SECTION – A (APTITUDE )

S.NO NAME PAGE NUMBER

1. NUMBER SYSTEM 1-10

2. LCM AND HCF 11-16

3. SQUARE AND CUBE ROOT 17-21

4. AVERAGE 22-25

5. AGES 26-29

6. RATIO AND PROPORTION 30-34

7. PERCENTAGES 35-41

8. MENSURATION 42-54

9. TIME AND WORK 55-61

10. TIME AND DISTANCE 62-68

11. SIMPLE INTEREST 69-71

12. COMPOUND INTEREST 72-78

13. PROFIT AND LOSS 79-85

14. ALGEBRA 86-103

15. GEOMETRY 104-114

16. TRIGNOMETRY 115-123

17. STATISTICS 124-126

18. DATA INTERPRETATION 127-131

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1.NUMBER SYSTEM
Numbers:

In Decimal number system, there are ten symbols namely 0,1,2,3,4,5,6,7,8

and 9 called digits. A number is denoted by group of these digits called as
numerals.

Types of Numbers:

Natural Numbers:
The numbers that are used for counting are called Natural numbers.
Ex: 1, 2, 3, …..

Whole Numbers:
All the natural numbers including zero are called whole numbers.
Ex: 0, 1, 2, 3, 4, ……. Are whole numbers.

Integers:
All the whole numbers and negative of natural numbers are called Integers.
Ex: ………-3, -2, -1, 0, 1, 2, 3 ………. Are integers

Prime Numbers:
A number which has exactly two factors, 1 and the number itself, is called a
prime number.
e.g.: 2, 3, 5, 7, ……….

Composite Numbers:
A number which has atleast one factor other than 1 and the number itself is
called a composite number.
e.g.: 4, 8, 12, ……
Note:
1 is neither a composite nor a prime number.

Co-Prime:
Two numbers are said to be co-prime numbers if they do not have any
comman factor other than 1.
e.g.: 2 and 3

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Real Numbers:
All the numbers that are present on the number line is called real numbers.
1
e.g.: -1, 2, , √3
3
Rational Numbers:
𝑝
A number that can be written in the form where p and q are integers, co-
𝑞
prime number and q≠0 is called rational number.
E.g.: 2, 3/7, 22/7 etc

Irrational Numbers:
𝑝
A number that can not be written in the form where p and q are integers
𝑞
and q≠0 is called rational number. They have an infinite number non-repeating
decimal places.
e.g.: √2, 𝜋, etc

Face Value:
Face value of a digit in a numeral is value of the digit itself. For example in
321, face value of 1 is 1, face value of 2 is 2 and face value of 3 is 3.

Place Value
Place value of a digit in a numeral is value of the digit multiplied by 10 n where
n starts from 0.
For example in 321:
0
 Place value of 1 = 1 x 10 = 1 x 1 = 1
1
 Place value of 2 = 2 x 10 = 2 x 10 = 20
2
 Place value of 3 = 3 x 10 = 3 x 100 = 300
0th position digit is called unit digit and is the most commonly used topic in
aptitude tests.

Divisibility

Following are tips to check divisibility of numbers.

1. Divisibility by 2 - A number is divisible by 2 if its unit digit is 0,2,4,6 or 8.

Example: 64578 is divisible by 2 or not?
Solution:

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Step 1 - Unit digit is 8.

Result - 64578 is divisible by 2.

Example: 64575 is divisible by 2 or not?

Solution:
Step 1 - Unit digit is 5.
Result - 64575 is not divisible by 2.

2. Divisibility by 3 - A number is divisible by 3 if sum of its digits is

completely divisible by 3.
Example: 64578 is divisible by 3 or not?
Solution:
Step 1 - Sum of its digits is 6 + 4 + 5 + 7 + 8 = 30
which is divisible by 3.
Result - 64578 is divisible by 3.

Example: 64576 is divisible by 3 or not?

Solution:
Step 1 - Sum of its digits is 6 + 4 + 5 + 7 + 6 = 28
which is not divisible by 3.
Result - 64576 is not divisible by 3.

3. Divisibility by 4 - A number is divisible by 4 if number formed using its last

two digits is completely divisible by 4.
Example: 64578 is divisible by 4 or not?
Solution:
Step 1 - number formed using its last two digits is 78
which is not divisible by 4.
Result - 64578 is not divisible by 4.

Example: 64580 is divisible by 4 or not?

Solution:
Step 1 - number formed using its last two digits is 80
which is divisible by 4.
Result - 64580 is divisible by 4.

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4. Divisibility by 5 - A number is divisible by 5 if its unit digit is 0 or 5.

Example: 64578 is divisible by 5 or not?
Solution:
Step 1 - Unit digit is 8.
Result - 64578 is not divisible by 5.

Example: 64575 is divisible by 5 or not?

Solution:
Step 1 - Unit digit is 5.
Result - 64575 is divisible by 5.

5. Divisibility by 6 - A number is divisible by 6 if the number is divisible by

both 2 and 3.
Example: 64578 is divisible by 6 or not?
Solution:
Step 1 - Unit digit is 8. Number is divisible by 2.
Step 2 - Sum of its digits is 6 + 4 + 5 + 7 + 8 = 30
which is divisible by 3.
Result - 64578 is divisible by 6.

Example: 64576 is divisible by 6 or not?

Solution:
Step 1 - Unit digit is 8. Number is divisible by 2.
Step 2 - Sum of its digits is 6 + 4 + 5 + 7 + 6 = 28
which is not divisible by 3.
Result - 64576 is not divisible by 6.

6. Divisibility by 8 - A number is divisible by 8 if number formed using its last

three digits is completely divisible by 8.
Example: 64578 is divisible by 8 or not?
Solution:
Step 1 - number formed using its last three digits is 578
which is not divisible by 8.
Result - 64578 is not divisible by 8.

Example: 64576 is divisible by 8 or not?

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Solution:
Step 1 - number formed using its last three digits is 576
which is divisible by 8.
Result - 64576 is divisible by 8.

7. Divisibility by 9 - A number is divisible by 9 if sum of its digits is

completely divisible by 9.
Example: 64579 is divisible by 9 or not?
Solution:
Step 1 - Sum of its digits is 6 + 4 + 5 + 7 + 9 = 31
which is not divisible by 9.
Result - 64579 is not divisible by 9.

Example: 64575 is divisible by 9 or not?

Solution:
Step 1 - Sum of its digits is 6 + 4 + 5 + 7 + 5 = 27
which is divisible by 9.
Result - 64575 is divisible by 9.

8. Divisibility by 10 - A number is divisible by 10 if its unit digit is 0.

Example: 64575 is divisible by 10 or not?
Solution:
Step 1 - Unit digit is 5.
Result - 64578 is not divisible by 10.

Example: 64570 is divisible by 10 or not?

Solution:
Step 1 - Unit digit is 0.
Result - 64570 is divisible by 10.

9. Divisibility by 11 - A number is divisible by 11 if difference between sum

of digits at odd places and sum of digits at even places is either 0 or is
divisible by 11.
Example: 64575 is divisible by 11 or not?
Solution:
Step 1 - difference between sum of digits at odd places

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and sum of digits at even places = (6+5+5) - (4+7) = 5

which is not divisible by 11.
Result - 64575 is not divisible by 11.

Example: 64075 is divisible by 11 or not?

Solution:
Step 1 - difference between sum of digits at odd places
and sum of digits at even places = (6+0+5) - (4+7) = 0.
Result - 64075 is divisible by 11.

Division Algorithm

When a number is divided by another number then

Dividend = (Divisor x Quotient) + Reminder

Series

Following are formulaes for basic number series:

1. (1+2+3+...+n) = (1/2)n(n+1)
2. (12+22+32+...+n2) = (1/6)n(n+1)(2n+1)
3. (13+23+33+...+n3) = (1/4)n2(n+1)2

Basic Formulae

These are the basic formulae:

(a + b)2 = a2 + b2 + 2ab

(a - b)2 = a2 + b2 - 2ab

(a + b)2 - (a - b)2 = 4ab

(a + b)2 + (a - b)2 = 2(a2 + b2)

(a2 - b2) = (a + b)(a - b)

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

(a3 + b3) = (a + b)(a2 - ab + b2)

(a3 - b3) = (a - b)(a2 + ab + b2)

(a3 + b3 + c3 - 3abc) = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)

1. Which of the following is the output of (578 x 578 x 578 + 432 x 432 x
432) / (578 x 578 - 578 x 432 + 432 x 432) ?
A - 2000 B - 4000 C - 3000 D - 1000
Answer - D
Explanation
(578 x 578 x 578 + 432 x 432 x 432) / (578 x 578 - 578 x 432 + 432 x 432)
Let's have a = 578, b = 432
Now expression is (a3 + b3) / (a2 - ab + b2)
=a+b
= 578 + 432
= 1000
We've used following formula here:
a3 + b3 = (a + b)(a2 - ab + b2).

2. Which of the following is the output of (141 x 141 x 141 - 58 x 58 x 58) /

(141 x 141 + 141 x 58 + 58 x 58) ?
A - 83 B - 100 C - 90 D - 73
Answer - A
Explanation
(141 x 141 x 141 - 58 x 58 x 58) / (141 x 141 + 141 x 58 + 58 x 58)
Let's have a = 141, b = 58
Now expression is (a3 - b3) / (a2 + ab + b2)
=a-b
= 141 - 58
= 83
We've used following formula here:
a3 - b3 = (a - b)(a2 + ab + b2).

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3. Which of the following is the output of 213 x 213 + 187 x 187 ?

A - 50338 B - 80338 C - 90338 D - 70338
Answer - B
Explanation
213 x 213 + 187 x 187
Let's have a = 213, b = 187
Now expression is a2 + b2
Using following formula, (a + b)2 + (a - b)2 = 2 x (a2 + b2)
2 x ( 213 x 213 + 187 x 187) = (213 + 187)2 + (213 - 187)2
2 x ( 213 x 213 + 187 x 187) = 4002 + 262
2 x ( 213 x 213 + 187 x 187) = 160000 + 676
213 x 213 + 187 x 187 = 160676 / 2
= 80338

4. Which of the following is the output of ((637 + 478)2 - (637 - 478)2) /(637
x 478) ?
A-4 B-6 C-8 D - 24
Answer - C
Explanation
((637 + 478)2 - (637 - 478)2)/(637 x 478)
Let's have a = 637, b = 478
Now expression is ((a + b)2 - (a - b)2) / ab
= (a2 + b2 + 2ab - (a2 + b2 - 2ab)) / ab
= (a2 + b2 + 2ab - a2 - b2 + 2ab) / ab
= 4ab / ab
=4

5. Which of the following is the output of ((964 + 578)2 + (964 - 578)2) /(964
x 964 + 578 x 578) ?
A-4 B-6 C-8 D-2
Answer - D
Explanation
((964 + 578)2 + (964 - 578)2) /(964 x 964 + 578 x 578)
Let's have a = 964, b = 578
Now expression is ((a + b)2 + (a - b)2) / (a2 + b2)
= (a2 + b2 + 2ab + (a2 + b2 - 2ab)) / (a2 + b2)
= (a2 + b2 + 2ab + a2 + b2 - 2ab) / (a2 + b2)

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= 2(a2 + b2) / (a2 + b2)

=2

6. On dividing a number by 342, 47 is the remainder. What will be

remainder if same number is divided by 18?
A - 11 B-6 C-8 D-2
Answer - A
Explanation
Let's quotient is a and given number be b.
b = 342a + 47
= (18 x 19)a + 36 + 11
= (18 x 19)a + (18 x 2) + 11
= 18 x (19a + 2) + 11
Thus, if same number is divided by 18, remainder will be 11.
We've used following formulae here:
Dividend = (Divisor x Quotient) + Reminder

7. What will be unit digit in (3157)754?

A-8 B-9 C-7 D-6
Answer - B
Explanation
unit digit in (3157)754 = unit digit in (7)754
= unit digit in (74)188 x 72
= unit digit in (1 x 49) = 9
Thus Unit digit in (3157)754 is 9.
We've used following formulae here:
Unit digit in 71 = 7
Unit digit in 72 = 9
Unit digit in 73 = 3
Unit digit in 74 = 1
Unit digit in 75 = 7
Unit digit in 76 = 9
Unit digit in 77 = 3
Unit digit in 78 = 1
So pattern is 7-9-3-1. This pattern works for all numbers. So Unit digit in
((7)4)n) will be 1.

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8. What will be unit digit in 658 x 539 x 436 x 312?

A-8 B-9 C-4 D-6
Answer - C
Explanation
Multiply unit digits of each number.
Unit digit in 658 x 539 x 436 x 312
= Unit digit in 8 x 9 x 6 x 2.
= Unit digit in 864.
= 4.

9. What will be unit digit in 357 x 641 x 763?

A-8 B-9 C-4 D-6
Answer - C
Explanation
357 = (34)14 x 3
So Unit digit in 357
= Unit digit in 1 x 3
=3
641 = (64)10 x 6
So Unit digit in 641
= Unit digit in 6 x 6
=6
763 = (74)15 x 73
So Unit digit in 761
= Unit digit in 1 x 343
=3
So Unit digit in 357 x 641 x 763
= Unit digit in 3 x 6 x 3
=4
We've used following formulae here:
Unit digit in 34 = 1
Unit digit in 64 = 6
Unit digit in 74 = 1
So Unit digit - in ((3)4)n) will be 1.
- in ((6)4)n) will be 6.
- in ((7)4)n) will be 1.

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2.HCF AND LCM

1. Factors and Multiples:
If number a divided another number b exactly, we say that a is a factor of b.
In this case, b is called a multiple of a.
Example: 4 x 5 = 20
Here, 4 and 5 are factors of 20 and 20 is a multiple of both 4 and 5

2. Highest Common Factor (H.C.F.) or Greatest Common Measure (G.C.M.) or

Greatest Common Divisor (G.C.D.):
The H.C.F. of two or more than two numbers is the greatest number that divides
each of them exactly.

There are two methods of finding the H.C.F. of a given set of numbers:
I. Factorization Method: Express the each one of the given numbers as the product
of prime factors. The product of least powers of common prime factors gives
H.C.F.
II. Division Method: Suppose we have to find the H.C.F. of two given numbers,
divide the larger by the smaller one.
Now, divide the divisor by the remainder.
Repeat the process of dividing the preceding number by the remainder last
obtained till zero is obtained as remainder.
The last divisor is required H.C.F.
Finding the H.C.F. of more than two numbers: Suppose we have to find the H.C.F.
of three numbers, then, H.C.F. of [(H.C.F. of any two) and (the third number)]
gives the H.C.F. of three given number.
Similarly, the H.C.F. of more than three numbers may be obtained.

