change in percentage is
𝒙𝒚 percentage of passed students of the total
Percentage change = (𝒙 + 𝒚 + )% 𝑥×𝑎+𝑦×𝑏
𝟏𝟎𝟎 students = ( )%
𝑥+𝑦

INCOME / EXPENDITURE
GEOMETRICAL FIGURES
PROBLEMS: RELATED PROBLEMS
• If the income of a person ‘A’ is ‘x%’ more • If the sides of an equilateral triangle, square,
than person ‘B’, then income of ‘B’ is less in
𝑥 rhombus or radius of a circle are increased
comparison to ‘A’ by ( × 100) % by a%, then its area is increased by
100 + 𝑥
• If the income of a person ‘A’ is ‘x%’ less than 𝑎2
person ‘B’, then income of ‘B’ is more in = (2𝑎 + )%
100
𝑥
comparison to ‘A’ by ( × 100) % • If the sides of an equilateral triangle, square,
100 − 𝑥
rhombus or radius of a circle decreased by
POPULATION / PRICE / a%, then its area is decreased by
𝑎2
QUANTITY RELATED PROBLEMS = (−2𝑎 +
100
)%
• If the length of a rectangle is increased by
If the present population of a town is ‘P’ and the
‘x%’ and breadth is increased by ‘y%’, then
population increases or decreases at rate of ‘x%’,
‘y%’ and ‘z%’ in the first, second and third year, the area of rectangle will increase by
𝑥𝑦
respectively. = (𝑥 + 𝑦 + )%
100

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 7 RATIO, MIXTURE & PROPORTION
CONTINUED PROPORTION milk at cost price, then profit percentage of
𝑦
the milkman = × 100
‘a’, ‘b’ and ‘c’ are said to be in continued 𝑥+𝑦

proportion if a : b = b : c or 𝑏 2 = 𝑎𝑐 • If mixture ‘A’ costing Rs. ‘x’ per kg is mixed

Here ‘b’ is the mean proportional to ‘a’ and ‘c’, with mixture ‘B’ costing Rs. ‘y’ per kg and
and ‘c’ is the third proportional to ‘a’ and ‘b’. the resultant mixture is sold for Rs. ‘z’ per
Fourth Proportion: If four quantities ‘a’, ‘b’, ‘c’
kg such that x < z < y, then the ratio in which
and ‘x’ are such that a:b :: c:x, then ‘x’ is called
the fourth proportion of ‘a’, ‘b’ and ‘c’. mixture ‘A’ and mixture ‘B’ are mixed can
Also, 𝑥 =
𝑏𝑐 be given by:
𝑎

PARTNERSHIP
If two partners entered into a business investing
Rs. ‘𝐼1 ’ and Rs. ‘𝐼2 ’ for ‘𝑡1 ’ months and ‘𝑡2 ’
months such that profit earned by them is Rs.
𝑃1 𝐼1 × 𝑡1
‘𝑃1 ’ and Rs. ‘𝑃2 ’, then =
𝑃2 𝐼2 × 𝑡2

MIXTURE AND ALLIGATION

𝑸𝒖𝒂𝒏𝒕𝒊𝒕𝒚 𝒐𝒇 𝒎𝒊𝒙𝒕𝒖𝒓𝒆 ‘𝑨’ 𝒚−𝒛
• ‘P’ litres of mixture contains ‘a%’ of liquid =
𝑸𝒖𝒂𝒏𝒕𝒊𝒕𝒚 𝒐𝒇 𝒎𝒊𝒙𝒕𝒖𝒓𝒆 ‘𝑩′ 𝒛−𝒙
‘A’. The quantity of water that must be • If a container contains ‘P’ units of pure
mixed in the mixture such that the resulting liquid (milk/kerosene/wine etc.) and is
mixture contains ‘b%’ of liquid ‘A’ altered by taking out ‘Q’ units of the pure
𝑎−𝑏
=𝑃 × liquid and replacing it by same quantity of
𝑏
another liquid and this process is repeated
• If a milkman mixes ‘y’ litres of water with ‘x’ ‘n’ times, then quantity of pure liquid in the
litres of pure milk and promises to sell the 𝑄 𝑛
find mixture will be = 𝑃 × {1 − }
100

8 PROFIT, LOSS & DISCOUNT

IMPORTANT FORMULAS SOME IMPORTANT RESULT
• If cost price = Selling price, then there will
(i) In case of Profit; Profit = SP – CP
be no profit no loss.
𝑝𝑟𝑜𝑓𝑖𝑡
(ii) Profit percentage = × 100 % • If the profit earned on selling an article for
𝑐𝑜𝑠𝑡 𝑝𝑟𝑖𝑐𝑒

(iii) In case of Loss; Loss = CP – SP Rs. ‘x’ is same as the loss incurred on selling
𝑙𝑜𝑠𝑠 the article for Rs. ‘y’, then cost price of the
(iv) Loss percentage = × 100 % 𝑥+𝑦
𝑐𝑜𝑠𝑡 𝑝𝑟𝑖𝑐𝑒
article =
2
(v) SP = (100 + k)% of CP (k = profit %)
• If two articles are sold at the same price such
(vi) SP = (100 – k)% of CP (k = loss %) that profit earned on selling one article is x%

6
 while loss incurred on selling the other When three successive discounts are given:
article is x% then there will be overall loss of Suppose there are three successive discounts of
𝑥2 ‘𝑟1 %’ , ‘𝑟2 %’ and ‘𝑟3 %’ is given on an article, then
%.
100 the selling price of the article after these
discounts will be:
DISCOUNT 𝑟3 𝑟2 𝑟1
= (1 − ) × (1 − ) × (1 − ) ×P
100 100 100
(i) Selling price = Marked price - Discount
DISHONEST SHOPKEEPER
(ii) Discount = Marked price - Selling price
(iii) Discount % = (
𝐷𝑖𝑠𝑐𝑜𝑢𝑛𝑡
) × 100 If a shopkeeper promises to sell his goods at
𝑀𝑎𝑟𝑘𝑒𝑑 𝑝𝑟𝑖𝑐𝑒
cost price but uses a faulty weight of ‘x’ gm
100 − 𝑟
(iv) Selling price = ( × marked price), instead of 1 kg or 1000 gm, then overall gain
100
where ‘r’ = discount percentage percentage of the shopkeeper
1000 − 𝑥
When two successive discounts are given: =( × 100) %
𝑥
Suppose two successive discounts of ′𝑟1 %’ and • If a shopkeeper sells his goods at ‘x%’ profit
′𝑟2 %’ are given on an article, then the selling
and uses a weight which is ‘y%’ less than the
price of the article after these discounts will be:
𝒓𝟐 𝒓𝟏 actual weight, then overall gain percentage
= (𝟏 − ) x (𝟏 − ) xP
𝟏𝟎𝟎 𝟏𝟎𝟎 of the shopkeeper
or 𝑥% + 𝑦%
𝒓𝟏 𝒓𝟐
={ × 100} %
100 − 𝑦%
Net discount percentage = (𝒓𝟏 + 𝒓𝟐 − )%
𝟏𝟎𝟎

9 SIMPLE & COMPOUND INTEREST

Simple Interest (SI) =
𝑃×𝑅×𝑇 Difference between CI and SI
100
Total Amount = Principal Amount + SI For the first compounding period (for first year
Compound Interest (CI) = A – P in general), the SI and CI are equal.
𝑹 𝑻
Total Amount = 𝑨 = 𝑷 × (𝟏 + ) • The difference between SI and CI for first
𝟏𝟎𝟎
Where: 𝑃𝑅 2
two years =
1002
‘A’ is the future value of the investment. • The difference between SI and CI for the first
‘P’ is the principal amount or initial investment. 𝑅 2 300 + 𝑅
three years = 𝑃 ( ) ( )
‘R’ is the rate of interest (in per annum). 100 100
1. Installment Paid With Simple Interest:
‘T’ is the time in years.
The annual payment that will discharge a debt
When the rate of of compound interest is
different for different years: of Rs. ‘P’ due in ‘T’ years at the rate of simple
100𝑃
𝐴 = 𝑃 × (1 +
𝑅1
) × (1 +
𝑅2
) × (1 +
𝑅3
) … × (1 +
𝑅𝑛
) interest of ‘R%’ per annum is: 𝑥 = 𝑅𝑇(𝑇 − 1)
100 100 100 100 100𝑇 +
2
Here, ‘A’ is the amount accumulated after ‘n’ 2. Installment Paid With Compound Interest:
years. To calculate the installments paid with
‘P’ = Principal amount invested compound interest, we use the following
And, R1,R2,.....,Rn is the rate of interest offered in formula: 𝑥 =
𝑃
𝑅 𝑅 2 𝑅 𝑛
first year, second year and nth year respectively. (1 +
100
) + (1 +
100
) + ...+ (1 +
100
)