3. Least Common Multiple (L.C.M.):

The least number which is exactly divisible by each one of the given numbers is
called their L.C.M.

There are two methods of finding the L.C.M. of a given set of numbers:

I. Factorization Method: Resolve each one of the given numbers into a product of
prime factors. Then, L.C.M. is the product of highest powers of all the factors.
II. Division Method (short-cut): Arrange the given numbers in a row in any order.
Divide by a number which divided exactly at least two of the given numbers and
carry forward the numbers which are not divisible. Repeat the above process till no

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two of the numbers are divisible by the same number except 1. The product of the
divisors and the undivided numbers is the required L.C.M. of the given numbers.
4. Product of two numbers = Product of their H.C.F. and L.C.M.
5. Co-primes: Two numbers are said to be co-primes if their H.C.F. is 1.
6. H.C.F. and L.C.M. of Fractions:
HCF of Numerators
HCF =
L. C. M of Denominators
LCM of Numerators
LCM =
HCF of Denominators
7. H.C.F. and L.C.M. of Decimal Fractions:
In a given numbers, make the same number of decimal places by annexing zeros in
some numbers, if necessary. Considering these numbers without decimal point,
find H.C.F. or L.C.M. as the case may be. Now, in the result, mark off as many
decimal places as are there in each of the given numbers.

8. Comparison of Fractions:
Find the L.C.M. of the denominators of the given fractions. Convert each of the
fractions into an equivalent fraction with L.C.M as the denominator, by
multiplying both the numerator and denominator by the same number. The
resultant fraction with the greatest numerator is the greatest.

Things to remember:
The HCF of two or more numbers is smaller than or equal to the smallest
number of the given numbers.

The LCM of two or more numbers is greater than or equal to the greatest
number of the given numbers.
Let a : b is ratio of two numbers, where a and b are relatively prime.

Then LCM =HCF x a x b.

If the HCF of the numbers a,b,c is k then a,b,c can be written as multiples of
k(kx, ky, kz, where x,y,z are some numbers)
LCM of some numbers is always divisible by HCF of the numbers.

If the HCF of the numbers a,b is k, then the numbers (a+b), (a-b) is also
divisible by k.
Product of two numbers = (HCF of the two numbers) x (LCM of the two numbers)

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PROBLEMS:
1. Find the greatest number that will divide 72, 96, and 120 leaving the same
remainder in each case.
Solution:
To find the greatest number that will divide these numbers leaving the
same remainder, we need to find the Highest Common Factor (HCF) of the
differences between the numbers. The differences are as follows:
96 – 72 = 24
120 – 96 = 24
HCF(24, 24) = 24
Therefore, the greatest number that will divide 72, 96, and 120 leaving the
same remainder in each case is 24.
2. If the HCF of two numbers is 12 and their LCM is 360, find the numbers.
Solution:
We know that the product of the HCF and LCM of two numbers is equal to
the product of the two numbers.
So, for two numbers a and b with HCF = 12 and LCM = 360:
HCF × LCM = a × b
12 × 360 = a × b
4320 = a × b
Now, we need to find two numbers whose product is 4320 and HCF is 12.
There can be multiple pairs of numbers that satisfy this condition, and one
such pair is:
a = 120 and b = 36
Because 120 × 36 = 4320 and the HCF of 120 and 36 is 12.
3. Find the HCF of 36, 48, and 72.
Solution:
Prime factorization of 36: 22 × 32
Prime factorization of 48: 24 × 3
Prime factorization of 72: 23 × 32
Common factors: 22 and 3 (take the minimum power)
So, the HCF of 36, 48, and 72 is 22 × 3 = 12.
4. What is the largest three-digit number that is exactly divisible by the HCF
of 24 and 36?
Solution:
The HCF of 24 and 36 is 12. To find the largest three-digit number that is
exactly divisible by 12, we need to find the largest multiple of 12 that is
less than 1000.

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The largest multiple of 12 less than 1000 is 996 (83 × 12 = 996).

So, the largest three-digit number exactly divisible by the HCF of 24 and
36 is 996.
5. The sum of two numbers is 1001, and their HCF is 7. Find the numbers.
Solution:
a + b = 1001 (Sum of the two numbers)
HCF(a, b) = 7
We know that if the HCF of two numbers divides their sum, then it also
divides the difference of the two numbers.
So, a – b is divisible by 7.
Now, we look for pairs of numbers whose difference is divisible by 7 and
whose sum is 1001.
One such pair is a = 504 and b = 497.
504 – 497 = 7 (divisible by the HCF)
504 + 497 = 1001
Therefore, the two numbers are 504 and 497.
6. Find the HCF of 32 and 14 by Listing Factors Method
Solution:
First, list down the factors of 32 and 14.
 The factors of 32 are: 1, 2, 4, 8, 16, 32
 The factors of 14 are: 1, 2, 7, 14
We can see that 1, 2 are the only common factors of 32 and 14. Whereas 2
is the greatest among all the common factors.
Hence, HCF of 32 and 14 is 2.
7. Find the HCF of 80 and 90 by Prime Factorization
Solution:
 The prime factors of 80: 2 * 2 * 2 * 2 * 5;
 The prime factors of 90: 2 * 3 * 3 * 5.
We can see that 2, 5 are the only common factors of 80 and 90, Now, the
HCF of 80 and 90 will be the product of the common prime factors, which
are 2 and 5.
Hence, HCF of 80 and 90 is 10.
8. The LCM of the two numbers is 360, and their HCF is 24. If one of the
numbers is 120, find the other number.
Solution:
We know that the product of the HCF and LCM of two numbers is equal to
the product of the two numbers.

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So, for two numbers a and b with HCF = 24 and LCM = 360:
HCF × LCM = a × b
24 × 360 = 120 × b [Given a = 120]
(24 × 360)/120 = b
24 × 3 = b
Thus, b = 72.
9. A factory manufactures products in batches of 16, 24, and 32 units. What is
the minimum number of units the factory needs to produce so that each
batch can be formed exactly?
Solution:
To find the minimum number of units the factory needs to produce so that
each batch size (16, 24, and 32) can be formed exactly, we need to find the
least common multiple (LCM) of these batch sizes.
The prime factorization of each batch size is as follows:
 16 = 24
 24 = 23 × 3
 32 = 25
To find the LCM, we take the highest power of each prime factor that
appears in any of the batch sizes:
 The highest power of 2 is 25.
 The highest power of 3 is 31.
So, the LCM of 16, 24, and 32 is 25 × 31 = 32 × 3 = 96.
10.Find LCM of two positive integers 120 and 300 by Prime Factorization
Method
Solution:
 The prime factorization of 120 are: 2*2*2*3*5 = 23*31*51
 The prime factorization of 300 are: 2*2*3*5*5 = 22*31*52
Now, find the product of only those factors that have the highest powers
among these. This will be, 23 * 31 * 52 = 8 * 3 * 25 = 600
Hence, LCM(120, 300) = 600

11.Find the greatest number which on dividing 70 and 50 leaves

remainders 1 and 4 respectively.
Solution:
The required number leaves remainders 1 and 4 on dividing 70 and 50
respectively.
This means that the number exactly divides 69 and 46.
So, we need to find the HCF of 69 = 3 x 23 and 46 = 2 x 23.

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HCF (69, 46) = 23 Thus, 23 is the required number.

12.Find the largest number which divides 64, 136, and 238 to leave the
same remainder in each case.
Solution:
To find the required number, we need to find the HCF of (136-64), (238-
136), and (238-64), i.e., HCF (72, 102, 174).
72 = 23 x 32
102 = 2 x 3 x 17
174 = 2 x 3 x 29
Therefore, HCF (72, 102, 174) = 2 x 3 = 6
Hence, 6 is the required number.

13.Find the least number which when divided by 5,7,9 and 12, leaves the
same remainder of 3 in each case.
Solution:
In these types of questions, we need to find the LCM of the divisors and
add the common remainder (3) to it.
So, LCM (5, 7, 9, 12) = 1260
Therefore, required number = 1260 + 3 = 1263

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3.SQUARE AND CUBE ROOT

Square of a number:

Square of a number is the produt of the number with itself. Square of a

number n is denoted as n2.

Example: Square of 4 = 4 x 4 = 16

Square root:

The square root of a number is that number the product of which iself gives
the given number

Example: The square root of 400 is 20, the square root of 625 is 25.

Prime factorization method:

When a given number is a perfect square, we resolve it into prime factors

and take the product of prime factors, choosing one out of every two.

1. Find the square root of 4356.

2 4356
2 2178
3 1089
3 363
11 121
11 11
1

4356 = 2 x 2 x 3x 3 x 11 x 11=22 x 32 x 112

√4356 = 2 × 3 × 11 = 66

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Table of Unit digit of Square

Unit digit of 0 1 2 3 4 5 6 7 8 9
Number

Unit digit of 0 1 4 9 6 5 6 9 4 1
square of the
number

Cube :

When we multiply a number by its square then we get the cube of that
number.

Example: 23 = 2 x 2 x 2 = 8 (8 is the cube of 2)

Cube Root:

The cube root of a number is that number the cube of which itself gives the given
number

Example:

The cube root of 64 is 4.

The cube root of a number is denoted by the symbol ∛.

3
The expression √8 is read as ‘cube eight’ or the ‘cube root of eight’.

2. Find the cube root of 3375.

3 3375
3 1125
3 375
5 125
5 25
5 5
1

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3375 = 3 × 3 × 3 × 5 × 5 × 5 = 33 × 53
3
√3375 = 3 × 5 = 15

Number Square Cube

1 1 1
2 4 8
3 9 27
4 16 64
5 25 125
6 36 216
7 49 343
8 64 512
9 81 729
10 100 1000

SURDS:

As surds can be expressed with fractional exponents, the law of indices are
𝑛 𝑛
( √𝑎 ) = 𝑎
𝑛 𝑛 𝑛
√𝑎𝑏 = √𝑎 × √𝑏

𝑚 𝑛
√ 𝑛√𝑎 = 𝑚𝑛√𝑎 = √ 𝑚√𝑎

𝑛
𝑛𝑎 √𝑎
√ =𝑛
𝑏 √𝑏
𝑛 𝑚 𝑛
( √𝑎) = √𝑎𝑚

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3. Find the value of

a. 0.3 b. 0.7 c. 0.09 d. None of these

View solution
Correct Option: (a)

Step 1: First find 0.000729

Step 2:
The value of

4. 36 + 9 + (?)2 = 9.4

a. 3.5 b. 4 c. 4.4 d. 5
Correct Option: (b)
Let the unknown number be x.
Step 1: Firstly, find the value of 19.36
19.36 = 4.4
Step 2:
4.4 + 9 + (x)2 = 9.4
9 + (x)2 = 5
Squaring both the sides, we get
9 + x2 = 25
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x2 = 16
x=4
a
5. If 0.09 × 0.9 × a = 0.009 × 0.9 × b, then is _______
b
a. 9 × 10–3 b. 9 × 10–5 c. 81 × 10–4 d. 81 × 10–5
View solution
Correct Option: (d)
a
The given question can be written in the form of
b
a 0.009 × 0.9
=
b 0.09 × 0.9
Squaring both sides, we get
a 0.009 × 0.9 × 0.009 × 0.9 0.0000729 0.00729
= = = 0.00081 = 81 × 10 –5
b 0.09 × 0.9 0.09 9

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4.AVERAGE
AVERAGE :

The average of n quantities of the same kind is equal to the sum of all the
quantities divided by the number of quantities.
sum of all quantities
Average (of all quantities) =
number of quantities

It is also called the mean or mean value of all the quantities.

Sum of all quantities=Average × no. of quantities

 Average of ‘n’ consecutive natural numbers

(n + 1)
=
2
 Average of squares of consecutive natural numbers up to n
(n + 1)(2n + 1)
=
6

 Average of cubes of consecutive natural numbers up to n

n(n + 1)2
=
4
CONSECUTIVE EVEN NUMBERS:

 Average of odd numbers till ‘n’

n+1
=
2
 Average of even numbers till n
n+2
=
2
 Average of squares of ‘n’ consecutive even numbers
2(n + 1)(2n + 1)
=
3

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 Average of squares of consecutive even numbers till n

(n + 1)(n + 2)
=
3
CONSECUTIVE ODD NUMBERS:

 Average of n consecutive odd numbers(from 1 to n) =n

 Average of consecutive odd numbers till
(n + 1)
n=
2
 Average of squares of consecutive odd numbers till n =
n(n + 2)
n=
3
 If average of n1 observations is x1 and average of n2 observations is x2 then
the average of total observation n1+n2 is :
n1 x1 + n2 x2
n1 + n2
 I f the average of ‘n’ consecutive numbers/odd numbers/even numbers is x
then the first number is [x-(n-1)]

In case of increase in average,the age of person left

= previous average + no. of persons at present × increase in average

 In case of decrease in average, the age of the person left = previous average
– no. of person at presents × decrease in average.
 Number of passed candidates
Total no. of candidates(total average – failed average )
=
(passed average − failed average)

Number of failed candidates

Total no. of candidates(total average – failed average )
=
(passed average − failed average)

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Total distance travelled

 Average speed = k/h
total time taken
(or)

(d1 + d2 + − − − − − − +dn)
k/hr
[(d1/s1) + (d2/s2) + − − − − − − − − − − (dn/sn)]

 If two equal distances are travelled at the rate of x k/h and y k/h then the
average speed during the whole journey
(2xy)
k/hr
(x + y)

PROBLEMS:

1. A student calculates the arithmetic mean of the following 5 numbers: 10,

15, 20, 25 and x He found the mean is 15. Find out the value of x?
(1) 3 (2) 7 (3) 17 (4) 5
Solution:
Mean = (10+15+20+25+X)/5=15
70+X = 75
X=5

2. The average of the marks obtained in a mock test by 8 boys was 50 and by
2 girls was 80. The average marks of all 10 students were –
(1)50 (2) 56 (3) 60 (4) 62
Solution:
Sum of total number of 8 boys in mock = 8 × 50 = 400
Sum of total number of 2 girls in exam = 2 × 80 = 160
Required average = (400+160)/10 = 560/10 = 56

3. If the average weight of 5 girls is 20 kg; that of 3 girls is 15 kg, and that of
the other 2 girls is 25 kg; then the average weight of all girls is-
(1) 15 kg (2) 11.5 kg (3) 16.5 kg (4) 19.5 kg
Solution:
Required Average
= (20 X 5+ 15 X 3+ 25X2)/10
= (100+45+50)/10
= 195/10
= 19.5 kg