7
 10 TIME, SPEED & DISTANCE
Distance = Speed × Time, TRAIN PROBLEMS
𝑫𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝑫𝒊𝒔𝒕𝒂𝒏𝒄𝒆
Time = and Speed =
𝒔𝒑𝒆𝒆𝒅 𝒕𝒊𝒎𝒆 • If a train of length ‘l’ metres travelling with
a speed of ‘x’ m/s crosses a
CONVERSION OF UNITS pole/man/tree/bus/truck/car in ‘t’
seconds, then it will cover a distance equal
• 1 km =1000 metres to its length i.e. l = s × t
• 1 hour = 60 minutes • If a train of length ‘l’ metres travelling with
• 1 minute = 60 seconds a speed of ‘s’ m/s crosses a ‘p’ metres long
• 1 m/s =
18
𝑘𝑚/ℎ platform/tunnel in ‘s’ seconds, then
5
l+p=s×t
AVERAGE SPEED • If two trains are travelling on parallel tracks
and in the same direction, then time taken
𝑻𝒐𝒕𝒂𝒍 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒕𝒓𝒂𝒗𝒆𝒍𝒍𝒆𝒅 by them cross each other
Average Speed =
𝑻𝒐𝒕𝒂𝒍 𝒕𝒊𝒎𝒆 𝒕𝒂𝒌𝒆𝒏 𝑺𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆𝒊𝒓 𝒊𝒏𝒅𝒊𝒗𝒊𝒅𝒖𝒂𝒍 𝒍𝒆𝒏𝒈𝒕𝒉𝒔
=
• If a certain distance is covered at a speed of 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝒃𝒆𝒕𝒘𝒆𝒆𝒏 𝒕𝒉𝒆𝒊𝒓 𝒔𝒑𝒆𝒆𝒅𝒔

‘x’ km/h and the same distance is covered • If both trains are travelling on parallel tracks
at a speed of ‘y’ km/h, then the average and in opposite directions, then time taken
𝟐𝒙𝒚 by them cross each other
speed for the whole journey = 𝒌𝒎/𝒉
𝒙+𝒚 𝑺𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆𝒊𝒓 𝒊𝒏𝒅𝒊𝒗𝒊𝒅𝒖𝒂𝒍 𝒍𝒆𝒏𝒈𝒕𝒉𝒔
=
• If a person or a motor car covers three equal 𝑺𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆𝒊𝒓 𝒔𝒑𝒆𝒆𝒅𝒔

distances at the speed of ‘x’ km/h, ‘y’ km/h A train of length ‘l’ metres, travelling with a
and ‘z’ km/h, respectively, then the speed of ‘x’ m/s crosses a man travelling with
average speed for the entire journey a speed of ‘y’ m/s, then
𝟑𝒙𝒚𝒛 • If the men is coming from opposite
= 𝒌𝒎/𝒉
𝒙𝒚 + 𝒚𝒙 + 𝒛𝒙 directions then time taken by the train to
• If a person covers ‘A’ km at a speed of ‘x’ cross the man =
𝑙
𝑥+𝑦
km/h, ‘B’ km at a speed of ‘y’ km/h and ‘C’
• If the man is travelling in the same direction
km at a speed of ‘z’ km/h, then average
as that of the train, then time taken by the
speed during the whole journey 𝑙
𝑨+𝑩+𝑪 train to cross the man =
= 𝑨 𝑩 𝑪 𝒌𝒎/𝒉 𝑥–𝑦
+ +
𝒙 𝒚 𝒛

BOAT AND STREAM

RELATIVE MOTION
Let the speed of a boat in still water be ‘b’ km/h
• When two bodies are moving in the same
and speed of stream = ‘s’ km/h. Then,
direction: If the speeds of bodies ‘A’ and ‘B’
• Upstream speed of boat = (b – s)
are 𝑆𝐴 and 𝑆𝐵 , then their relative speed is
• Downstream speed of boat = (b + s)
(𝑆𝐴 − 𝑆𝐵 ) or ( 𝑆𝐵 − 𝑆𝐴 ).
• Speed of boat in still water
• When two bodies are moving in the (𝐷𝑜𝑤𝑛𝑠𝑡𝑟𝑒𝑎𝑚 𝑠𝑝𝑒𝑒𝑑 + 𝑈𝑝𝑠𝑡𝑟𝑒𝑎𝑚 𝑠𝑝𝑒𝑒𝑑)
opposite direction: If the speeds of bodies =b=
2
‘A’ and ‘B’ are ′𝑆𝐴 ′ and ′𝑆𝐵 ′ , then their • Speed of stream
(𝐷𝑜𝑤𝑛𝑠𝑡𝑟𝑒𝑎𝑚 𝑠𝑝𝑒𝑒𝑑 − 𝑈𝑝𝑠𝑡𝑟𝑒𝑎𝑚 𝑠𝑝𝑒𝑒𝑑)
relative speed is (𝑆𝐴 + 𝑆𝐵 ). =s=
2

8
CIRCULAR RACE If the two persons are running with speed of ‘a’
m/s and ‘b’ m/s, then calculate the simplest ratio
If ‘A’ and ‘B’ started running on a circular track
of ‘L’ metres at a speed of ‘p’ m/s and ‘q’ m/s, of a:b and the following results can be used,
respectively, then number of distinct meeting Case 1: When the runners are running in the
points on the circular track can be calculated as:
opposite direction, then number of distinct
Time taken by Time taken by
‘A’ and ‘B’ to ‘A’ and ‘B’ to meeting points = a + b (where ‘a’ and ‘b’ are the
meet for the meet for the value from simplest ratio)
first time first time at the
anywhere on starting point Case 2: When the runners are running in the
the track
‘A’ and ‘B’ are same direction, then number of distinct
running in the 𝐿 𝐿 𝐿 meeting points = |𝑎 − 𝑏| (where ‘a’ and ‘b’ are
same direction LCM of {𝑝 , 𝑞}
|𝑝 − 𝑞|
the value from simplest ratio)
‘A’ and ‘B’ are If the difference between ‘a’ and ‘b’ is odd, then
running in the 𝐿 𝐿 𝐿
opposite LCM of {𝑝 , 𝑞} the runners will never meet at diametrically
𝑝 + 𝑞
direction
opposite points.

11 TIME & WORK

1. If ‘M’ persons can complete a piece of ‘𝑊2 ’ women can complete the work in ‘𝐷2 ’
work in ‘N’ days while ‘P’ persons can days, and efficiency of each men is ‘m’
complete the same work in ‘Q’ days and units per day while that of each women is
efficiency of each person is the same, then ‘w’ units per day, then
Total work = M × N = P × Q (𝑀1 × m + 𝑊1 × w) × 𝐷1
2. If ‘𝑀1 ’ workers working ‘𝐻1 ’ hours per day = (𝑀2 × m + 𝑊2 × w) × 𝐷2
can complete a piece of work in ‘𝐷1 ’ days 5. If ‘𝑀1 ’ workers working ‘𝐻1 ’ hours per day
while ‘𝑀2 ’ workers working ‘𝐻2 ’ hours per can complete of ‘𝑊1 ’work in ‘𝐷1 ’ days
day can complete a piece of work in ‘𝐷2 ’ while ‘𝑀2 ’ workers working ‘𝐻2 ’ hours per
days and efficiency of every man is the day can complete ‘𝑊2 ’ of work in ‘𝐷2 ’ days
same, then and efficiency of every man is the same,
Total Work = 𝑀1 × 𝐷1 × 𝐻1 = 𝑀2 × 𝐷2 × 𝐻2 then
3. If ‘M’ men or ‘W’ women can complete a 𝑀1 × 𝐷1 × 𝐻1
=
𝑀2 × 𝐷2 × 𝐻2
𝑊1 𝑊2
piece of work in ‘X’ days such that
6. A certain number of 𝑀1 men finish a task
efficiency of a man and a woman is ‘m’
𝑊1 in 𝐷1 days, putting in 𝐻1 hours of work
units per day and ‘w’ units per day,
each day, and they earn a wage of Rs. 𝑅1 .
respectively, then;
Similarly, another group of 𝑀2 men
Total work = M × m × X = W × w × X
𝑚 𝑊
complete a task 𝑊2 in 𝐷2 days, working 𝐻2
Or, = hours per day, and they earn Rs. 𝑅2 as
𝑤 𝑀
𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝑜𝑓 𝑎 𝑚𝑎𝑛 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑜𝑚𝑒𝑛
Or, = their wage.
𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝑜𝑓 𝑎 𝑤𝑜𝑚𝑎𝑛 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑒𝑛
𝑀1 × 𝐷1 × 𝐻1 𝑀2 ×𝐷2 × 𝐻2
4. If ‘𝑀1 ’ men and ‘𝑊1 ’ women can complete =
𝑊1 ×𝑅1 𝑊2 ×𝑅2
a work in ‘𝐷1 ’ days while ‘𝑀2 ’ men and