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4. Total daily salary of the writers in geeks for geeks is 15000. The average
daily salary of a writer is 750. Find the number of writers in geeks for
geeks is –
(1) 16 (2)14 (3) 20 (4)12
Solution:
Number of writers in geeks for geeks
= 15000/750 = 20

5. In 2021, the average monthly income of an employee was 3,000. For the
first nine months of the year, his average monthly income was 3100 and for
the last four months, it was 4,500. His income in the ninth month of the
year was –
(1) 16000 (2) 5080 (3) 8200 (4) 9900
Solution:
Employee’s income in the eighth month
= (3100 × 9 + 4 × 4500 – 12 × 3000)
= (27900 + 18000 – 36000)
= (45900 – 36000)
= 9900
6. The average yearly expenditure of a company for the first five years is
25700, for the next two years 24900 and for the last five years 30300. If the
company is saved 53200 during 12 years, the average yearly earnings of
the company are (approx value, not exact value)-
(1) 30000 (2) 31850 (3) 31917 (4) 35807
Solution:
Total expenditure of the company
= (5 × 25700 + 2 × 24900 + 5 × 30300)
= (128500 + 49800 + 151500)
= 329800
Total earning = (329800 + 53200) = 383000
Required average yearly earning = 383000/12 = 31917

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5.AGES
Important Formulae on "Problems on Ages" :
1. If the current age is x, then n times the age is nx.
2. If the current age is x, then age n years later/hence = x + n.
3. If the current age is x, then age n years ago = x - n.
4. The ages in a ratio a : b will be ax and bx.
𝟏 𝐱
5. If the current age x, then of the age is
𝐧 𝐧

PROBLEMS:
1. Seven years ago, the average age of a family of four members was 30 years.
Two babies having been born, the average age of the family remains the same
today. If the ages of babies differ by 2 years, what is the present age of the
younger baby?
a) 7 years b) 9 years c) 15 years d) 16 years
Solution:
7 years ago, total ages of 4 members = 30 × 4 = 120 years
Sum of the present ages of 4 members = 120 + (4 × 7) = 148 years
Sum of the present ages of 6 members = 30 × 6 = 180 years
Now, sum of the present age of 2 babies = 180 – 48 = 32 years
Again, let the present age of younger baby be x years and the present age of the
older baby be (x + 2) years.
Then, x + x +2 = 32
2x +2 = 32 or, 2x = 30
𝟑𝟎
x= ∴ x = 15 years
𝟐
Hence, the present age of the younger baby = 15 years.

2. The average age of Ram and Salmon is 20 years. If Mohan were to replace
Ram, the average would be 19 years. And if Mohan were to replace Salmon,
the average would be 21 years. What is the age of Mohan?
a) 30 years b) 15 years c) 35 years d) 20 years
Solution:
Let the ages of Ram, Salmon and Mohan be R, S and M.

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(𝟒𝟎 + 𝟑𝟖 + 𝟒𝟐)
∴𝐑+𝐒+𝐌= = 𝟔𝟎
𝟐
∴ Ages of Mohan = 60 – 40 = 20 years.

3. Rupali’s age 6 years hence will be equal to Sudha’s age 8 years ago. If the
respective ratio between Sudha’s present age and Harshad’s present age is 3:4
and Harshad’s present age is 36 years, what is Rupali’s present age? (in years)
a) 19 b) 2 c) 24 d) 13 e) 28
Solution:
𝐒𝐮𝐝𝐡𝐚 ∶ 𝐇𝐚𝐫𝐬𝐡𝐚𝐝 = 𝟑: 𝟒 = 𝟐𝟕: 𝟑𝟔
∴ 𝐏𝐫𝐞𝐬𝐞𝐧𝐭 𝐚𝐠𝐞 𝐨𝐟 𝐒𝐮𝐝𝐡𝐚 = 𝟐𝟕 𝐲𝐞𝐚𝐫𝐬

𝐒𝐮𝐝𝐡𝐚′ 𝐬 𝐚𝐠𝐞 𝟖 𝐲𝐞𝐚𝐫𝐬 𝐚𝐠𝐨 = 𝟐𝟕 − 𝟖 = 𝟏𝟗 𝐲𝐞𝐚𝐫𝐬

∴ 𝐑𝐮𝐩𝐚𝐥𝐢′ 𝐬 𝐩𝐫𝐞𝐬𝐞𝐧𝐭 𝐚𝐠𝐞 = 𝟏𝟗 − 𝟔 = 𝟏𝟑 𝐲𝐞𝐚𝐫𝐬

4. The age of Mohan’s father was 40 years at his birth. When Mohan becomes 8
his sister becomes 12. Mohan’s mother was 30 at his sister’s birth. What is the
ratio of the present age of Mohan’s mother to his father’s, if Mohan is 20 years
old now?
a) 9:10 b) 10:9 c) 11:10 d) 10:11
Solution:
Present age of Mohan = 20 years
Present age of his father = (40 +20) = 60 years
Difference between the age of Mohan’s sister and Mohan = (12 - 8) = 4years
Now, Mohan’s sister’s age = (20 + 4) = 24 years
Present age of her mother = (24 + 30) = 54 years
∴ Required ratio = 54:60 = 9:10

5. The ratio of A’s and B’s age is 7:5 and the sum of their ages is 36 years. What
will be the ration of their ages after 9 years?
a) 6:5 b) 5:4 c) 5:3 d) 4:3 e) 7:5
Solution:
Let the ages of person A and B be x and y respectively.

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𝐱 𝟕
𝐓𝐡𝐞𝐧, = 𝐨𝐫, 𝟓𝐱 = 𝟕𝐲 … . (𝐢)
𝐲 𝟓
𝐚𝐧𝐝 𝐱 + 𝐲 = 𝟑𝟔 … … . (𝐢𝐢)
Solving eqn (i) and (ii)
X = 21 years and y = 15 years
∴ Ratio after 9 years
𝐱 + 𝟗 𝟑𝟎 𝟓
= = = = 𝟓: 𝟒
𝐲 + 𝟗 𝟐𝟒 𝟒
6. The product of the ages of A and B is 240. If twice the age of B is more
than A’s age by 4 years, what was B’s age 2 years ago?
Solution :
Let A’s present age be x years. Then, B’s present age = 240 / x years
So, according to question
2 (240 / x ) – x = 4
480 – x2 = 4 x
x2 + 4 x – 480 = 0
(x + 24) (x – 20) = 0
x = 20
B’s present age = 240 / 20 = 12 years
Thus, B’s age 2 years ago = 12 – 2 = 10 years
7. The present age of a mother is 3 years more than three times the age of her
daughter. Three years hence, the mother’s age will be 10 years more than
twice the age of the daughter. Find the present age of the mother.
Solution :
Let the daughter’s present age be ‘n’ years.
=> Mother’s present age = (3n + 3) years
So, according to the question
(3n + 3 + 3) = 2 (n + 3) + 10
3n + 6 = 2n + 16
n = 10
Hence, mother’s present age = (3n + 3) = ((3 x 10) + 3) years = 33 years
8. The ratio of present ages of A and B is 6 : 7. Five years hence, this ratio
would become 7 : 8. Find the present age of A and B.
Solution :
Let the common ratio be ‘n’.
A’s present age = 6 n years
B’s present age = 7 n years
So, according to the question
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(6 n + 5) / (7 n + 5) = 7 / 8
48 n + 40 = 49 n + 35
n=5
Thus, A’s present age = 6 n = 30 years and B’s present age = 7 n = 35 years
9. A father is currently three times as old as his son. Five years ago, he was
four times as old as his son. Find the current ages of the father and the son.
Solution:
Let the son’s current age be x.
Then, the father’s current age is 3x.
Five years ago:
Son’s age = x − 5
Father’s age = 3x − 5
According to the problem, five years ago, the father was four times as old
as the son: 3x − 5 = 4(x − 5)
Solve for x: 3x − 5 = 4x − 20 ⇒ x = 15
Current ages:
Son’s age = 15 years
Father’s age = 3 × 15 = 45 years
Therefore, the son is 15 years old, and the father is 45 years old.

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6.RATIO AND PROPORTION

Ratio:

The ratio of two quantities of x and y is the comparision between the two
quantities in the same unit by division. It is denoted b y x:y and given as
x
x: y =
y

Proportion:

When the two ratio a:b and c:d are equal then the quantities a,b,c and d are
called in proportions. It is denoted as a:b::c:d or a:b = c:d

Here a,b, c and d are called 1st proportion 2nd proportion, 3rd proportion and
4th proportion respectively. Also a and d are known as extremes and b and c care
known as means.

Direct proportion:

Two quantities a and b are said to be in direct p[roportion if the ratio of the
two quantities is a constant.
a
= k, where k is a constant.
b

Indirect proportion:

Two quantities a and b are said to be in indirect proportion if the product of

the two quantities is a constant .

Let ab and b are indirect proportion i.e.,

1
a ∝ then ab = k, where k is a constant.
b

If the two numbers are in the ratio a : b and the sum of these numbers is x,
then these numbers will be

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ax bx
and
a+b a+b
Mean Proportion = √ab

Reciprocal ratio of a and b = b: a or 1⁄a : 1⁄b

Duplicate ratio of a and b = a2 : b2

Triplicate ratio of a and b = a3 : b3

Sub – Duplicate ratio= √a: √b

3 3
Sub triplicate ratio = √a: √b

Compounded ratio of (a:b)(c:d)(e:f) = (ace:bdf)

a c
Comparision of ratio (a:b) > (c:d) = >
b d

1. If a: b = 5: 9 and b: c = 7: 4, then find a: b: c.

Solution:
Here, we make the common term ‘b’ equal in both ratios.
Therefore, we multiply the first ratio by 7 and the second ratio by 9.
So, we have a : b = 35 : 63 and b : c = 63 : 36
Thus, a : b : c = 35 : 63 : 36
2. Divide Rs. 981 in the ratio 5: 4
Solution:
The given ratio is 5: 4
Sum of numbers in the ratio = 5 + 4 = 9
We divide Rs. 981 into 9 parts.
981 / 9 = 109
Therefore, Rs. 981 in the ratio 5: 4 = Rs. 981 in the ratio (5 / 9) : (4 / 9)
=> Rs. 981 in the ratio 5 : 4 = (5 x 109) : (4 x 109) = 545 : 436
3. A bag contains 50 p, 25 p, and 10 p coins in the ratio 2: 5 : 3, amounting to
Rs. 510. Find the number of coins of each type.
Solution:

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Let the common ratio be 100k.

Number of 50 p coins = 200 k
Number of 25 p coins = 500 k
Number of 10 p coins = 300 k
Value of 50 p coins = 0.5 x 200 k = 100 k
Value of 25 p coins = 0.25 x 500 k = 125 k
Value of 10 p coins = 0.1 x 300 k = 30 k
Total value of all coins = 100 k + 125 k + 30 k = 255 k = 510 (given)
k=2
Therefore, Number of 50 p coins = 200 k = 400
Number of 25 p coins = 500 k = 1000
Number of 10 p coins = 300 k = 600
4. Milk and water are mixed in a vessel A in the proportion of 5:2, and in
vessel B in the proportion of 8:5. In what proportion should the quantities be
taken out of these two vessels to form a new mixture by mixing the
remaining quantities in the vessels, so that the proportion of milk to water
becomes 9:4?
a) 3:2 b) 7:2 c) 7:3 d) 2:7

Solution:
5 8
In vessel A, Milk = of the weight of mixture. In vessel B, milk = of
7 13
the weight of mixture
By Alligation method:
5 8
7 13
↘ ↙
9
13
↗ ↖

1 2
13 91
1 2
Required proportion is : = 91: 26 = 7: 2
13 91
5. 465 coins consist of rupee, 50 paise and 25 paise coins. Their values are in
the ratio 5:3:1. Find the number of each coin.
Solution:
The ratio of number of coins

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100 100 100

=5× :3 × :1 × = 5: 6: 4
100 50 25
∴ The number of one – rupee coins
465
= × 5 = 155
5+6+4
The number of 50 P coins
465
= × 6 = 186
5+6+4
The number of 25 P coins
465
= × 4 = 124
5+6+4
6. A mixture contains sugar solution and colored water in the ratio of 4 : 3. If
10 liters of colored water is added to the mixture, the ratio becomes 4: 5.
Find the initial quantity of sugar solution in the given mixture.
Solution:
The initial ratio is 4 : 3.
Let ‘k’ be the common ratio.
=> Initial quantity of sugar solution = 4 k
=> Initial quantity of colored water = 3 k
=> Final quantity of sugar solution = 4 k
=> Final quantity of colored water = 3 k + 10
Final ratio = 4 k : 3 k + 10 = 4 : 5
=> k = 5
Therefore, the initial quantity of sugar solution in the given mixture = 4 k =
20 liters
7. Two friends A and B started a business with an initial capital
contribution of Rs. 1 lac and Rs. 2 lacs. At the end of the year, the
business made a profit of Rs. 30,000. Find the share of each in the
profit.
Solution:
We know that if the time period of investment is the same, profit/loss is
divided by the ratio of the value of the investment.
Ratio of value of investment of A and B = 1,00,000 : 2,00,000 = 1 : 2
Ratio of share in profit = 1 : 2
Share of A in profit = (1/3) x 30,000 = Rs. 10,000
Share of B in profit = (2/3) x 30,000 = Rs. 20,000
8. Three friends A, B, and C started a business, each investing Rs. 10,000.
After 5 months A withdrew Rs. 3000, B withdrew Rs. 2000 and C

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invested Rs. 3000 more. At the end of the year, a total profit of Rs.
34,600 was recorded. Find the share of each.
Solution:
We know that if the period of investment is not uniform, the gains/losses
from the business are divided in the ratio of their inputs, where input is
calculated as the product of an amount of investment and the time period of
investment.
So, input = value of investment x period of investment, and here, the period
of investment would be broken into parts as the investment is not uniform
throughout the time period.
A’s input = (10,000 x 5) + (7,000 x 7) = 99,000
B’s input = (10,000 x 5) + (8,000 x 7) = 1,06,000
C’s input = (10,000 x 5) + (13,000 x 7) = 1,41,000
=> A : B : C = 99000 : 106000 : 141000
=> A : B : C = 99 : 106 : 141
=> A : B : C = (99 / 346) : (106 / 346) : (141 / 346)
Thus, A’s share = (99 / 346) x 34600 = Rs. 9900
B’s share = (106 / 346) x 34600 = Rs. 10600
C’s share = (141 / 346) x 34600 = Rs. 14100

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7.PERCENTAGES
Percentage:
The term percent comes from the latin word ‘per centum’ which means ‘per
hundred’ or ‘for every hundred’.
5
= 5%
100
1. A and B are two numbers.
𝐴
𝐴 𝑎𝑠 𝑎 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑜𝑓 𝐵 = × 100
𝐵
𝐵
𝐵 𝑎𝑠 𝑎 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑜𝑓𝐴 = × 100
𝐴
To convert a decimal fraction to percentage multiply it by 100
0.05 = 0.05 × 100 = 5%
If A is x% more than that of B, then B is less than that of A by
𝑋
[ × 100] %
100 + 𝑋
2. If A is x% less than that of B, then B is more than that of A by
𝑋
[ × 100] %
100 − 𝑋
3) If A is x% of C and B is y% of C, then
𝑥
𝐴 = × 100% 𝑜𝑓 𝐵
𝑦
If two numbers are respectively x% and y% more than a third number, then the
first number is
100 + 𝑥
[ × 100] %𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑎𝑛𝑑
100 + 𝑌

100 + 𝑌
𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑠 [ × 100] %𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡.
100 + 𝑋
5) If two numbers are respectively x% and y% more than a third number, then the
first number is
100 − 𝑥
[ × 100] %𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑎𝑛𝑑
100 − 𝑌

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100 − 𝑌
𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑠 [ × 100] % 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡
100 − 𝑋
6) If the price of a commodity increases by N%, then the reduction in consumption
so as not to increase the expenditure is
𝑁
[ × 100] %
100 + 𝑁
7) If the price of a commodity decreases by N%, then the increase in consumption
so as not to decrease the expenditure is
𝑁
[ × 100] %
100 − 𝑁
8) If a number is changed (increased / decreased) successively by x% and y% then
net % change is given by
𝑋𝑌
[𝑋 + 𝑌 + ]%
100
which represents increase or decrease in value according as the sign is +ve or –
ve.
Note:
If x and y indicates decrease in percentage then put –ve sign before x and y else
+ve sign.
9) If the population of a town (or the length of a tree) is P and its annual increase
is r% then
(i)Population (or length of tree) after n years
𝑟 𝑛
= 𝑃[1 + ]
100
(ii)Population (or length of tree) n years ago
𝑃
= 𝑟 𝑛
[1 + ]
100
10) If the population (or value of machine in rupees) is P and annual decrease (or
depreciation) is r%, then
𝑟 𝑛
(i) Population (or value of machine) after n years =P [1 − ]
100
(ii) Population (or value of machine) n years ago
𝑃
= 𝑟 𝑛
[1 − ]
100
11) If a number K is increased successively by x% followed by y% and z%, then

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the final value of K will be

𝑋 𝑌 𝑍
𝐾[1 + ] [1 + ] [1 + ]
100 100 100
12) In an examination, the minimum pass percentage is x%. If a student scores y
marks and fails by z marks, then the maximum marks in the examination is
100(𝑦 + 𝑧)
𝑥
13)In an examination a% and b% students respectively fail in two different
subjects while c% students fail in both the subjects , then the percentage of
students who pass in both the subjects will be (100 – (a+b-c))%
PROBLEMS:
1. A defect-finding machine rejects 0.085% of all cricket bats. Find the number
of bats manufactured on a particular day if it is given that on that day, the
machine rejected only 34 bats.
Solution :
Let the total number of bats on that day be n.
0.085 % of n = 34
(0.085 / 100) x n = 34
n = 34 x (100 / 0.085)
n = 40,000
Therefore, total number of bats manufactured on the day = 40,000
2. 25 % of a number is 8 less than one-third of that number. Find the number.
Solution:
Let the number be n.
(n / 3) – 25 % of n = 8
(n / 3) – (n / 4) = 8
n / 12 = 8
n = 96
Thus, 96 is the required number.
3. The production of a company experiences ups and downs every year. The
production increases for two consecutive years consistently by 10% and in the
third year it decreases by 5%. Again, in next two years, it increases by 10%
each year, and decreases by 5% in the third year. If we start counting from the
year 2008 what will be the effect on the production of the company in the year
2012?

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a) 35% decrease b) 42% increase

c) 26% decrease d) 31% decrease
Solution:
Let the production of the company in the year 2008 be x.
∴ Production in the year 2012
110 110 95 110
=x× × × ×
100 100 100 100
= 1.26445x = 1.26x
0.26x
∴ % increase = × 100 = 26%
x
4. If the first number is 20% less than the third number and the second number is
24% less than the third number, then the second number is what percent less
than the first number?
a) 8% b) 7% c) 9% d) 5%
Solution:
Let the third number be 100
Then 2nd number = 76 and 1st number = 80
4
∴ 𝑅𝑒𝑞𝑑 % = × 100 = 5%
80
5. The bacteria in a culture grow by 5% in the first hour, decrease by 10% in the
second hour and again increase by 20% in the third hour. If the original count
of bacteria in a sample is 5000, find the count of bacteria at the end of 3 hours.
a) 5230 b) 5435 c) 6260 d) 5670 e) none of these
Solution:
10 90 120
5000 × × × = 5670
100 100 100
6. In an examination, the percentage of students qualified with respect to the
number of students appeared from school A is 80%. The number of students
appeared from school B is 20% more than the number of students appeared
from school A, and the number of students qualified from school B is 40%
more than the number of students qualified from school A. What percentage of
students qualified among those who appeared from school B?
2 4 1 1
a) 91 % b) 90 % c) 92 % d) 93 % e) 32%
3 5 3 3
Solution:
Suppose 100 students appeared from school. Now,

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Appeared Qualified
A → 100 80
B → 120 80+ 40 % of 80 = 112
112
𝑅𝑒𝑞𝑑 % = × 100
120
11200 1
= = 93 %
120 3

7. Difference of two numbers ‘x’ and ‘y’ (x > y) is 100. Also, 10 % of ‘x’ is
equal to 15 % of ‘y’. Find the numbers.
Solution :
We are given that x – y = 100 and
10 % of x = 15 % of y
x – y = 100 and (10 / 100) x = (15 / 100) y
x – y = 100 and 10 x = 15 y
x – y = 100 and 2 x = 3 y
x – y = 100 and x = 1.5 y
1.5 y – y = 100
0.5 y = 100
y = 200
x = 1.5 y = 300
Thus, the required numbers are 300 and 200.
8. In a gaming event, 75 % of the registered participants actually turned up. Out
of those, 2 % were declared unfit for participation. The winner defeated 9261
participants which are 75 % of the total valid participation. Find the number
of registered participants.
Solution:
Let the number of registered participants be n.
Number of participants who actually turned up = 75 % of n
Number of valid participations = 98 % of (75 % of n)
Number of participants defeated by the winner
= 75 % of 98 % of (75 % of n) = 9261
0.75 x 0.98 x 0.75 x n = 9261
0.55125 x n = 9261
n = 16800
Therefore, number of registered participants = 16800

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9. In a class of 90 students, amongst 50% of the students each student got

number of sweets that are 20% of the total number of students and the amongst
remaining 50% of the students each student got number of sweets that are 10%
of the total number of students. How many sweets were distributed among 90
students?
a) 1620 b) 1215 c) 1350 d) can’t be determined
Solution:
50
50% of 90 students = 90 × = 45 students
100
Total number of sweets to 45 students
20
= 45 × 90 ×
100
= 45 × 18 = 810
Total number of sweets to remaining 45 students
10
= 45 × 90 ×
100
= 45 × 9 = 405
So total number of sweets to all 90 students = 810 + 405 = 1215
10.Mrs. Poonam invests 15% of her monthly salary, i.e., Rs.4, 428 in Mutual
funds. Later she invests 18% of her monthly salary on pension policies; also
she invests another 9% of her salary on Insurance policies. What is the total
monthly amount invested by Mrs.Poonam
a) Rs.1, 13,356.8 b) Rs.12, 398.40
c) Rs.56, 678.4 d) Rs.13, 855.6
Solution:
Monthly income of Poonam
4428×100
= 𝑅𝑠. 29520
15
Total invested amount
(15+18+9)
= 29520 ×
100
42
= 29520 ×
100
1239840
=
100
= 𝑅𝑠. 12398.40
11.Gopal gave 40% of his retirement money to his wife. He also gave 20 % of
the remaining amount to each of his 3 sons. 50 % of the amount now left was
spent on miscellaneous items and the remaining amount of Rs. 1,20,000 was
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deposited in the bank. How much money did the geek get as retirement
money?
Solution:
Let the retirement money be Rs. 100 n
Money given to wife = 40 % of 100 n = 40 n,
Balance = 60 n
Money given to 3 sons = 3 x (20 % of 60 n) = 3 x 12 n = 36 n,
Balance = 24 n
Money spent on miscellaneous items = 50 % of 24 n = 12 n,
Balance = 12 n
Now, this remaining 12 n is the money deposited in the bank,
i.e., Rs. 1,20,000 12 n = 1,20,000
n = 10,
Therefore, Gopal’s retirement money = 100 n = Rs. 10,00,000

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8.MENSURATION

Mensuration is the branch of mathematics that deals with the measurement

of various geometric figures and shapes.

This includes calculating areas, volumes, and perimeters of two-

dimensional

shapes like squares, rectangles, circles, and triangles, as well as three-dimensional

figures

like cubes, cylinders, spheres, and cones.

These shapes can exist in 2 ways:

Two-Dimensional Shapes – circle, triangle, square, etc.

 Three-Dimensional Shapes – cube, cuboid, cone, etc.

Difference Between 2D and 3D Shapes
2-Dimensional vs 3-Dimensional Shapes

2D Shape 3D Shape

Any shape is 2D if it is bound by

A shape is a three-dimensional shape if there
three or more straight lines in a
are several surfaces or planes around it.
plane.

In contrast to 2D forms, these are sometimes

There is no height or depth in these
known as solid shapes and have height or
shapes.
depth.

Since they have depth (or height), breadth, and

These shapes just have length and
length, they are referred to as three-dimensional
width as their dimensions.
objects.

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2-Dimensional vs 3-Dimensional Shapes

2D Shape 3D Shape

Their volume, curved surface area, lateral

We can calculate their perimeter
surface area, or total surface area can all be
and area.
calculated.

Perimeter:
The sum of all the sides of a two-dimensional figure.

Area:
Area is the amount of space occupied by a two-dimensional figure.

Volume:
Volume is the measurement of amount of space occupied by a three
dimensional shape.

Curved Surface Area (C.S.A)

This is the area of only the circular parts of a solid. It does not include the
area of the bases of the solid.

Lateral Surface Area (L.S.A)

This is the area of only side faces of the solid.
It does not include the area of the top and bottom faces of the solid.
Total surface Area (T.S.A)
The total surface area of a solid is the sum of the areas of all the fces or
surfaces that enclose the solid.
It is the sum of curved / lateral surface area and the area of the base (s) of the
solid.

Mensuration Formula For 2D Shapes

Mensuration Formulas for 2-Dimensional Figures
Shape Area Perimeter
Circle πr² 2πr

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Square (side)² 4 × side

Rectangle length × breadth 2 (length + breadth)
√[s(s−a)(s−b)(s−c),
Scalene Triangle a+b+c (sum of sides)
Where, s = (a+b+c)/2

Isosceles Triangle ½ × base × height 2a + b (sum of sides)

Equilateral Triangle (√3/4) × (side)² 3 × side
Right Angled A + B + hypotenuse, where the
½ × base × hypotenuse
Triangle hypotenuse is √A²+B²
Parallelogram base × height 2(l+b)
½ × diagonal1 ×
Rhombus 4 × side
diagonal2
½ h(sum of parallel
Trapezium a+b+c+d (sum of all sides)
sides)

Mensuration Formulas for 3-D Figures

Under 3-D Figures, we can calculate the total surface area which is equal to
curved surface area+ area of top and bottom.

Curved surface area is also known as lateral surface area, and is measured in
square units.

Total surface area is also measured in square units whereas volume is

measured

in cubic units. So, find out the mensuration formulas for 3-D figures such as
cone, cylinder, cube, cuboid, sphere, etc.

Mensuration Formulas for 3-Dimensional Figures

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Curved Surface Area

(CSA)/ Total Surface Area
Shape Area
Lateral Surface Area (TSA)
(LSA)
Cone (1/3) π r² h πrl πr (r + l)
Cube (side)³ 4 (side)² 6 (side)²
length × breadth ×
Cuboid 2 height (length + breadth) 2 (lb +bh +hl)
height
Cylinder π r² h 2π r h 2πrh + 2πr²
Hemisphere (2/3) π r³ 2 π r² 3 π r²
Sphere 4/3πr³ 4πr² 4πr²

Differences between 2-Dimensional and 3-Dimensional Figures

There are two types of figures in geometry, one is 2-Dimensional and 3-

Dimensional figures. Check out the below table to know and understand the 2-
Dimensional and 3-Dimensional figures and the differences between them.

2-Dimensional Figures 3-Dimensional Figures

As the name suggests, a 2D shape means that it Here, a 3D shape means that this figure
will have only 2 dimensions which are length will have 3 dimensions, which are
and breadth. length, breadth, and height.
For 3D shapes, we can calculate their
For 2D shapes, we can calculate 2 things i.e.
volume, total surface area, and
area and perimeter.
curved/lateral surface area.
2D shapes are flat as they do not have depth 3D shapes contain a depth so they can
and also, these cannot be held physically be held physically and are not flat like
because of the lack of depth. 2D shapes.
Example: Square, Rectangle, Triangle, Circle, Example: Cone, Cylinder, Sphere,
etc. Cube, Prism, Pyramid, etc.

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Mensuration maths formulas for all 2-D and 3-D figures

1. Cube Formula

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Cuboid Formula

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3. Prism Formula

4. Sphere Formula

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5. Hemisphere Formula

6. Pyramids Formula

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7. Right circular cone:

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8.Right Circular Cylinder Formula

1. Find the volume of a cone if the radius of its base is 1.5 cm and its perpendicular
height is 5 cm.

Solution:

Radius of the cone, r = 1.5 cm

Height of the cone, h = 5 cm

∴ Volume of the cone, V = 13πr2h=13×227×(1.5)2×5= 11.79 cm3

Thus, the volume of the cone is 11.79 cm3.

2. The dimensions of a cuboid are 44 cm, 21 cm, 12 cm. It is melted and a

cone of height 24 cm is made. Find the radius of its base.

Solution:

The dimensions of the cuboid are 44 cm, 21 cm and 12 cm.

Let the radius of the cone be r cm.

Height of the cone, h = 24 cm

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It is given that cuboid is melted to form a cone.