9
12 PERMUTATION & COMBINATION
FACTORIAL NOTATION • The number of combinations of ‘n’ distinct
things taken ‘r’ at a time when ‘p’
The continued product of first ‘n’ natural particular things are always included
numbers is called ‘n-factorial’ and is denoted = (𝒏 – 𝒑) 𝑪(𝒓 − 𝒑)
as n!. • The number of combinations of ‘n’ distinct
n! = n (n – 1) (n – 2)….…3.2.1 things taken ‘r’ at a time when ‘p’
particular things are always excluded
Points to Remember:
= (𝒏 – 𝒑) 𝑪𝒓
• 0! = 1
The following results must be remembered:
• Factorials of only natural numbers are
• Given a set of n points in a plane, with the
defined. condition that no three of them are
• n! = n × (n – 1)! collinear:
i.e. 6! = 6 × 5! or 8! = 8 × 7! (a) Number of straight lines that can be
• Remember : 0! = 1, 1! = 1, 2! = 2, 3! = 6, formed: Number of ways by which
4! = 24, 5! = 120, 6! = 720 we can select any two points gives the
total number of straight lines = 𝒏 𝑪𝟐 .
PERMUTATIONS (b) Number of triangles that can be
formed: Number of ways by which we
The number of permutations of n different can select any three non-collinear
things taken r at a time is 𝑛 𝑃𝑟 , where points gives total number of triangles
𝑛 𝒏!
𝑃𝑟 = , where ‘n’ is the total number of = 𝒏 𝑪𝟑
(𝒏 − 𝒓)!
items, ‘r’ is the number of items you're (c) The total number of diagonals in a
arranging. polygon with n sides:
Number of diagonals = Total lines –
Circular Permutation Formula: The number of
(number of sides of polygon)
ways to arrange 'n' distinct objects in a circle is
= 𝒏 𝑪𝟐 – n
given by (n - 1)! when rotations are considered
• Given a set of n points in a plane, with the
identical. Remember that here clockwise and condition that m points of them are
anti-clockwise arrangements are considered collinear:
different. (a) Number of straight lines that can be
• If we arrange flowers or garland beads in a formed:
necklace, then there is no distinction Number of ways by which we can
between clockwise & anticlockwise select any two points gives the total
(𝒏 − 𝟏)!
direction. So the formula becomes number of straight lines
𝟐
= 𝒏 𝑪𝟐 – 𝒎 𝑪𝟐 + 𝟏
COMBINATIONS (b) Number of triangles that can be
formed:
Selecting ‘r’ things out of ‘n’ distinct things is
Number of ways by which we can
denoted as : 𝑛 𝐶𝑟 or C (n, r )and
select any three non-collinear points
𝒏 𝒏! 𝒏(𝒏 − 𝟏)(𝒏 − 𝟐)....(𝒏 − 𝒓 + 𝟏)
𝑪𝒓 =
𝒓!(𝒏 − 𝒓)!
=
𝒓!
gives total number of triangles
= 𝒏 𝑪𝟑 – 𝒎 𝑪𝟑

10
GAP METHOD permutations with repetition. The formula for
permutations of alike objects is:
If you want to consider arrangements of some 𝑛!
items when no two of some specified items are P(n; 𝑛1 , 𝑛2 , 𝑛3 ,........., 𝑛𝑘 ) =
𝑛1 ! 𝑛2 ! 𝑛3 !………𝑛𝑘 !
together.
where P represents the number of
To arrange ‘n’ different things in such a way
permutations, n is the total number of objects
that no two of the ‘r’ things are together, then
we arrange the ‘r’ things in gaps created by in the sets 𝑛1 , 𝑛2 , 𝑛3 ,........., 𝑛𝑘 of similar type.
arranging remaining (n - r) things.
SUM OF NUMBERS
PERMUTATION OF ALIKE Formula for the Sum (S):
OBJECTS The formula to find the sum of all numbers formed
When you want to find the number of by permuting the digits of a set is given by:
permutations of a set of objects, some of which S = (n – 1)! × (100 + 101 + 102 +…. + 10n - 1) ×
are alike, you can use the concept of (Sum of numbers),
where ‘n’ is the number of digits in the set.

13 PROBABILITY
Probability =
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑎𝑣𝑜𝑢𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 • Rolling a single die
𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
Total number of possible outcomes = 6
Sample Space = {1, 2, 3, 4, 5, 6}
TOSSING OF COINS • Rolling of two dice
When working with coins in probability, we Total number of possible outcomes
often consider a fair coin, which is a coin with = 62 = 36
two equally likely outcomes: Heads (H) and PROBABILITY RELATED WITH A
Tails (T).
DECK OF WELL SHUFFLED CARD
• Tossing a Single Coin:
Total number of possible outcomes = 2
A standard deck has 52 playing cards which
Sample Space = {H, T} consists of four suits (Hearts, Diamonds,
• Tossing Two Coins: Clubs, and Spades), each with 13 ranks
Total number of possible outcomes = 22 = 4 [numbers from 2 to 10, three face cards (Jack,
Sample Space = {HH, HT, TH, TT} Queen, King), and an Ace].
• Tossing Three Coins:
Total number of possible outcomes = 23 = 8
Sample Space = {HHH, HHT, HTH, HTT,
THH, THT, TTH, TTT}

ROLLING OF DICE

Standard six-sided dice are commonly used in • Total number of cards = 52

probability scenarios, and they have six • Number of Red Cards = Number of Black
equally likely outcomes: 1, 2, 3, 4, 5, and 6. Cards = 26 each

11
• Hearts = Diamond = Spade = Clubs = 13 words, it quantifies the likelihood of an event
cards each happening under a specific condition or
• Total number Kings = Total number of constraint. Conditional probability is denoted
Queens = Total number of Jacks = Total as P(A | B), where "P(A | B)" represents the
number of Aces = 4 probability of event ‘A’ occurring given that
• Number of Face Cards = King (4) + Queen event ‘B’ has occurred.
(4) + Jack (4) = 12 𝑛(𝐴∩𝐵)
𝑃(𝐴 ∩ 𝐵) 𝑛(𝑆) 𝑛(𝐴∩𝐵)
ADDITION THEOREM P(A|B) = = 𝑛(𝐵) =
𝑃(𝐵) 𝑛(𝐵)
𝑛 (𝑆)

• P(A ∪ B) = P (A) + P(B) – P(A ∩ B) MULTIPLICATION THEOREM

• P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B)
– P(B ∩ C) – P(C ∩ A) + P(A ∩ B ∩ C) The Multiplication Theorem is used to
calculate the probability of the joint occurrence
CONDITIONAL PROBABILITY
of two or more events.
(A ∩ B) = A × B = A and B denotes the
Conditional probability deals with the
simultaneous occurrence of A and B.
probability of an event occurring given that
another event has already occurred. In other • P (A ∩ B) = P (A) × P(B/A) = P(B) × P(A/B)