∴ Volume of metal in cone = Volume of metal in cuboid

⇒(1/3)πr2h=44×21×12

(Volume of cuboid=Length×Breadth×Height)

⇒(1/3)×(22/7)×r2×24=44×21×12

⇒r= √(44×21×12×21) / (22×24)

=21 cm

Thus, the radius of the base of cone is 21 cm.

3.Find the area of a rhombus whose diagonals are of lengths 9 cm and 7.2 cm.

Solution:

Now, d1=9 cm, d2=7.2 cm where d1 and d2 are the lengths of diagonals of a
rhombus.

Area of rhombus formula is ½ (d1xd2)

Therefore, area of the given rhombus = ½ x 9 x 7.2 cm² = 32.4 cm²

Note: Here, the value of π is either considered as 22/7 or 3.14, r means radius, and
h means height.

4. A rectangular paper of a width of 10 cm is rolled along its width and a cylinder

of a radius of 18 cm is formed. Find the volume of the cylinder. (Take π=22/7)

Solution: A cylinder is formed by rolling a rectangle of 10 cm about its width.

The radius of the cylinder is 18 cm and the width of the paper becomes the height.

Height of the cylinder, h = 10 cm and Radius of the cylinder, r = 18 cm

Now, volume of the cylinder, V = πr²h = (22/7) x (18)² x 10

= 10,183 cm³

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Hence, the volume of the cylinder is 10183 cm³.

5. A circle has a radius of 21 cm. Find its circumference and area. (Use π = 22/7)

Solution:

We know, Circumference of circle = 2πr = 2 x (22/7) x 21 = 2 x 22 x 3 = 132 cm

Area of circle = πr² = (22/7) x (21)² = 22/7 x 21 x 21 = 22 x 3 x 21

Area of circle with radius, 21cm = 1386 cm²

6.The length, width, height of a cuboidal box are 30 cm, 25 cm and 20 cm,
respectively.

Find its area.

Solution: Here, l- 30 cm, b= 25 cm, and h= 20 cm

Total surface area = 2 (lb +bh +hl)

= 2 (30 × 25 + 25 × 20 + 20 × 35)

TSA = 2( 750 + 500 + 700) = 3900 cm²

7.A rectangular piece of paper 11 cm × 4 cm is folded without overlapping to make

a cylinder of the height of 4 cm. Find the volume of the cylinder.

Solution: The length of the paper will be the perimeter of the base of the cylinder
and

the width will be its height.

Circumference of base of cylinder = 2πr = 11 cm

2 x 22/7 x r = 11 cm

r = 7/4 cm

Volume of cylinder = πr²h = (22/7) x (7/4)² x 4 = 38.5 cm³

8.The total surface area of a hemisphere is 166.32 cm², find its curved surface
area?

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Solution: Let radio of hemisphere = r cm

Tital surface area of hemisphere = 3 π r² = 166.32 ……………….. (i)

Multiplying equation (i) by ⅔

=> ⅔ x 3 π r² = ⅔ x 166.32

=> Curved surface area of hemisphere = 2 π r² = 2 x 55.44 = 110.88 cm²

9. If G is the centroid and AD, BE, and CF are three medians of the triangle with
72 cm², then the area of triangle BDG is:

Solution: The area of the triangle formed by any two vertices and centroid is ⅓)
times the area of △ABC.

Also, the median divides the triangle into two equal areas.

So, the area of △BDG = 1/6 times of △ABC

=> 1/6 x 72

=> 12

10.The perimeters of two similar triangles ABC and PQR are 156 cm and 46.8 cm
respectively. If BC= 19.5 cm and QR= x cm, then the value of x is?

Solution: (△ABC) Perimeter/ (△PQR) Perimeter) = BC/QR

=> 15.6/46.8 = 19.5/x

=> x = 46.8 x 19.5 / 156

=> 5.85 cm

11. A solid metallic sphere of radius 10 cm is melted and recast into spheres of
radius 2 cm each. How many such spheres can be made?

Solution: As, V = 4/3 π r³

Volume of big sphere = n x volume of small sphere
=> 4/3 π (10)³ = n x 4/3 π (2)³
=> n = 125

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9.TIME AND WORK

“Time and Work” topic deals with the time taken by an individual or a
group to complete a job or a piece of work and its efficiency.
The formulas can completely help you to find a solution as soon as you read
the question. Thus, it makes the solution and the related calculations simpler.
Some of Important Formulas and Concepts on Time and Work are:
1. Work Done by a person = Time Taken × Rate of Work
2. Rate of Work of a person = 1 / Time Taken by him
3. Time Taken by him = 1 / Rate of Work
4. If a piece of work or a job is done in n number of days, then the work done
in one day = 1/n
5. Total Work Done = Number of Days × Efficiency
6. Efficiency of work done and Time are inversely proportional to each other.
7. M : W is the ratio of the number of men and women which are required to
complete a piece of work, then the ratio of the time taken by them to complete
the work will be W : M.
xy
8. If A can do a job in ‘x’ days and B in ‘y’ days together they will finish job in
x+y

days.
9. If A can do a job in ‘x’ days B in ‘y’ days and C in ‘z’ days, the number of days
xyz
they will take to complete work = days.
xy+yz+xz

10. If A+B can complete a work in ‘x’ days B+C in ‘y’ days and A+C in ‘z’ days
2xyz
then (A+B+C) can complete the same work in days.
xy+yz+xz

2xyz 2xyz 2xyz

A alone can do it in , B in and C in days.
xy+yz−zx yz+zx−xy zx+xy−yz

11. When the quantity of work done is the same,

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m1 d1 = m2 d2

m1 d1 h1 =m2 d2 h2 where m – no. of men, d – no. of days, h-no. of

hours, w1 w2 – work done

12.When the quantities of work done are different, then

m1 d1 m2 d2
=
w1 w2

m1 h1 m2 h 2
=
w1 w2

m1 d1 h 1 m2 d2 h 2
= where m – no. of men, d – no. of days, h-no. of
w1 w2
hours ,w1 w2 – work done

Work and Wages

Wages are distributed in direct proportion to the work done and in inverse
proportion to the time taken by the individual.

i.e., Wages are distributed in proportion to the one day’s work/ efficiency.

If A, B and C can do a work in 6,8,12 days respectively, their wages are in the
1 1 1
ratio. : :
6 8 12

i.e., 4: 3: 2

13.If A and B working together can complete a work in ‘X’ days and B is ‘K’
times efficient than A, then

a) The time taken by A working alone, to complete the work is (k+1) x.

k+1
b) The time taken by B working alone, to complete the work is ( )x
k

c) The time taken by both A and B working together to complete the work when A
X
completes the work in x days and B is K times as efficient as A = days.
(1+K)

14.If A working alone takes ‘a’ days more than A and B working together .If B
worked alone he takes ‘b’ days more than A and B working together to

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complete the work, then both A and B working together can finish the job in
√ab days.
a c
15.If A can complete part of a work in ‘X’ days then part of work will be
b d
X×b×c
done in days.
a×d
16.If ‘a’ men and ‘b’ women can do a piece of work in ‘n’ days, then ‘c’ men
nab
and ‘d’ women can do the work in [ ] days.
bc+ad
17.If A is ‘K’ times more efficient than B and hence able to finish the work in
‘l’ days than B, then
l
a) A working alone can finish the work in days.
k−1

kl
b) B working alone can finish the work in days.
k−1

kl
c) A and B working together can complete the work in days.
k2 −1

PROBLEMS:

1. A man can do a work in 20 days and a woman in 15 days. If they work on it

together for 5 days, then the fraction of the work that is left is :
A. 1/12 B. 1/10 C. 5/12 D. 7/15
Solution:
Man’s 1 day’s work = 1/20
Woman’s 1 day’s work = 1/15
(Man + woman)’s 1 day’s work = (1/20 + 1/15) = 7/60
(Man + woman)’s 5 day’s work = (7/60 × 5) = 7/12
Thus, Remaining work = 1 – 7/12 = 5/12

2. L can finish a work in 16 days and M can do the same work in 12 days.
With help of N, they did the work in 4 days only. Then, N alone can do the
work in how many days.
A. 48/5days B.48/7days C.48/11days D.10days
Solution:
(L + M + N)’s 1 day’s work =1/4
L’s 1 day’s work = 1/16
M’s 1 day’s work = 1/12
Therefore, N’s 1 day’s work

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= 1/4 – (1/16 + 1/12)

= 1/4 – 7/48 = 5/48
So, N alone can do the work in 48/5 days.
3. P, Q and R can do a job in 20, 30 and 60 days respectively. In how many
days can P do the job if he is assisted by Q and R every third day?
A. 11days B. 15days C. 17days D. 16days
Solution:
P’s 2 day’s work = 2/20 = 1/10
(P + Q + R)’s 1 day’s work
= (1/20 + 1/30 + 1/60)
= 6/60 = 1/10
Job done in 3 days = (1/10 + 1/10) = 1/5
Now, 1/5 jobs is done in 3 days
Whole job will be done in (3 x 5) = 15 days.

4. M’s efficiency is three times N’s efficiency. M can finish a job in 60 days less
than N. If they work together, then in how many days the job will be done.
A. 20daysB. 22.5 days C. 25 daysD. 30 days
Solution:
Ratio of times taken by M and N = 1 : 3 (Since the efficiency of M is three times
to N)
Time difference is (3 – 1) = 2 days, while N take 3 days and M takes 1 day.
 2 units = 60 days
 1 unit = 30 days
So, M takes 30 days to do the job.
And N takes (30×3) = 90 days to do the job.
M’s 1 day’s work = 1/30
N’s 1 day’s work = 1/90
(M + N)’s 1 day’s work =( 1/30 + 1/90) = 2/45
M and N together can do the job in 45/2 days = 22.5 days.

5. Ankit alone can do a piece of work in 6 days and Bishal alone in 8 days.
Ankit and Bishal undertook to do it for Rs. 4800. With the help of Dinesh,
they completed the work in 3 days. How much is to be paid to Dinesh?
A. Rs.1375 B. Rs.1400 C. Rs. 1600 D. Rs. 600
Solution:
Ankit’s 1day work = 1/6
Bishal’s 1 day work = 1/8
(Ankit + Bishal + Dinesh)’s 1 day work =1/3
Dinesh’s 1 day work = 1/3 – (1/6+1/8) = 1/24
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So, Dinesh’s 3 day work = 3 * 1/24= 1/8

If Dinesh contributed 8th part of work then he will receive 8th part of total
payment
i.e., 4800 × 1/8 = 600

6. Vicky completes a job in 45/2 days. What part of the job will he do in 2
days?
A. 4/45 B.1/45 C.2/45 D.8/45
Solution:
We know, if a person does a job in n days, then his 1-day work = 1/n
Here, n = 45/2
Vicky’s 1-day work = 2/45
Thus, Vicky’s 2 days work = 2× 2/45 = 4/45

7. Karan completes 1/15 part of a certain job in 1 day. In how many, he will
complete the full job.
A) 30days B)15days C) 20 days D) 5 days
Solution:
Here, 1/n = 1/15
So, n = 15
Thus, the required days = 15

8. In a factory, 20 people can make 20 toys in 15 days working 10 hours per

day. Then, in how many days can 25 persons make 30 toys working 20 hr
per day?
A) 6days B) 9 days C) 10 days D) 12 days
Solution:
Here, M1 = 20, M2 = 25, D1 = 15, D2 =? , T1 = 10 , T2= 20, W1 = 20 and W2 =
30.
We know,
M1 × D1 × T1 × W2 = M2 × D2 × T2 × W1
20 × 15 × 10 × 30 = 25 × D2 × 20 × 20
D2 = 9
Thus, the required day = 9 days

9. P, Q and R together can complete a work in 16 days and R alone complete

the work in 20 days. If P, Q and R started the work together and after 10
days P and Q left the work, in how many days R alone complete the
remaining work?

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A) 12½days B) 20½ days C) 4 days D) 7½ days

Solution:
P + Q + R = 16 days
R =20 days
work (LCM of 16 and 20) = 80
( P + Q +R) ‘s work
= 80 /16 = 5 unit
Work done by R = 80 /20 = 4 unit
(P + Q + R) ‘s 10 days work
= 5 × 10 = 50 unit
Remaining work
= (80 – 50)
= 30 unit
Remaining work done by R
= 30/4
= 7½ days

10.M did a piece of work in 5 days. That piece of work was done by N in 9
days. If M and N worked together, they got total wages of Rs 4200. Find the
share of N.
A) 1500 B) 2000 C) 1000 D) 1200
Solution:
M : N
Time = 5 : 9
Efficiency= 9 : 5
(Time and efficiency are inversely proportional)
N gets = 4200 × 5/14
= 1500
Thus, N gets the wages of Rs 1500.

11.P and Q can do a job in 3 days. Q and R can do the same job in 9 days,
while R and P can do it in 12 days. In how many days the job will be
finished when P, Q and R working together.
A) 72/19 days B) 83/10 days C) 61/3 days D) 67/4 days
Solution:
Here, (P + Q)’s 1 day’s work = 1/3
(Q + R )’s 1 day’s work= 1/9
and (R + P)’s 1 day’s work = 1/12
Now, 2 (P + Q + R)’s 1 day’s work = (1/3+1/9+1/12) = 19/36
(P + Q + R)’s 1 day’s work = 19/(36 ×2)=19/72
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Hence, (P + Q + R) complete the work in= 72/19 days

12.E and F are two friends working together to finish a work in 24 days and E
alone can do the same work in 36 days. In how many days can F alone
complete the work?
A) 36 days B) 24 days C) 72 days D) 48 days
Solution:
(E+F)’s 1 days work = 1/24
E’s 1-day work = 1/36
F’s 1 day work = (1/24 – 1/36) = 1/72
Thus, the time is taken by F to finish the work alone= 72 days

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10.TIME AND DISTANCE

Time:

Time is the measure of passing of events from the past, through the present,
and into the future.

It helps an understand when things happen and how they are connected.

It is usually denoted as T and it is usually measured in seconds and hours.

Distance:

A path covered by a moving object irrespective of the direction of motion is

called distance.

Speed:

The rate of distance covered by a moving object in a certain period of time is

called the speed of the object.

It is denoted by S and it is usually measured in meters per second, kilometers

per hour.
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝐷 𝑘
𝑆𝑝𝑒𝑒𝑑 = 𝑖. 𝑒. , 𝑆 = 𝑜𝑟 𝑚/𝑠
𝑇𝑖𝑚𝑒 𝑇 ℎ𝑟
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝐷
𝑇𝑖𝑚𝑒 = 𝑖. 𝑒. , 𝑇 = ℎ𝑟𝑠 𝑜𝑟 𝑠𝑒𝑐𝑠
𝑆𝑝𝑒𝑒𝑑 𝑆

Distance = Speed × Time 𝑖. 𝑒. , 𝐷 = 𝑆 × 𝑇 𝑘𝑚𝑠 𝑜𝑟 𝑚𝑠.