14 SEQUENCE & SERIES

ARITHMETIC PROGRESSION • Geometric Mean (GM) for any ‘n’ positive
numbers a1, a2,..., an is
• nth term of an AP is given by = 𝑛√𝑎1 × 𝑎2 × 𝑎3 × 𝑎4 ×. . . . . . . . . . . . . .× 𝑎𝑛
Tn = a + (n – 1)d
where ‘a’ is first term and ‘d’ is common HARMONIC PROGRESSION
difference
1 1
• Sum of the first ‘n’ terms of an AP • General HP can be considered as , ,
𝒏 𝒏 𝑎 𝑎+𝑑
𝑺𝒏 = × [𝟐𝒂 + (𝒏 – 𝟏) 𝒅] = [a + Tn] ..... where ‘a’ and ‘d’ are first term and
𝟐 𝟐
= Number of terms × (Average of first and common difference of corresponding AP.
𝟏
last term) nth term of the HP, Tn =
𝒂 + (𝒏 − 𝟏)𝒅
• Arithmetic Mean (AM) for any ‘n’ numbers
𝑎 + 𝑎 + 𝑎 ..........+ 𝑎𝑛
a1, a2,..., an is = 1 2 3 RELATIONSHIP BETWEEN AM,
𝑛

GM AND HM
GEOMETRIC PROGRESSION
If AM, GM, HM are Arithmetic Mean,
• nth term of a GP is given by Tn = a rn – 1
Geometric Mean and Harmonic Mean between
• The sum of the first ‘n’ terms of a geometric
progression (GP), often denoted as Sn is: any two numbers, then
Sn =
𝒂(𝟏 – 𝒓𝒏 )
, when r < 1 • AM ≥ GM ≥ HM (equality holds when all
𝟏–𝒓
𝒂( 𝒓𝒏 – 𝟏) terms are equal)
Sn = , when r > 1
𝒓–𝟏 • (GM)2 = (AM) × (HM) (i.e., AM, GM, HM
𝒂
• Sum of infinite terms of a GP = , are in GP)
𝟏–𝒓
when r < 1

12
15 STATISTICS
ARITHMETIC MEAN (AVERAGE) • If ‘n’ is even then, the median is the average
of the two middle values.
If x1, x2,.....xn be n observations, then their i.e., Median (M)
arithmetic mean is given by 𝑛 𝑛+1
𝑉𝑎𝑙𝑢𝑒 𝑜𝑓 ( 2 )𝑡ℎ 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛 + 𝑉𝑎𝑙𝑢𝑒 𝑜𝑓 ( 2 )𝑡ℎ 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛
=
𝑥1 +𝑥2 +....+𝑥𝑛 2
=
𝑛
MODE
MEDIAN
This is the number that appears most
To calculate the median of ‘n’ number of frequently in a group of numbers. It's like the
observations most popular or common value in the set.
• Arrange the observations in order, either
from smallest to largest or largest to RELATION BETWEEN MEAN,
smallest.
MEDIAN AND MODE
• If ‘n’ is odd then, median is the middle
value.
𝑛+1 The following empirical relationship exists
i.e., Median (M) = Value of ( ) 𝑡ℎ
2
between the Mean, Median and Mode:
observation
2 × Mean + Mode = 3 × Median

16 GEOMETRY
• When two parallel lines are cut by a Consecutive Interior The pair of consecutive
transversal line: Angles interior angles are
∠4 and ∠6 & ∠3 and ∠5 supplementary, that is,
∠4 + ∠6 = 180°, and ∠3 + ∠5 =
180°.

• Exterior Angle Property of a Triangle:

If a side of a triangle is
produced, the exterior
angle so formed is equal
to the sum of two interior
Corresponding angles The pair of corresponding opposite angles. In the
are: angles are equal figure:
∠1 & ∠5, ∠2 & ∠6, ∠1 = ∠5, ∠2 = ∠6,
∠3 & ∠7, ∠4 & ∠8 ∠3 = ∠7 & ∠4 = ∠8 ∠4 = ∠2 + ∠3
∠5 = ∠1 + ∠3
Alternate Interior The pair of alternate interior ∠6 = ∠1 + ∠2
Angles are: angles are equal in measure,
∠3 and ∠6 & ∠4 and ∠5 that is, • Pythagoras Theorem: It states that in a
∠3 = ∠6, and ∠4 = ∠5 right angled triangle, the square of the
length of the hypotenuse (the side opposite
Alternate Exterior The pair of alternate exterior
Angles angles are equal in measure,
the right angle) is equal to the sum of the
∠1 and ∠8 & ∠2 and ∠7 that is, squares of the lengths of the other two
∠1 = ∠8, and ∠2 = ∠7 sides.

13
 CIRCLES
In ΔPQR, ∠Q = 90𝑜
So, PR2 = PQ2 + RQ2 Arc: An arc is a portion of the circle's
boundary. The length of an arc is proportional
QUADRILATERALS to the measure of the central angle that it
subtends.
A quadrilateral is a four-sided polygon. Sector: A sector is the region
Thus, it has four vertices, four of the circle enclosed by two
sides, four angles and two radii and the arc between
diagonals. them.
The sum of angles of AXB is an arc of the circle.
quadrilateral is AOB is a sector of the circle.
∠A + ∠B + ∠C + ∠D = 360° • The length of the arc of the sector of the
𝜃
Rectangle: circle (l) = × 2πr (where r is the radius of
360
A rectangle is a parallelogram with all angles the circle)
as right angles (90°). • The perimeter of the sector of the circle
Properties: = l + 2r
𝜃
• All properties of a • The area of a sector of a circle= × πr2
360
parallelogram.
• All angles are 90°.
BASIC THEOREMS & RESULTS
• Diagonals are equal in length. OF CIRCLES
• Diagonals bisect each other.
If two chords of a circle are of equal
Square: length, they will also subtend equal
A square is a rectangle with all sides central angles at the centre of the circle.
Its converse is also true.
equal all angles equal (each 90°).
If AB = CD, then angle AOB = angle COD
Properties: If angle AOB = angle COD, then AB = CD
Properties of square is same as that of
the parallelogram apart from the fact that The perpendicular from the centre of a
circle to a chord bisects the chord. Its
diagonals are equal and bisect each other at 90°.
converse is also true.
Trapezium: If OM is perpendicular to AB,
A trapezium, also known as a trapezoid, is a then AN = MB

quadrilateral (a four-sided polygon) If AM = MB, then OM is perpendicular to

AB.
with one pair of parallel
sides i.e. AB II CD while AD Two equal chords of a circle are
equidistant from the centre. Its converse
and CB are not parallel.
is also true.
Properties: If AB = CD, then OL = OM
If OL = OM, then AB = CD
• All sides are not equal.
• Opposite sides are not equal.
• Opposite angles are not equal. The angle subtended by a chord at the
• Diagonals are not equal. centre of the circle is twice the angle
subtended by that chord on the
• Diagonals do not bisect each other. circumference of the circle.
• Sum of adjacent angles is not 180°

14
 The angles subtended by a chord in the Equilateral Triangle:
same segment of a circle are equal.

The angle subtended by a diameter in a

semi circle is a right angle.