Units of speed are km per hour or m/s

5 𝑚
1 𝑘/ℎ𝑟 =
18 𝑠𝑒𝑐
18 𝑘
1 𝑚/𝑠𝑒𝑐 =
5 ℎ𝑟

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Average Speed:

When the travel consists of various Speeds, then the concept of average
Speeds comes into play
𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑐𝑜𝑣𝑒𝑟𝑒𝑑
1. 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 =
𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒 𝑇𝑎𝑘𝑒𝑛 𝑓𝑜𝑟 𝑡ℎ𝑒 𝐽𝑜𝑢𝑟𝑛𝑒𝑦
𝐷1 + 𝐷2 + 𝐷3 + ⋯ 𝐷𝑛 𝑘 𝑚
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 = 𝑜𝑟
𝑇1 + 𝑇2 + 𝑇3 + ⋯ 𝑇𝑛 ℎ𝑟 𝑠𝑒𝑐
𝑘 5 𝑚
2. 𝑥 =𝑥 ×
ℎ𝑟 18 𝑠𝑒𝑐
𝑚 18 𝑘
3. 𝑥 =𝑥 ×
𝑠𝑒𝑐 5 ℎ𝑟
If an object Covers a distance 𝐷1 Kms at 𝑥 𝑘/ℎ𝑟 and 𝐷2 Kms at 𝑦 𝑘/ℎ𝑟, then
𝑥𝑦(𝐷1 + 𝐷2 ) 𝑘 𝑚
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 = 𝑜𝑟
𝑥𝐷2 + 𝑦𝐷1 ℎ𝑟 𝑠𝑒𝑐

Average Speed:

a) When an object covers two equal distances at two different speeds, i.e.

If a certain distance is covered at 𝑥 𝑘/ℎ𝑟 and the same distance is covered at

𝑦 𝑘/ℎ𝑟

OR

If two equal distances are covered at 𝑥 𝑘/ℎ𝑟 and 𝑦 𝑘/ℎ𝑟 respectively, then
the
2𝑥𝑦 𝑘
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 =
𝑥+𝑦 ℎ𝑟

b) If three equal distances are covered at 3 different speeds viz., 𝑥 𝑘/ℎ𝑟 , 𝑦 𝑘/ℎ𝑟
and 𝑧 𝑘/ℎ𝑟

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Then the
2𝑥𝑦𝑧 𝑘
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 =
𝑥𝑦 + 𝑦𝑧 + 𝑥𝑧 ℎ𝑟

c) If a man travels a certain distance at 𝑢 𝑘/ℎ𝑟 reaching ‘p’ hrs late and when he
travels at ′𝑣′ 𝑘/ℎ𝑟 reaching ‘q’ hrs early then the distance travelled is given by
𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑡𝑤𝑜 𝑠𝑝𝑒𝑒𝑑𝑠
𝐷= × 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑎𝑟𝑟𝑖𝑣𝑎𝑙 𝑡𝑖𝑚𝑒𝑠
𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑓 2 𝑠𝑝𝑒𝑒𝑑𝑠
𝑢𝑣
𝐷= × (𝑝 + 𝑞) 𝑘𝑚𝑠
𝑣−𝑢
7. If a man goes from A to a place B at ′𝑢′ 𝑘/ℎ𝑟 and returns to A at ′𝑣′ 𝑘/ℎ𝑟 taking
a total of ‘T’ hours for the whole journey, then the distance between the two points
A and B
𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑡𝑤𝑜 𝑠𝑝𝑒𝑒𝑑𝑠
𝐷= × 𝑇𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
𝑆𝑢𝑚 𝑜𝑓 2 𝑠𝑝𝑒𝑒𝑑𝑠
𝑢𝑣
𝐷= × 𝑇 𝑘𝑚𝑠
𝑢+𝑣
8. If a certain distance is covered with a speed of 𝑥 𝑘/ℎ𝑟 and another distance
with a speed of 𝑦 𝑘/ℎ𝑟 but time interval (time taken) for both the journeys being
the same, then the average speed for the whole journey is given by
𝑥+𝑦
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 = ( ) 𝑘/ℎ𝑟
2
9. For 3 distances and three speeds 𝑥 𝑘/ℎ𝑟, 𝑦 𝑘/ℎ𝑟 , 𝑧 𝑘/ℎ𝑟, the time interval for
the 3 journeys being the same, then the whole journey
𝑥+𝑦+𝑧
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 = ( ) 𝑘/ℎ𝑟
3
10. If the ratio of speeds A and B is𝑥: 𝑦, then the ratio of times taken to clear the
same distance is
1 1
:
𝑥 𝑦

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𝑥
11. If the new speed is of the original speed, then the change in time taken to
𝑦
cover the same distance is given by
𝑦
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑡𝑖𝑚𝑒 = ( − 1) × 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑡𝑖𝑚𝑒
𝑥
12. A and B start at the same time from two points P and Q towards each other
and after crossing, they take 𝑇1 and 𝑇2 hours in reaching Q and P respectively, then

𝐴′ 𝑠 𝑆𝑝𝑒𝑒𝑑 √𝑇2
=
𝐵′ 𝑠 𝑆𝑝𝑒𝑒𝑑 √𝑇1

13. A body covers a distance d in time 𝑡1 with speed 𝑠1 and covers the same
distance when it travels for time 𝑡2 with speed 𝑠2 , then
𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑆𝑝𝑒𝑒𝑑𝑠 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑓 𝑆𝑝𝑒𝑒𝑑𝑠 𝑠1 ~𝑠2
= =
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒(𝑑) 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑓 𝑡𝑖𝑚𝑒 𝑡1 ~𝑡2

PROBLEMS:

1. A runner can complete a 750 m race in two and a half minutes. Will he be able
to beat another runner who runs at 17.95 km/hr?
Solution:
We are given that the first runner can complete a 750 m race in 2 minutes and 30
seconds or 150 seconds.
Speed of the first runner = 750 / 150 = 5 m / sec
We convert this speed to km/hr by multiplying it by 18/5.
Speed of the first runner = 18 km / hr
Also, we are given that the speed of the second runner is 17.95 km/hr.
Therefore, the first runner can beat the second runner.

2. A man decided to cover a distance of 6 km in 84 minutes. He decided to cover

two-thirds of the distance at 4 km/hr and the remaining at some different speed.
Find the speed after the two-thirds distance has been covered.
Solution:
We are given that two-thirds of the 6 km was covered at 4 km/hr.
4 km distance was covered at 4 km/hr.

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Time taken to cover 4 km = 4 km / 4 km / hr = 1 hr = 60 minutes

Time left = 84 – 60 = 24 minutes
Now, the man has to cover the remaining 2 km in 24 minutes or 24 / 60 = 0.4
hours
Speed required for remaining 2 km = 2 km / 0.4 hr = 5 km / hr

3. A postman traveled from his post office to a village in order to distribute mail.
He started on his bicycle from the post office at a speed of 25 km/hr. But, when
he was about to return, a thief stole his bicycle. As a result, he had to walk back
to the post office on foot at the speed of 4 km/hr. If the traveling part of his day
lasted for 2 hours and 54 minutes, find the distance between the post office and
the village.
Solution :
Let the time taken by postman to travel from post office to village=t minutes.
According to the given situation, distance from post office to village, say
d1=25/60*t km {25 km/hr = 25/60 km/minutes}
And
distance from village to post office, say d2=4/60*(174-t) km {2 hours 54
minutes = 174 minutes}
Since distance between village and post office will always remain same i.e. d1 =
d2
25/60*t = 4/60*(174-t)
t = 24 minutes.
Distance between post office and village = speed*time
25/60*24 = 10km

4. Walking at the speed of 5 km/hr from his home, a geek misses his train by 7
minutes. Had he walked 1 km/hr faster, he would have reached the station 5
minutes before the actual departure time of the train. Find the distance between
his home and the station.
Solution:
Let the distance between his home and the station be ‘d’ km.
Time required to reach the station at 5 km / hr = d/5 hours
Time required to reach the station at 6 km/hr = d/6 hours
Now, the difference between these times is 12 minutes = 0.2 hours. (7 minutes
late – 5 minutes early = (7) – (-5) = 12 minutes)
Therefore, (d / 5) – (d / 6) = 0.2
d / 30 = 0.2

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d=6
Thus, the distance between his home and the station is 6 km.

5. Two trains start simultaneously from Lucknow and Jaipur towards each other
with speeds of 70km/hr and 90km/hr respectively. When they met each other it
was observed that one of them had covered 350km more than the other. Find the
distance between Lucknow and Jaipur.
A) 2800 km b) 3200km c) 1800 km d) 2600 km
Solution:
Both trains meet at the same time, so we can conclude that the distance travelled
by the trains is directly proportional to their speeds.
If the distance travelled by the slower train is x km then the distance travelled by
the faster train is (x+350) km
90
Now,
70
𝑜𝑟, 𝑥 = 1225 𝑘𝑚
∴ 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝐿𝑢𝑐𝑘𝑛𝑜𝑤 𝑎𝑛𝑑 𝐽𝑎𝑖𝑝𝑢𝑟 = 2𝑥 + 350
= 2450 + 350 = 2800𝑘𝑚

6. A train running at 72 km/hr takes 15 seconds to cross a platform and 9 seconds

to pass a man walking in the same direction at the speed of 6 km/hr. find the
length of the platform.
a) 165 metres b) 145 metres c) 135 metres d) 160 metres
Solution:
Let the length of platform be x metres and the length of the train = y metres
Then, according to the question,
𝑥+𝑦
15 =
5
72 ×
18
𝑜𝑟, 𝑦 = 55 × 3 = 165 𝑚𝑒𝑡𝑟𝑒𝑠.

Hence, the length of the train is 165 metres. Therefore, the length of the platform

𝑦 = 300 − 165 = 135 𝑚𝑒𝑡𝑟𝑒𝑠

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1
7. The average speed of a train is 4 times the average speed of a tractor. The
2
tractor covers 384 kms. In 16 hours. How much distance will the train cover in
12 hours?
a) 1396 kms b) 1296 kms. c) 1406 kms d) 1460 kms
Solution:
Average speed of tractor
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑐𝑜𝑣𝑒𝑟𝑒𝑑
=
𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
384
= = 24 𝑘𝑚𝑝ℎ
16
9
∴ 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑡𝑟𝑎𝑖𝑛 = × 24 = 108 𝑘𝑚𝑝ℎ
2
∴ Distance covered in 12 hours= 𝑠𝑝𝑒𝑒𝑑 × 𝑡𝑖𝑚𝑒
= 108 × 12 = 1296 𝑘𝑚.

8. The average speed of a bus is 67 kms per hour. The bus was scheduled to start at
12 pm. It was scheduled to reach a destination 335 kms away from its starting
point at 7 pm and a halt was scheduled on the way. For how long was the halt
scheduled?
a) 3 hours b) 2 hours c) 1 hour d) cannot be determined

Solution:
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 335
𝑇𝑖𝑚𝑒 = = =5
𝑆𝑝𝑒𝑒𝑑 67
∴ Time of halt = 7-5 =2 hours.

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11.SIMPLE INTEREST
SIMPLE INTEREST

Interest:

Interest is the money paid by the borrower to the lender for the use of money.

Simple Interest:

If the interest is calculated on the original principal at any rate of interest for any
period on time, then it is called simple interest.

Formulae:

A -Amount= Principal + Interest

I - Simple Interest P-Principal

n- Number of years r- Rate of interest (%)

𝑃𝑛𝑟
1. 𝐼=
100
100𝐼
2. 𝑃=
𝑛𝑟
100𝐼
3. 𝑛=
𝑃𝑟
100𝐼
4. 𝑟=
𝑃𝑛
𝑛𝑟
5. 𝐴𝑚𝑜𝑢𝑛𝑡 𝐴 = 𝑃( 1 + )
100

Example Question Based on S.I. Formula

1. Calculate the Simple Interest if the principal amount is Rs. 2000, the time period
is 1 year and the rate is 10%. Also, calculate the total amount at the end of 1
year.
Solution:
According to the formula of simple interest we have,
S.I. = [(Principal (P) × Time (T) × Rate (r)) / 100]
So, from the above values,

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S.I. = [(2000 × 1 × 10)] / 100

= 20000/100
=200
So, the simple interest at the end of 1 year will be Rs. 200.
For the amount after 1 year,
A = P + S.I.
So, A = 2000+200 = 2200
Hence, the total amount at the end of the given tenure (i.e. 1 year) will be Rs. 2200.

2. Rishav takes a loan of Rs 10000 from a bank for a period of 1 year. The rate
of interest is 10% per annum. Find the interest and the amount he has to
pay at the end of a year.
Solution:
Here, the loan sum = P = Rs 10000
Rate of interest per year = R = 10%
Time for which it is borrowed = T = 1 year
Thus, simple interest for a year, SI = (P × R ×T) / 100 = (10000 × 10 ×1) / 100 =
Rs 1000
Amount that Rishav has to pay to the bank at the end of the year = Principal +
Interest = 10000 + 1000 = Rs 11,000

3. Namita borrowed Rs 50,000 for 3 years at the rate of 3.5% per annum. Find
the interest accumulated at the end of 3 years.
Solution:
P = Rs 50,000
R = 3.5%

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T = 3 years
SI = (P × R ×T) / 100 = (50,000× 3.5 ×3) / 100 = Rs 5250
4. Mohit pays Rs 9000 as an amount on the sum of Rs 7000 that he had
borrowed for 2 years. Find the rate of interest.
Solution:
A = Rs 9000
P = Rs 7000
SI = A – P = 9000 – 7000 = Rs 2000
T = 2 years
R=?
SI = (P × R ×T) / 100
R = (SI × 100) /(P× T)
R = (2000 × 100 /7000 × 2) =14.29 %
Thus, R = 14.29%

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12.COMPOUND INTEREST
The addition of interest to the principal at the end of each year or some fixed
time period is known as compounding. In the case of compound interest, the
interest of each year or some fixed is added to the principal and the new amount
becomes the principal for the next year and interest is calculated on the
increased amount for the next year.