• Perimeter (P) of an equilateral triangle

The tangent line at any point of a cirlce = 3 × (side) = 3a
is perpendicular to the radius through • Area (A) of an equilateral triangle
the point of contact. √3 √3
= × (side)2 = × a2
4 4
Cyclic Quadrilateral: • Altitude (h) of an equilateral triangle
A cyclic quadrilateral, also known as a √3 √3
= × (side) = ×a
concyclic quadrilateral, is a four-sided 2 2
polygon in which all four vertices lie on • Area (A) of an equilateral triangle
the circumference of a single circle. In a (𝑎𝑙𝑡𝑖𝑡𝑢𝑑𝑒)2
=
cyclic quadrilateral, the sum of the √3
measures of the opposite angle is always Rectangle:
180 degrees i.e.
∠A + ∠C = 180° or ∠B + ∠D = 180°

AREA AND PERIMETER FOR

DIFFERENT SHAPES

Triangle: • Perimeter (P) = 2 × (length + breadth)

Formulae given below are valid for all types = 2 × (l + b)
• Area (A) = length × breadth = l × b
of triangles.
• Length of Diagonal = √𝑙 2 + 𝑏 2
Perimeter = (a + b + c)
1
Area = × (base × height)
1
2
Square:
= ×a×h
2
Heron’s Formula:
• A = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
1
where s = × (a + b + c) or semi-perimeter
2
of the triangle
Right Angled Triangle:
In a right angled triangle, by • Perimeter (P) = 4 × (side) = 4a
Pythagoras theorem • Area (A) of a square = (side)2 = a2
(Hypotenuse)2 = sum of • Length of diagonal = √2 × 𝑆𝑖𝑑𝑒 = 𝑎√2
squares of sides i.e. (𝐷𝑖𝑎𝑔𝑜𝑛𝑎𝑙)2 𝑑2
• Area (A) of a square = =
ℎ2 = 𝑎 2 + 𝑏 2 2 2
(𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟)2 𝑃2
1
Area = A = × (base × height) = × a × b
1 • Area (A) of a square = =
2 2 16 16

15
Trapezium (Trapezoid): [where, A1 = area of base or top;
Area (A) of a trapezium
1
A2 = area of one side face, and A3 = area of
= × (sum of parallel sides) × other side face]
2
(perpendicular distance • Total Surface Area = 2(lb + bh + lh)
between the parallel sides)
1
= (l + b + h)2 – d2
= × (a + b) × h • Area of four walls of a room = Lateral
2

Circles: Surface area = 2(l + b)h

If r is the radius of a circle, then Cube:
• Perimeter or Circumference = 2πr or πd, If ‘a’ be the edge of a cube, then
where d = 2r is the diameter of the circle. • Body diagonal of the cube =
𝜋𝑑 2
• Area = πr2 or d = √3𝑎
4
𝜋𝑟 2 • Volume of the cube =
• Area of semi-circle =
2 (edge)3 = a3
𝜋𝑟 2
• Area of a quadrant of a circle = • Total surface area of the cube = 6 (edge)2 = 6a2
4
Area enclosed by two Concentric Circles: • Lateral Surface area = 4 (edge)2 = 4a2
If R and r are radii of two con- Right Circular Cylinder:
centric circles, then: If ‘r’ is the radius of base and ‘h’ is the
Area enclosed by the two height of the cylinder
circles (Area of ring) Volume of cylinder = Area of the
base × height = π r2 × h = π r2 h
= Area of outer circle – Area of
Curved surface area = Circumference of
inner circle
the base × height = 2 π r × h = 2πrh
= πR2 – πr2 = π (R2 – r2) = π (R + r) (R – r) • Total surface area = Area of the curved
Rotations and Revolutions: surface + Area of the two circular ends
• Distance moved by a rotating wheel in one = 2πrh + 2πr2 = 2πr (h + r)
revolution is equal to the circumference of Hollow Right Circular Cylinder:
the wheel. Let external radius = R, Internal radius = r and
• The number of revolutions completed by a height = h
rotating wheel in time ‘t’
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑚𝑜𝑣𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑤ℎ𝑒𝑒𝑙 𝑖𝑛 𝑡𝑖𝑚𝑒 ′𝑡′ • Outer curved surface area = 2πRh
= • Inner curved surface area = 2πrh
𝐶𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑤ℎ𝑒𝑒𝑙
• Total curved surface area = Outer
SURFACE AREA AND VOLUME curved surface area + Inner curved
surface area
Cuboid: = 2πRh + 2πrh = 2πh(R + r)
If ‘l’, ‘b’ and ‘h’ denote the • Area of upper and lower cross-section
length, breadth and = πR2 – πr2 = π(R2 – r2)
height of the cuboid and • Total surface area = C.S.A. of hollow
‘d’ denotes the body cylinder + Area of 2 circular end rings.
diagonal, then: = 2πh(R + r) + 2π(R2 – r2)
• Volume = Volume of outer cylinder –
• d = √𝑙 2 + 𝑏 2 + ℎ2 Volume of inner cylinder
• Volume = l × b × h = √𝐴1 × 𝐴2 × 𝐴3 = πR2h – πr2 h = π(R2 – r2)h

16
Right Circular Cone: called the frustum. The frustum of a cone has
If r = radius of base, h = height, two parallel and congruent bases (circles).
l = slant height ‘r’ = radius of top, ‘R’ = radius of bottom,
• l = slant height = √ℎ2
+ 𝑟2 ‘L’ = slant height and H = Height
• Volume of cone So, Curved surface area = 𝛱𝐿(𝑅 + 𝑟)
1
= × area of the base × height Total surface area = 𝜋𝐿(𝑅 + 𝑟) + 𝜋𝑟 2 + 𝜋𝑅 2
3
1
= × πr2h Volume =
𝜋𝐻
× (𝑟 2 + 𝑅 2 + 𝑅𝑟)
3
3
• Area of curved surface = π r l
Also, L = √𝐻2 + (𝑅 − 𝑟)2
• Total surface area of cone = Area of the
base + area of the curved surface Sphere:
= πr2 + πrl = πr(r + l) If ‘r’ = radius of the sphere, then
Frustum of a Cone: • Volume of sphere = π𝑟 3
4
3
A frustum is a three-
• Surface area = 4πr 2
dimensional geometric
shape that results from Hemisphere:
2
cutting a cone into two parts • Volume of hemisphere = π𝑟 3
3
with a plane parallel to its • Curved surface Area = 2πr2
base. The portion that • Total surface area = 3πr2
remains after the cut is

17 DATA INTERPRETATION
100 5
A pie chart either represent the distribution And, 1𝑜 = ( ) % = ( )%
360 18
100% of the data in terms of percentage or 360𝑜
So, 𝑥% = 𝑥 × 3.6𝑜
of the data in terms of degrees. 5
So, 100% = 360𝑜 And, 𝑥 𝑜 = ( ) × 𝑥%
18

So, 1% = 3.6𝑜

18 SERIES
The Number Series asked in CSAT exam Series based on addition of odd multiples
primarily follows the following patterns: of 10: 10, 20, 50, 100, 170
• Series following constant addition Series based on addition of even multiples
/subtraction of 20: 20, 40, 80, 140, 220
Series based on constant addition of ‘8’: • Series following multiplication and division.
12, 20, 28, 36, 44, 52 Series based on multiplication:
8, 16, 48, 192, 460
Series based on constant subtraction of ‘6’:
Series based on division: 360, 60, 12, 3, 1
36, 30, 24, 18, 12, 6
Series based on addition of prime numbers:
• Series following addition and subtraction
12, 25, 42, 61, 84
of multiples of a certain number Series based on addition of squares of
Series based on addition of consecutive consecutive numbers:
multiple of 12 : 20, 32, 56, 92, 140 100, 125, 161, 210, 274

17
 Series based on addition of cubes of
consecutive numbers: 10, 74, 199, 415, 758
• In some cases the pattern can be seen after
calculating the double difference and the
pattern in the double difference can follow
any of the above given patterns.
CONTINUOUS ALPHABETICAL
SERIES

In continuous series, firstly count the total

given letters and try to divide the series into
equal parts to find the possible patterns in the
given series. For example, a series with 15
• Some series do contain two series which letters can be divided into smaller parts of 3 or
contain different patterns. You need to find 5 letters to find the pattern in the series.
the pattern followed in the given two series cccbb_aa_cc_bbbaa_
separately and then find the missing or the ⇒ ccc / bb_ / aa_ / cc_ / bbb / aa_
odd one out number. ⇒ ccc / bbb / aaa / ccc / bbb / aaa

19 CODING & DECODING

POSITION OF ALPHABETS Opposite letters
A B C D E F G H I J K L M
In forward order:
Z Y X W V U T S R Q P O N
A B C D E F G H I J K L M

1 2 3 4 5 6 7 8 9 10 11 12 13
Tricks to remember the position of
Alphabet:
Remembering the position of each alphabet
N O P Q R S T U V W X Y Z can be quite challenging, so we often create
simple tricks to help us recall their place values
14 15 16 17 18 19 20 21 22 23 24 25 26
easily. Some of them are discussed below:
In backward order Trick 1: CFILORUX and EJOTY Formula:
A B C D E F G H I J K L M C F I L O R U X

26 25 24 23 22 21 20 19 18 17 16 15 14 3 6 9 12 15 16 18 21

Things will become easier if you remember

N O P Q R S T U V W X Y Z and use these formulas. In the 1st formula, we
have the position values of C, F, I, L, O, R, U,
13 12 11 10 9 8 7 6 5 4 3 2 1
and X, which are the multiples of THREE,
Backward position of letter = 27 - forward which will make it easier to remember the
position of letter.