FORMULAE:

A – Amount after ‘n’ years ; P – principal ;

n – No. Of years; r – rate of interest (%)

𝑟
1. 𝐴 = 𝑃[1 + ]𝑛
100
(Interest is compound annually)

𝑟 3
If n = 3 2⁄5 yrs, 𝐴 = 𝑃 (1 + )
100

2. 𝐶. 𝐼 = 𝐴 − 𝑃

𝑟 𝑛
= 𝑃[1 + ] −𝑃
100

𝑟 𝑛
= 𝑃[(1 + ) −1
100
S.I and C.I for first year at given rate of interest per annum are always equal.
1
𝐴 𝑛
3. Rate of interest 𝑟= 100 [( ) − 1] % 𝑝. 𝑎
𝑃

4. If the interest is compounded half-yearly (2 times a year) then,

𝑟 2𝑛
𝐴 = 𝑃[1 + ]
200
𝑟
𝐶. 𝐼 = 𝑃[(1 + )2𝑛 − 1
200
1
𝐴 ×2
𝑟 = 2 × 100[( ) 𝑛 − 1]% 𝑝. 𝑎.
𝑃

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𝑟
5. If compounding is done quarterly (4 times a year) 𝐴 = 𝑃[1 + ]4𝑛
400
𝑟
𝐶. 𝐼 = 𝑃[(1 + )4𝑛 —1]
400

𝐴 1
𝑟 = 4 × 100[( )𝑛 ×4 1−]% 𝑝. 𝑎.
𝑃

6. If it takes place ‘n’ times a year,

𝑟
𝐴 = 𝑃[1 + ]𝑛×𝑛
𝑛 × 100

𝑟
𝐶. 𝐼 = 𝑃[(1 + )𝑛×𝑛 − 1
𝑛 × 100

𝐴 1
𝑟 = 𝑛 × 100[( )𝑛 ×𝑛 − 1]%𝑝. 𝑎
𝑃
7. If the rate of interest is different for different years r1 %, r2 %, r3 % for 1st, 2nd
and 3rd years respectively, then
𝑟1 𝑟2 𝑟3
𝐴 = 𝑃(1 + ) (1 + ) (1 + )
100 100 100
8. The difference between C.I and S.I on a sum for 2 years at r% p.a is

𝑟 2
𝐷2 = 𝑃( ) 𝑖𝑓 𝑃 𝑎𝑛𝑑 𝑟 𝑎𝑟𝑒 𝑔𝑖𝑣𝑒𝑛
100

9. The difference between the C.I and S.I on a sum for 3 years at r% p.a is

𝑟 2 𝑟
𝐷3 = 𝑃( ) ( + 3)
100 100

5. What is the interest for the third year if Rs.26600 is lent out, at the rate of
interest 11% p.a, the interest being compounded annually? (in approximate
value)

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a) Rs.3277.38 b) Rs.3287.38 c) Rs.3288.38 d) Rs.3277.83

Solution:

The amount at the end of the second year will be the principal for the third year.

Hence, the amount at the end of the second year would be

11 2
= 26600 × (1 + )
100
= 26600 × (1.11)2 = 32773.86

Therefore the interest for the 3rd year would be

11
= 32773.86 ×
100
= 327.7386 × 11

6. Prem Shankar took a loan of Rs.75000 at a simple rate of interest 14% per
annum for 3 years and invested the money thus obtained at the rate of 14%
per annum for the same period, compounding annually. How much money did
he gain or lose in the whole transaction?

a) 4615.80 gain b) 4615.80 loss c) 3615 gain d) 4214 loss

Solution:

Interest paid on the loan

𝑃 × 14 × 3
= = 0.42𝑃
100
Now, the interest after investment
14 3
= 𝑃 [(1 + ) − 1]
100
= 𝑃[(1.14)3 − (1)]

= 𝑃[1.481544 − 1] = 0.481544 𝑃

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Now P=75000

Interest earned after investing = 0.481544 × 75

= 481544 × 75

= Rs. 36115.8

And the interest before investing = 0.42 P

= 0.42 × 75000 = Rs.31500

Hence, 36115.8 — 31500 = Rs.4615.80 is the profit he earned after investing.

7. The difference between the simple interest calculated yearly and the compound
interest calculated half yearly at same rate of interest for one year on a sum of
Rs.1200 is Rs.432. What is the rate of interest?
a) 10% b) 18% c) 15% d) 12%
Solution:
𝑃𝑟 2 1200 × 𝑟 2
𝐶𝐼 − 𝑆𝐼 = 𝑜𝑟 4.32 =
(100)2 100 × 100
432
𝑜𝑟, = 𝑟2
12

𝑟 = √36 = 6
But the rate of interest is half-yearly
Therefore 𝑟 = 6 × 2 = 12%
8. At a simple rate of interest money becomes it’s 5.5 times in 12.5 years. What
will Rs.13000 become after 2 years when compounded annually at the same rate
of interest?
a) Rs.24044.8 b) Rs.23621.58 c)Rs.22618.48 d) Rs.21958.48
Solution:
Let the amount be x.
𝑇ℎ𝑒𝑛 5.5𝑥 − 𝑥 = 4.5𝑥
4.5𝑥 × 100 4500
𝑁𝑜𝑤 𝑟𝑎𝑡𝑒 = =
12.5 × 𝑥 125

= 36%

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Therefore the rate of interest is 36% per annum. Now,

36 2
𝐴 = 13000 (1 + )
100
= 13000 × 1.36 × 1.36

Amount after two years = Rs.24044.8

9. What will be the difference in simple and compound interest at 8% per annum
on the sum of Rs.960 after 2 years?
a) Rs.6.144 b) Rs. 24.04 c) Rs.20.224 d) Rs.31.5

Solution:

Required difference

𝑟 2
= 𝑃( )
100
8 2
= 960 ( ) [𝑠𝑖𝑛𝑐𝑒 𝑃 = 960, 𝑟 = 12%]
100
8×8
= 960 × = 6.144
100 × 100
10.Divide Rs. 17943 between Peter and Wilson in such a way that the share of
Peter at the end of 32 years equals that of Wilson at the end of 34 years at
compound rate of interest 18% per annum. Find the share of Peter.
a) 1043 b) 7500 c) 3268 d) 12443
Solution:
Let Peter’s share be Rs.x and Wilson’s share be Rs. x
Then,
18 32 18 34
𝑥 (1 + ) = 𝑦 (1 + )
100 100

𝑥 118 2 13924
𝑜𝑟, = ( ) =
𝑦 100 10000
3481
𝑜𝑟, 𝑜𝑟, 𝑥: 𝑦 = 3481: 2500
2500
Therefore share of Peter

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3481
= × 17943 = 10443
5981
11.The difference between compound interest and simple interest on a certain sum
of money at 10% per annum for 2 years is Rs.700. Find the sum when the
interest is compounded annually.
a) 40000 b) 45000 c) 55000 d) 70000
Solution:
𝑝𝑟 2
𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 =
(10)2

700 × 100 × 100

𝑜𝑟, 𝑃 = = 𝑅𝑠 70000
10 × 10
12.A sum of Rs. 1260 was taken as a loan. This is to be repaid in two equal annual
instalments. If the rate of interest be 10% compounded annually, then the value of
each instalment is
a) Rs. 514 b) Rs. 718 c) Rs. 726 d) Rs. 748
Solution:
Let the value of each instalment be Rs. x then
𝑥 𝑥
+ 2 = 1260
10 10
(1 + 100) (1 +
100)
10𝑥 100𝑥
+ = 1260
11 121
110𝑥 + 100𝑥 = 1260 × 121
210𝑥 = 152460
𝑥 = 726
1
13.A tree increases annually by 𝑡ℎ of its height. By how much will it increase after
8
1
2 𝑦𝑒𝑎𝑟𝑠. If it stands today 16m high?
4
a) 20.88 m b) 16.87 m c) 20 m d) 17.58m
Solution:
1 1
𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒 % = × 100 = 12 %
8 2
1
Height after 2 𝑦𝑒𝑎𝑟𝑠
4

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2
25 25
= 16 × (1 + ) × (1 + )
2 × 100 8 × 100

9 9 33
= 16 × × × ≅ 20.88𝑚
8 8 32

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13.PROFIT AND LOSS

IMPORTANT FACTS

Cost Price:
The price, at which an article is purchased, is called its cost price,
abbreviated as C.P.

Selling Price:
The price, at which an article is sold, is called its selling prices, abbreviated
as S.P.

Profit or Gain:
If S.P. is greater than C.P., the seller is said to have a profit or gain.

Loss:
If S.P. is less than C.P., the seller is said to have incurred a loss.

IMPORTANT FORMULAE

1. Gain = (S.P.) - (C.P.)

2. Loss = (C.P.) - (S.P.)

3. Loss or gain is always reckoned on C.P.

4. Gain Percentage: (Gain %)

S.P−C.P
5. profit or gain % = × 100
C.P

C.P−S.P
6. loss % = × 100
C.P
7. Marked Price: (M.P)
This is basically the price of a product labeled by shopkeepers on the
product for sale.
8. Discount:
The reduction in the marked price of an object is called discount.
discount = marked price − selling price
discount
discount percentage = ( ) × 100
M. P

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9. When there is profit,

100 + Profit%
SP = × CP
100
10. When there is loss,
100 − LOSS%
SP = × CP
100
100
11. CP = SP × [if there is profit]
100+profit%
100
12. CP = SP × [if there is loss]
100−loss%
13. Cost price is always 100%
14. Assume cost price as Rs.100, when it is unknown Profit% - profit on
Rs.100 Loss% - loss on Rs.100
15. If profit is x%, then the SP = 100+x %
16. If loss is x%, then the SP = 100 – x%
17. If CP of two articles is the same, one is sold at again of x% and the other at
loss of x% then the result is no gain / no loss.
18. If SP of two articles is the same, one is sold at a gain of x% and the other at
x2
loss of x% then there is always loss equal to %
100

Dishonest dealer using false weight error

= × 100
% gain true value−error

true weight−false weight

= × 100
false weight

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PROBLEMS:

1. A person buys a pen from a wholesaler at Rs. 10 for 20 pens. He sells those
pens at Rs. 10 for 15 pens. Find his profit or loss percent.
Solution:
CP for each pen = 10 / 20 = Rs. 0.50
SP for each pen = 10 / 15 = Rs. 2 / 3
Profit = SP – CP = Rs. (2 / 3) – 0.50 = Rs. 1 / 6
Therefore, profit percent = [ (1/6) / (0.50) ] x 100 = 33.334%
2. A dealer incurs a loss of 5 % if he sells an article for Rs. 1805. What price
must he sell the article so as to gain 5 % on that article?
Solution:
Let the cost price of the article be Rs. C
SP = CP – Loss
1805 = C – 0.05 C
0.95 C = 1805
C = 1900
Therefore, to gain 5 %, SP = 1900 + (0.05 x 1900) = 1900 + 95 = Rs. 1995
3. If the cost price of an article is 67 % of the selling price, what is the profit
percent?
Solution:
Let the selling price of the article be Rs. S
Cost price of the article = 67 % of S = 0.67 S
Profit = SP – CP = 0.33 S
Therefore, profit percent = (0.33 S / 0.67 S) x 100 = 49.25 %
4. A man bought a tonga and a horse for 15,000. He sold the tonga and the
horse at 10% and 25% profit respectively, thereby making 15% profit on the
whole. Find the cost price of the horse.
a) 7000 b) 7500 c) 5000 d) 8900
Solution:
By Alligation method
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Tonga 10% Horse 25%

↘ ↙
15%

↗ ↖

10 5

Ratio = 2:1
1
𝐶. 𝑃 𝑜𝑓 ℎ𝑜𝑟𝑠𝑒 = × 15000 = 5000
3

5. A seller claims to sell at cost price but gives 750 gm for each KG. Find his
gain percent.
a) 20.3 b) 33.33 c) 42.3 d) 23.33
Solution:
Profit percent = [ (True Value – Given Value) / Given Value ] x 100 %
Here,
True Value = 1 KG = 1000 gm
Given Value = 750 gm
Therefore, profit percent = [ (1000 – 750) / 750 ] x 100
= (250 / 750) x 100
= 33.334 %
6. A man bought goods worth 4096 and sold one – sixth of them at 12%, one
third of them at 9% and one – fourth of them at 10% profit. Then at what
loss percent must he sell the remaining goods so as to get no profit and no
loss on the whole transaction?
a) 30% b) 28% c) 27.6% d) 29.6%
Solution:
1 1 1 1
(12%) + (9%) + (10%) + (−𝑥%) = 0
6 3 4 4

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1 1 1 1
𝑅𝑒𝑚𝑎𝑖𝑛𝑖𝑛𝑔 𝑔𝑜𝑜𝑑𝑠 = 1 − − − =
6 3 4 4
𝑥
𝑜𝑟, 2 + 3 + 2.5 − =0
4
𝑜𝑟, 𝑥 = 7.5 × 4

𝑥 = 30%
7. A shopkeeper gives two successive discounts of 20 % and 10 % on surplus
stock. Further, he also gives a 5 % extra discount on cash payments. If a
person buys a shirt from the surplus stock and pays in cash, what overall
discount percent will he get on the shirt?
a) 32.50 b) 42.5 c) 31.60 d) 36.50
Solution:
Let the marked price of the shirt be Rs. 1000
Price after first discount = Rs. 1000 – 20 % of Rs. 1000
= Rs. 1000 – 200 = Rs. 800
Price after second discount = Rs. 800 – 10 % of Rs. 800
= Rs. 800 – 80 = Rs. 720
Price after cash discount = Rs. 720 – 5 % of Rs. 720
= Rs. 720 – 36 = Rs. 684
Therefore, total discount = Rs. 1000 – 684 = Rs. 316
Overall discount percent = (316 / 1000) x 100 = 31.60 %
8. A shopkeeper gives 10% discount on the market price and gives 4 articles
free of cost for buying every 20 articles and thus gains 25%. The market
price is how much percent more than the cost price?
2
a) 50% b) 45% c) 166 % d) 32%
3
Solution:
Let the market price be Rs.100
According to the question, SP = Rs. 90
For 20 articles SP = 90 × 20 = Rs.1800
But in place of 20 he provides 24 and still gains 25%