18
position value of the alphabet in the forward Trick 2: To find the position value in reverse
direction order, we can use the 1st formula by reversing
the alphabet as shown in the figure below:
E J O T Y

5 10 15 20 25

The position values of E, J, O, T and Y, which

are the multiples of FIVE.

20 CLOCK
ANGLE BETWEEN MINUTE AND POSITION OF CLOCK HANDS
HOUR HAND AT ANY POINT OF
Clock Hands Overlap: The hands of a clock
TIME
overlap (form a zero-degree angle) 11 times in
𝟏𝟏
12 hours, occurring once every hour, except for
If M > H; 𝜃 = × 𝑴 − 𝟑𝟎 × 𝑯 the overlap at exactly 12 o'clock between 12:00
𝟐
Here, M = minutes and H = hour and 1:00.
𝟏𝟏
If M < H; 𝜃 = 𝟑𝟎 × 𝑯 − × 𝑴 Clock Hands Form a Right Angle: The hands
𝟐
form a 90-degree angle 22 times in 12 hours,
Here, M = minutes and H = hour
typically twice every hour, except between 2
o'clock and 4 o'clock and 8 o'clock and 10
TIME GAINED OR LOST
o'clock, when they form a right angle three
times.
If the clock is fast then incorrect time showed
by the clock = correct time + fastness Clock Hands Are Opposite (180° apart): The
hands are 180° apart 11 times in 12 hours,
If the clock is slow then incorrect time showed
occurring once every hour, except at 6 o'clock
by the clock = correct time – slowness
when they are exactly opposite during both 5
o'clock and 7 o'clock.
MIRROR IMAGE OF CLOCK
Angle Angle Angle
between between between
Mirror image of time minute and minute and minute and
= (11 hr : 60 min) – (Given time) hour hand hour hand hour hand
will be 0𝑜 will be 90𝑜 will be 180𝑜
Example: If the time in a clock is 5 hr. 45 min.
then what time does it show on the mirror? In 12 hours 11 22 11
period
Explanation: Mirror image of time
= (11 hr : 60 min) – (Given time) In 24 hours 22 44 22
period
= (11 hr : 60 min) – (5 hr : 45 min)= 6 hr : 15 min

19
21 CALENDAR
• Ordinary Year: An ordinary year is a year Odd days in Century:
that has 365 days (52 weeks + 1 extra/odd Year Ordinary Years Leap Years Odd Days
day). In an ordinary year, the month of 100 76 24 5
February has 28 days.
200 152 48 3
• Leap Year: A leap year is a year that has
300 228 72 1
366 days (52 weeks + 2 extra/odd days).
There is an extra day, 29th February, in 400 303 97 0
addition to the usual 28 days in February. In an ordinary year, 1 day is gained when we
move forward by one year.
ODD DAYS In an ordinary year, 1 day is lost when we
move backwards by one year.
• A normal year is considered a leap year if
In a leap year, 2 days are gained when we
it is divisible by 4. e.g., 2020, 2024, etc.
proceed forward by one year. (When we cross
• For a century year (multiple of 100) to be a
leap february)
leap year, it should be divisible by 400 also,
e.g., 2000, 2400, etc. In an leap year, 2 day are lost when we move
backwards by one year. (When we cross leap
Months Odd days Months Odd days
february)
January 3 July 3
0 (ordinary year)
Repetition of calender year:
February August 3
1 (leap year) Year Repetition after years
March 3 September 2 Leaps year 28
April 2 October 3 Leap year + 1 6
May 3 November 2 Leap year + 2 11
June 2 December 3 Leap year + 3 11

22 AGE BASED PROBLEMS

• If present age of a person is ‘x’ years, then 𝑀+𝑛
=
𝑝
𝑁+𝑛 𝑞
Age after ‘n’ years = (x + n) years
Age ‘n’ years ago = (x – n) years • If ratio of present ages of ‘A’ and ‘B’ is m:n,
• If present average age of a group of ‘n’ respectively and after ‘t’ years the ratio of
persons is ‘m’ years and ‘p’ persons having their ages will become a:b, then
average age ‘x’ years joins the group, then
new average age of the group First, we are required to suppose that the
=
𝑛×𝑚+𝑝×𝑥
years present ages of ‘A’ and ‘B’ is ‘mx’ years and
𝑛+𝑝
If present ages of ‘A’ and ‘B’ is ‘M’ years ‘nx’ years, respectively. So,
𝑚𝑥 + 𝑡 𝑎
and ‘N’ years, respectively and the ratio of =
𝑛𝑥 + 𝑡 𝑏
their ages after ‘n’ years becomes p:q, then

20
23 DIRECTION & DISTANCE
MAIN DIRECTIONS Angle formed between two main directions is 90°
and the angle formed between the main direction
and the intermediate direction is 45°. Following
figure will help us to understand it more.

SHORTEST DISTANCE

To find the shortest distance we need to know

about Pythagoras theorem:
INTERMEDIATE DIRECTIONS
Here, AB = perpendicular,
There are 4 Intermediate BC = Base
directions: AC = Hypotenuse
North - East (NE), Angle B = 90°
North - West (NW), Now, by Pythagoras theorem shortest distance
South - West (SW) between point A and C
South - East (SE) 𝐴𝐶 = √(𝐴𝐵)2 + (𝐵𝐶)2

24 SITTING ARRANGEMENT

LINEAR ARRANGEMENT CIRCULAR ARRANGEMENT

In the Linear Arrangement type of puzzles,
you need to arrange objects or people in a
straight line. This line can be either horizontal
(side by side) or vertical (one above the other).
The following facts are necessary to solve
these kinds of questions:
If A, B, C and D are facing toward the south
direction and P, Q, R and S are facing toward
the north direction in a line then positions at
their left and the right will be as shown in the
following figure:

21
25 RANKING
POSITION TEST • Maximum number of persons in a row
= (Position of Person 1 from Left/Top)
• Rank/Position from Left/Top + (Position of Person 2 from Right/Bottom)
= Total – Rank from Right/Bottom + 1 + (Middle Places)
• Rank/Position from Right/ Bottom • Minimum number of persons in a row
= Total – Rank fromLeft/Top + 1 = (Position of Person 1 from Left/Top)
• Total Number of persons in Queue/Row + (Position of Person 2 from Right/Bottom)
= Position from Left + Position from Right – 1 – (Middle Places) – 2

26 VENN DIAGRAM

Venn Diagram Relationship

This diagram shows that a group is totally inserted into other but are
not equal.
For example: The relationship between tree and banyan tree.

This diagram indicates that two groups have no connection, and

there are no shared elements between them.
For example: Relationship between trees and animals.

This diagram illustrates that no single group is completely within

another, but they are connected to each other to some extent.
For example: Teachers and Authors.

This diagram illustrates that three groups have no connection, and

there are no shared elements between them.
For example: Men, Animals and Birds.

This diagram illustrates that two groups are somewhat connected to

the third group but are independent of each other.
For example: Employed, Graduate and Unemployed.

22
 This diagram shows three separate groups partly related to each
other.
For example: Persons who speak Hindi, English and Punjabi.

This diagram demonstrates that two distinct things are both part of a
larger group.
For example: Human, Doctors and Engineers.

This diagram indicates that two sets are part of a larger group, and
there are shared items between these two sets within the larger
group.
For example: Human, Graduate and Employed.

If one group is inside another group, and a third group is completely

separate from the first two, you can illustrate this relationship with a
diagram.
For example: Male, Boys and Female.

When one group is part of another group, and a third group is related
to both of them to some extent, you can represent this relationship
with this diagram.
For example: Triangle, Right angle triangle and Isosceles triangle.

VENN DIAGRAM FOR 2 VARIABLES Total number of people in the village (N)
= Number of people who like only Android
There are ‘N’ number of people in a village + Number of people who like only Apple
such that each of them like at least one of the + Number of people who like Android and
two phones i.e. android or apple. The Apple
representation of this type of data can be given
n(A) = Total number of people who like Apple
as follows:
Number of people
n(A) = ( ) + n(A ∩ B)
who like only Apple
n(B) = Total number of people who like Android
Number of people
n(B) = ( ) + n(A ∩ B)
who like only Android
n(A ∩ B) = Number of people who like both
Apple and Android
n(A U B) = Total number of people in the
village = N
n(A U B) = n(A) + n(B) – n(A ∩ B)

23
VENN DIAGRAM FOR 3 VARIABLES • n(C ∩ A) = Number of
students who like both
Maths and English
A survey was conducted in a school of ‘Z’
• n(A ∩ B ∩ C) = Number
students such that each student in the school
of students who like all
like one of the three subjects i.e. English, Hindi three subjects.
or Maths. The given figure shows the pictorial • ‘a’ = Number of students
representation of the same. who like only English
• ‘b’ = Number of students who like only Hindi
• ‘c’ = Number of people who like only Maths
• ‘d’ = Number of students who like both
English and Hindi but not Maths
• ‘e’ = Number of students who like both
English and Maths but not Hindi
• ‘f’ = Number of students who like both
n(A) n(A ∩ B) Maths nd Hindi but not English
+ + • ‘g’ = Number of students who like all three
n (A ∪ B ∪ C) = n(B) − n(B ∩ C) + n(A ∩ B ∩ C) subjects.
+ +
Total number of students surveyed
[ n(C) ] [ n(C ∩ A)]
= Z = (a + b + c) + (d + e + f) + g
Here,
Here,
• n(A ∪ B ∪ C) = Z
• (a + b + c) = Number of students who like
= Total number of students in the school.
exactly one subject.
• n(A) = Number of students who like
• (d + e + f) = Number of students who like
English
exactly two subjects
• n(B) = Number of students who like Hindi
• Total number of students surveyed
• n(C) = Number of students who like Maths
Number of Number of Number of
• n(A ∩ B) = Number of students who like students students students
bothEnglish and Hindi = who like + who like + who like
• n(B ∩ C) = Number of students who like exactly exactly all
both Hindi and Maths [ one subject] [ two subjects] [ three subjects. ]

27 INEQUALITIES
Priority of the inequality symbol is as: Statement Conclusion
>, ≥ and = A=B≥C A ≥ C or A > C or A = C
Statement and Conclusion: A≥B≥C A ≥ C or A > C or A = C
Statement Conclusion A>B<C NO CONCLUSION
A>B>C A>C
A≥B<C NO CONCLUSION
A≥B>C A>C
A>B≥C A>C A>B≤C NO CONCLUSION
A=B>C A>C A<B>C NO CONCLUSION
A>B=C A>C A≤B>C NO CONCLUSION
A≥B=C A ≥ C or A > C or A = C A<B≥C NO CONCLUSION

24
28 SYLLOGISM
Statement Venn Diagram Statement Venn Diagram

All ‘B’ are ‘A’ and All ‘C’

Some ‘A’ are ‘B’
are ‘A’

Some ‘A’ are ‘B’ and

No ‘A’ are ‘B’
Some ‘B’ are ‘C’

All ‘A’ are ‘B’ and

All ‘A’ are ‘B’ No ‘B’ are ‘C’

All ‘A’ are ‘B’

All ‘A’ are ‘B’ and and
All ‘B’ are ‘C’ No ‘A’ are ‘C’

Some ‘A’ are ‘B’ and Some ‘A’ are ‘B’ and
All ‘B’ are ‘C’ No ‘B’ are ‘C’

All ‘A’ are ‘B’ and Some ‘A’ are ‘B’ and
Some ‘B’ are ‘C’ No ‘A’ are ‘C’

29 CUBES & DICE

If a larger cube of edge length ‘M’ units is • Number of cubes with exactly two faces
painted from all sides and is cut into ‘n’ painted = Number of cubes on the edges
identical smaller cubes of edge length ‘N’ = 12(n – 2)
units, then • Number of cubes with exactly three faces
𝑴×𝑴×𝑴 painted = Number of cubes on the vertices
n=
𝑵×𝑵×𝑵 =8
• Total number of cubes = 𝒏𝟑
• Number of cubes with ‘0’ face painted CUTTING OF A CUBOID
= Cubes which are in second layer or inside
= (𝒏 − 𝟐)𝟑 • When a cuboid of the L × B × H, is cut into
smaller cubes (edge length ‘a’) of equal
• Number of cubes with exactly one face
volume, then total number of cubes formed
painted = Cubes on the surface or face L×B×H
=
= 𝟔(𝒏 − 𝟐)𝟐 3𝑎

25
If a cuboid of dimension L × B × H is painted When digits move clockwise.
on all sides and is then cut into smaller cubes
of dimension 1 × 1 × 1, then
Number of cubes with ‘0’ face painted =
Cubes which are in second layer or inside
• Case 2: Two Common Digits
layer = (L – 2) × (B – 2) × (H – 2) When two numbers are common in two
• Number of cubes with exactly one face positions of the dice, the remaining
painted = Cubes on the surface or face uncommon numbers are opposite to each
= 2 × [(L – 2) × (B – 2) + (B – 2) × (H – 2) other.
+ (L – 2) × (H – 2)] For example: If 5 and 6 are common in two
• Number of cubes with exactly two faces positions, and the remaining numbers are
painted = Number of cubes on the edges 3 and 4, then 3 is opposite to 4.
= 4 × (L + B + H – 6)
• Number of cubes with exactly three faces
painted = Number of cubes on the vertices = 8

DICE When dice is cut open the alternate places

Finding Digits / Dots / Words / Letters / will be opposite to each other
Figures/Symbols on Opposite Faces of a Dice
• Case 1: Common Digit on Different Faces
When the same number appears on
different faces in two dice positions, rotate
the dice from the common number in a
clockwise direction.
For example: If 3 is the common digit, and Here the opposites are 2-4, 1-6 and 3-5
rotating clockwise shows 1 and 2 on one
dice and 4 and 5 on another, then 2 is
opposite 5, and 3 is opposite 6.

30 READING COMPREHENSION
HOW TO APPROACH READING understanding the main idea and the
supporting details.
COMPREHENSION IN CSAT?
• Highlight or underline important
Solving Reading comprehension in the CSAT information: When you come across
can be challenging, but here are some tips that important information, such as dates, names,
you can use to improve your performance and key concepts, highlight or underline
• Improve your reading speed and them. This will help you to find them quickly
comprehension skills: Reading is a skill when you need to answer the questions.
that you can improve with practice. Try to • Identify the type of question being asked:
read as much as possible and pay attention There are different types of questions that
to the structure of the passages. Focus on can be asked, such as factual questions,
inference questions, and vocabulary

26
 questions.Understanding the type of OPTION ELIMINATION IN
question being asked will help you to
approach the question correctly.
READING COMPREHENSION
• Practise previous year question papers: • In any exam which tests your reading
Practising previous year question papers comprehension, including UPSC-CSAT, there
will give you an idea of the type of are five primary criterias which helps you
questions that can be asked in the exam. decide that a particular option is incorrect.
This will also help you to develop your • Often, it is observed in UPSC-CSAT that a
time management skills. candidate is able to eliminate two options
• Look for context clues: Sometimes, comfortably. But UPSC gives very close
unfamiliar words or phrases may appear in choices for the remaining two options
the passage. If you come across such which makes it really difficult to eliminate
words, look for context clues such as the third option to arrive at the answer, in
synonyms or antonyms, to help you a limited time frame.
understand the meaning of the word or • In such situations having predefined
phrase. criteria as to in which all ways an option
• Focus on the structure of the passage: The can be incorrect will give candidates much
structure of the passage can give you needed clarity of thought and ultimately
important clues about the author's purpose increase their speed in solving reading
and main ideas. Look for transition words, comprehension based questions.
such as "however" and "therefore," to
• However, candidates should keep in mind
understand the relationships between
that first they should always attempt to
different ideas.
arrive at the right answer without thinking
• Make educated guesses: If you are unsure about wrong options. If that does not work,
about an answer, try to make an educated then only one should resort to elimination
guess by eliminating obviously incorrect
techniques.
answers and making an educated guess
based on the information you have. FIVE CRITERIA TO ELIMINATE
• Practice active reading: When reading a OPTIONS
passage, try to actively engage with the
text by asking yourself questions and 1. Out of the Scope of Passage
making connections to your own • This refers to options, which are not even
experiences and knowledge. This will help mentioned in the passage explicitly or
you to understand the passage better and implicitly, that is, you find no direct or
indirect reference of the option in the passage.
retain the information more effectively.
Let us consider following example to
• Manage your time wisely: Time understand this:
management is crucial in the UPSC CSE
Passage
CSAT exam. Make sure to allocate your
Humans often seek change due to a
time wisely, and don't spend too much
combination of innate curiosity, a desire for
time on any one question. If you get stuck improvement, and a coping mechanism for
on a question,move on and come back to it discomfort or dissatisfaction. Evolutionarily,
later if you have time. adaptability has been crucial for survival,
Apart from these tips, as a matter of last driving the inclination to modify
resort; elimination techniques, as discussed circumstances. In this ever-evolving journey of
below, also help candidates to arrive at the life, the Japanese philosophy of Uketamo
correct answer in a limited time period. emerges as a beacon of light. Uketamo,
translated loosely as “to accept.”

27
The philosophy teaches that all things, Q. Based on the above passage, which of the
whether positive or challenging, are fleeting. It following assumption(s) is/are valid?
encourages individuals to acknowledge and (a) AI will ultimately enable renewable
receive both the joys and sorrows of life technology to replace fossil fuels
without resistance, fostering a profound sense completly in the coming future.
of equilibrium. (b) Data processing requires extensive
Q. Which of the following statements most computer infrastructure powered by fossil
accurately represents the central idea of the fuels.
passage?
(c) Manufacturing of hardware components
(a) Changes in human lives are often desirable. for AI systems heavily pollutes the
(b) Changes should be avoided to bring about environment.
the sense of equilibrium. (d) Both (b) & (c)
(c) Acceptance towards life’s ups and downs Answer: (d)
can lead to peace.
Here, option (a) can be eliminated since it makes
(d) Depression is a modern day phenomenon, an extreme assumption. At First, the extreme
which was not even there in the past. assumption is that AI will enable renewable
Answer: (c) energy to replace fossil fuels. We cannot
Here, one can easily eliminate option (d) since conclude this based on the information given in
it is out of the scope of the passage. Passage the passage. Passage only talks about AI
talks about change and how Uketamo helps making renewable energy technology more
individuals to deal with the changes. While efficient.
option (d) talks about depression, which finds Secondly, Way too long logic is applied here in
no direct or indirect mention in the passage, assuming whether renewable energy will
hence this option is out of the scope of the replace fossil fuels or not. This is nowhere
passage. explicitly mentioned in the passage. Hence,
2. Extreme options this option is an extreme option.
• This refers to an option which has been 3. Partially Correct
arrived at after making extreme • This refers to options which may appear
assumptions or very long inferences with true at first place, but only a part of the
respect to the content given in the passage. option, say a word or two makes this
• To put it in simpler terms too one needs to option incorrect.
apply way too many logics in addition to • Hence to eliminate such options a careful,
the information given in the passage to see word to word reading is required. Part,
this particular option as true. which makes this option incorrect, can be
Let us consider following example to an extreme word or something not
understand this: mentioned in the passage even implicitly.
Passage Let us consider following example to
Recent advancements in artificial intelligence understand this:
(AI) have led to significant progress in the field Passage
of renewable energy. AI algorithms are helping In the landmark Keshavananda Bharti Case,
researchers design more efficient solar panels, the Basic Structure Doctrine emerged as a
predict wind patterns for better turbine judicial doctrine safeguarding the core
placement, and develop smart grids that principles of the Constitution. The judiciary
optimise energy distribution. While the asserted that while Parliament has the power
potential of AI for tackling climate change is to amend the Constitution, it cannot alter its
undeniable, some experts warn that its basic structure. This doctrine, articulated by
increasing dependence on fossil fuels for data the Supreme Court of India, ensures the
processing and hardware production could preservation of essential features such as
ultimately undermine its environmental democracy, rule of law, and individual
benefits. liberties. Keshavananda Bharti's case marked a

28
pivotal moment in constitutional Aurelius. Just as each organ in our body has its
jurisprudence, establishing a framework that own individual function, but always towards
guards against arbitrary amendments the wellbeing of the whole; each of us has a
threatening the foundational integrity of the
role, and any action that is not in the benefit of
Constitution.
the collective, ultimately cannot benefit the
Q. Which of the following assumptions is
implicit in the passage? individual. Only when we truly learn to
(a) The Basic Structure Doctrine may require recognise that we are not separate from nature
periodic review to adapt to evolving but a part of this one life, can we positively
constitutional principles. alter the way we consume,
(b) In the landmark Keshavananda Bharti Q. In the context of the passage, what is the
Case, the Basic Structure Doctrine emerged most logical inference that can be drawn?
as the only judicial doctrine safeguarding
the core principles of the Constitution. (a) Recent emphasis on individual rights goes
(c) The preservation of essential features like against the idea of collective benefits.
democracy and individual liberties is a (b) Collective benefits should be placed over
universally uncontested aspect of
individual benefits.
constitutional jurisprudence.
(d) The Basic Structure Doctrine, as articulated (c) Each one of us has a role which ultimately
in the Keshavananda Bharti Case, is leads to the benefit of the collective
immune to legal challenges.
(d) We are separate from nature but we must
Answer: (a)
act in sync with nature.
Here, we can eliminate option (b) since it is a
rotten fruit. At first it appears correct since it is Answer: (b)
explicitly mentioned in the opening remarks of Here, at first instance option (c) may appear
the passage. But only one word ‘only’ makes correct since it is explicitly mentioned in the
this option an incorrect one. passage. But upon careful reading of the
Based on the information given in the passage, question we recognize that the question is
we can not infer whether it is the ‘only’ judicial
asking for the most logical inference.
doctrine safeguarding core principles of the
constitution or not. There may be other such Option (c) is a straight fact stated based on
doctrines as well. information given in the passage. Inference is
4. Contextually Wrong Options something which we derive based on logical
• This refers to options which are true in and rational reasoning and is not mentioned
itself based on the information in the directly in the passage. Hence, option (c) is
passage, when examined independently. contextually wrong.
But they do not specifically answer the
question. 5. True in Real World but Incorrect as
• For instance, the question is asking the per passage
underlying tone of the passage, but the • This refers to options which are true in real
option simply states one of the facts world situations and normal
mentioned in the passage and not the circumstances. These options are true
underlying tone.
based on common sense and general
Let us consider following example to
knowledge.
understand this:
Passage • But based on the information given in the
Philosophers through the ages have constantly passage and the question that follows,
reminded us of this underlying universal these options are wrong. Hence one needs
principle that we are a small but integral part to be always aware of the fact that answers
of this web of life. “That which isn’t good for need to be based on the passage solely in
the hive, isn’t good for the bee,” said Marcus order to eliminate such options.

29
Let us consider following example to (c) Balancing affordability, renewable
understand this: integration, and climate goals will define
Passage the CEA's future effectiveness.
As India's Central Electricity Authority (CEA) (d) CEA advises government on policy
marks its 50th year, its vital role in shaping the matters and formulates plans for
nation's energy backbone shines brightly. development of electricity systems
From fostering grid expansion to ensuring grid Answer: (c)
stability, the CEA has facilitated affordable
Here, if we rely on pre-existing general
electricity access for millions. Its technical
knowledge that we have based on our study of
expertise continues to be pivotal, even as the
GS subjects, options (a) and (d) are true
landscape shifts towards renewables and
climate concerns. Balancing these new But based on the information given in the
demands with its traditional strengths will be passage and the question that follows, , we
crucial for the CEA's future impact. simply can not determine whether these
Q. What is the most logical and crucial options are correct or not.
message conveyed by the passage? While attempting the CSAT, in a limited time
(a) CEA is a statutory organisation constituted frame, in a hurry it is a possibility that one
under Electricity Supply Act, 1948 marks one of these options as correct.
(b) Transitioning to renewables necessitates Especially if you are just skimming through the
dismantling the CEA's traditional passage and not reading both passage and the
framework and starting anew. question carefully.

30
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