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1800
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝐶𝑃 =
1.25 × 24
𝐶𝑃 𝑝𝑒𝑟 𝑎𝑟𝑡𝑖𝑐𝑙𝑒 = 𝑅𝑠. 60
100 × 100 2
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 % = = 166 %
60 3
2
Hence, the market price of an article is 66 % above the cost price.
3
9. The profit earned after selling an article for Rs.536 is the same as loss
incurred after selling the article for Rs.426. What is the cost price of the
article?
a) Rs.448 b) Rs. 470 c) 481 d) Rs.500
Solution:
536+426 962
𝐶𝑜𝑠𝑡 𝑝𝑟𝑖𝑐𝑒 𝑜𝑓 𝑎𝑛 𝑎𝑟𝑡𝑖𝑐𝑙𝑒 = = = 𝑅𝑠. 481
2 2
10.A dealer wants to mark the price of an article such that by offering a 5 %
discount, he is able to get 33 % profit. Find the percent of CP above which
the article should be marked.
Solution:
Let the cost price of the article be Rs. 100
Selling price of the article = Rs. 100 + 33% of CP = Rs. 133
Let the marked price be Rs. M
Selling price = Marked Price – Discount
133 = M – 0.05 M
133 = 0.95 M
M = 140
M – CP = 140 – 100 = 40
Therefore, percent of CP above which the article should be marked
= (40 / 100) x 100 = 40 %
11.The profit earned after selling an article for Rs. 625 is the same as the loss
incurred after selling the article for Rs. 435. What is the cost price of the
article?
a) 530 b) 490 c) 540 d) 550
Solution:
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Here, 625 – x = 435 + x

Or, 2x = 625 – 435
190
∴𝑥= = 95
2
Hence the cost price of the article = 435 + 95 = Rs. 530
12.The average profit earned by a trader in the first three days of a week is Rs.
550 per day. The average loss for the next two days is Rs. 400 per day. What
should be the loss or gain per day for the last 2 days so that there is no loss
or gain in the entire week?
a) 400 gain b) 425 gain c) 425 loss d) 400 loss
Solution:
𝑃𝑟𝑜𝑓𝑖𝑡 𝑓𝑜𝑟 3 𝑑𝑎𝑦𝑠 = 550 × 3 = 𝑅𝑠. 1650
𝐿𝑜𝑠𝑠 𝑓𝑜𝑟 2 𝑑𝑎𝑦𝑠 = 400 × 2 = 𝑅𝑠. 800
Let the profit or loss for the last two days be Rs.x each.
Then, 1650 – 800 – 2x = 0
or, x = - Rs.425

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14.ALGEBRA
Polynomial:
A polynomial is defined as an expression which is composed of variables,
constants and exponents that are combined using mathematical operations such as
addition, subtraction, multiplication and division (No division operation by a
variable).
Based on the number of terms present in the expression, it is classified as
monomial, binomial, and trinomial.
Examples of constants, variables and exponents are as follows:

 Constants. Example: 1, 2, 3, etc.

 Variables. Example: g, h, x, y, etc.
 Exponents: Example: 5 in x5 etc.

Standard Form of a Polynomial

p(x) = an x n + an−1 x n−1 + ⋯ … . . +a1 x + a0 where an, an-1, an-2, ……………………,

a1, a0 are called coefficients of xn, xn-1, xn-2, ….., x and constant term respectively
and it should belong to real number (⋲ R).

Notation
The polynomial function is denoted by P(x) where x represents the variable. For
example,

P(x) = x2-5x+11

If the variable is denoted by a, then the function will be P(a)

Degree of a Polynomial

The degree of a polynomial is defined as the highest exponent of a

monomial within a polynomial. Thus, a polynomial equation having one variable
which has the largest exponent is called a degree of the polynomial.

Polynomial Degree Example

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Zero polynomial Not defined 6

Constant 0 P(x) = 6

Linear polynomial 1 P(x) = 3x+1

Quadratic polynomial 2 P(x) = 4x2+1x+1

Example: Find the degree of the polynomial P(x) = 6s4+ 3x2+ 5x +19

Solution:

The degree of the polynomial is 4 as the highest power of the variable 4.

Terms of a Polynomial

The terms of polynomials are the parts of the expression that are generally
separated by “+” or “-” signs.

So, each part of a polynomial in an expression is a term.

For example, in a polynomial, say, 2x2 + 5 +4, the number of terms will be 3

The classification of a polynomial is done based on the number of terms in

it.

Polynomial Terms Degree

P(x) = x3-2x2+3x+4 x3, -2x2, 3x and 4 3

Types of Polynomials

Depending upon the number of terms, polynomials are divided into the following
categories:

 Monomial

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 Binomial
 Trinomial
 Polynomial containing 4 terms (Quadronomial)
 Polynomial containing 5 terms (pentanomial ) and so on …
These polynomials can be combined using addition, subtraction, multiplication,
and division but is never divided by a variable. A few examples of Non
Polynomials are: 1/x+2, x-3

Monomial
A monomial is an expression which contains only one term. For an expression
to be a monomial, the single term should be a non-zero term. A few examples of
monomials are:

 5x
 3
 6a4
 -3xy

Binomial
A binomial is a polynomial expression which contains exactly two terms. A
binomial can be considered as a sum or difference between two or more
monomials. A few examples of binomials are:

 – 5x+3,
 6a4 + 17x
 xy2+xy

Trinomial
A trinomial is an expression which is composed of exactly three terms. A few
examples of trinomial expressions are:

– 8a4+2x+7
2
 4x + 9x + 7

Monomial Binomial Trinomial

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One term Two terms Three terms

Example: x, 3y, 29, x/2 Example: x2+x, x3-2x, Example: x2+2x+20
y+2

Properties

Some of the important properties of polynomials along with some important

polynomial theorems are as follows:

Property 1: Division Algorithm

If a polynomial P(x) is divided by a polynomial G(x) results in quotient Q(x) with
remainder R(x), then,

P(x) = G(x) • Q(x) + R(x)

Where R(x)=0 or the degree of R(x) < the degree of G(x)

Property 2: Remainder Theorem

If P(x) is divided by (x – a) with remainder r, then P(a) = r.

Property 3: Factor Theorem

A polynomial P(x) divided by Q(x) results in R(x) with zero remainders if and only
if Q(x) is a factor of P(x).

Property 6
The addition, subtraction and multiplication of polynomials P and Q result in a
polynomial where,

Degree(P ± Q) ≤ Degree(P or Q)

Degree(P × Q) = Degree(P) + Degree(Q)

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Property 7
If a polynomial P is divisible by a polynomial Q, then every zero of Q is also a
zero of P.

Property 8
If a polynomial P is divisible by two co-prime polynomials Q and R, then it is
divisible by (Q • R).

Property 9
If P(x) = a0 + a1x + a2x2 + …… + anxn is a polynomial such that deg(P) = n ≥ 0
then, P has at most “n” distinct roots.

Property 10: Descartes’ Rule of Sign

The number of positive real zeroes in a polynomial function P(x) is the same
or less than by an even number as the number of changes in the sign of the
coefficients. So, if there are “K” sign changes, the number of roots will be “k” or
“(k – a)”, where “a” is some even number.

Polynomial Equations

Polynomial equations are those expressions which are made up of multiple

constants and variables. The standard form of writing a polynomial equation is to
put the highest degree first and then, at last, the constant term. An example of a
polynomial equation is:

0 = a4 +3a3 -2a2 +a +1

Polynomial Functions

A polynomial function is an expression constructed with one or more terms

of variables with constant exponents. If there are real numbers denoted by a, then
function with one variable and of degree n can be written as:

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f(x) = a0xn + a1xn-1 + a2xn-2 + ….. + an-2x2 + an-1x + an

Solving Polynomials

Any polynomial can be easily solved using basic algebra and factorization
concepts. While solving the polynomial equation, the first step is to set the right-
hand side as 0. The explanation of a polynomial solution is explained in two
different ways:

 Solving Linear Polynomials

 Solving Quadratic Polynomials

1. Quadratic Equations:

(i) An equation of the type ax2 + bx + c =0 is called the quadratic equation.

(ii) The highest power of the variable is called the degree of an equation.

(iii) An equation will have as many solutions as its degree. If an equation is of n

degree, it will have 'n' solutions.

(iv) The solution of an equation is the value by which equation is satisfied. The
values of the solution of an equation are also called the roots of the equation. This
quadratic equation has two solutions.

2. Solving Quadratic Equations:

Any quadratic equation can be either solved by the factor method or by formula.

(i) By the factor method: First find the factors of the given equation making the
right-hand side equal to zero and then by equating the factors to zero, we get the
values of the variable.

(ii) By Formula: Consider a quadratic equation ax2 + bx + c = 0,for finding the

roots of the equation, we use the following formula:

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Here + and - in the above formula is used to get the two values of x. Here the
quantity b2 - 4ac is called the discriminant.

6. Linear Equations:

A statement of equality that contains an unknown quantity or variable is called an

equation. In a linear equation the pattern of numbers increases or decreases by the
same amount every step of the way, so the graph of a linear equation is always a
straight line.

1. If there are 2 apples in a bag and x apples are added to it, making a
total of 10 apples, how many apples were added?
Solution:
Let the number of apples added be x.
The total number of apples is 2 (initial) + x (added) = 10.
So, 2 + x = 10.
Subtract 2 from both sides to find x: x = 10 - 2.
Therefore, x = 8 apples were added.

2. A pen costs x dollars. If 5 pens cost $15, what is the cost of one pen?
Solution:
Let the cost of one pen be x dollars.
The total cost for 5 pens is 5x = $15.
Divide both sides by 5 to find the cost per pen: x = 15 / 5.
Therefore, each pen costs $3.

3. A train travels x miles in 2 hours. If it traveled 100 miles, how many

miles does it travel in one hour?
Solution:
Let the distance traveled in one hour be x miles.
In 2 hours, the train travels 2x miles.
We know 2x = 100 miles.
Divide both sides by 2 to find x: x = 100 / 2.
Therefore, the train travels 50 miles in one hour.

4. If x people can complete a task in 4 hours, and 8 people can complete

the same task in 2 hours, how many people were originally there?

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Solution:
Let the original number of people be x.
The work done is the same, so x people in 4 hours is equal to 8 people in 2
hours.
So, 4x = 2 * 8.
Simplify to find x: x = (2 * 8) / 4.
Therefore, x = 4 people were originally there.

5. A rectangle's length is twice its width. If the width is x feet and the
area is 50 square feet, what is the width of the rectangle?
Solution:
Let the width be x feet.
The length is 2x feet.
The area of the rectangle is length * width, so 2x * x = 50.
Simplify to find x: x2 = 50 / 2.
Solving for x, we get x = sqrt(25) = 5 feet.

6. Sarah has x dollars. After buying a book for $10, she has $30 left. How
much money did Sarah have initially?
Solution:
Let Sarah's initial amount of money be x dollars.
After spending $10, she has x - 10 dollars left.
We know x - 10 = $30.
Add 10 to both sides to find x: x = 30 + 10.
Therefore, Sarah initially had $40.

7. A school has x students in each class. If there are 5 classes and a total
of 150 students, how many students are in each class?
Solution:
Let the number of students in each class be x.
Total students in 5 classes is 5x.
We know 5x = 150.
Divide both sides by 5 to find x: x = 150 / 5.
Therefore, there are 30 students in each class.

8. A baker makes x cookies from a batch of dough. If he makes 120

cookies and divides them into 6 boxes, how many cookies are in each
box?
Solution:
Let the number of cookies in each box be x.
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The total number of cookies is 6x.

We know 6x = 120.
Divide both sides by 6 to find x: x = 120 / 6.
Therefore, each box contains 20 cookies.

9. A baker used x cups of flour to make 3 cakes. If he used 15 cups in

total, how many cups did he use for each cake?
Solution:
Let the cups of flour used for each cake be x.
The total flour used for 3 cakes is 3x cups.
So, 3x = 15.
Dividing both sides by 3, we get x = 15 / 3.
Therefore, x = 5 cups of flour per cake.

10. In a garden, there are x roses and twice as many tulips. If there are 30
flowers in total, how many roses are there?
Solution:
Let the number of roses be x.
Then, the number of tulips is 2x.
The total number of flowers is x (roses) + 2x (tulips) = 30.
So, 3x = 30.
Dividing both sides by 3, x = 30 / 3.
Therefore, there are x = 10 roses.

ARITHMETIC PROGRESSION AND GEOMETRIC PROGRESSION

Definition 1:
A mathematical sequence in which the difference between two consecutive
terms is always a constant and it is abbreviated as AP.
Definition 2:
An arithmetic sequence or progression is defined as a sequence of numbers
in which for every pair of consecutive terms, the second number is obtained by
adding a fixed number to the first one.
The fixed number that must be added to any term of an AP to get the next
term is known as the common difference of the AP. Now, let us consider the
sequence, 1, 4, 7, 10, 13, 16,…

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It is considered as an arithmetic sequence (progression) with a common

difference 3.

Notation in Arithmetic Progression

In AP, we will come across some main terms, which are denoted as:

 First term (a)

 Common difference (d)
 nth Term (an)
 Sum of the first n terms (Sn)
All three terms represent the property of Arithmetic Progression. We will learn
more about these three properties in the next section.

First Term of AP
The AP can also be written in terms of common differences, as follows;
a, a + d, a + 2d, a + 3d, a + 4d, ………. ,a + (n – 1) d

where “a” is the first term of the progression.

Common Difference in Arithmetic Progression

In this progression, for a given series, the terms used are the first term, the
common difference and nth term. Suppose, a1, a2, a3, ……………., an is an AP,
then; the common difference “ d ” can be obtained as;
d = a2 – a1 = a3 – a2 = ……. = an – an – 1

Where “d” is a common difference. It can be positive, negative or zero.

General Form of an AP

Consider an AP to be: a1, a2, a3, ……………., an

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Position of terms Representation of terms Values of term

1 a1 a = a + (1-1) d

2 a2 a + d = a + (2-1) d

3 a3 a + 2d = a + (3-1) d

4 a4 a + 3d = a + (4-1) d

. . .

n an a + (n-1)d

Arithmetic Progression Formulas

There are two major formulas we come across when we learn about Arithmetic
Progression, which is related to:

 The nth term of AP

 Sum of the first n terms
Let us learn here both the formulas with examples.
nth Term of an AP
The formula for finding the n-th term of an AP is:
an = a + (n − 1) × d

Where
a = First term
d = Common difference
n = number of terms
an = nth term
Example:

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Find the nth term of AP: 1, 2, 3, 4, 5…., an, if the number of terms are 15.
Solution: Given, AP: 1, 2, 3, 4, 5…., an
n=15
By the formula we know, an = a+(n-1)d
First-term, a =1
Common difference, d=2-1 =1
Therefore, an = a15 = 1+(15-1)1 = 1+14 = 15
Note: The behaviour of the sequence depends on the value of a common
difference.

 If the value of “d” is positive, then the member terms will grow towards
positive infinity
 If the value of “d” is negative, then the member terms grow towards
negative infinity

Types of AP

Finite AP: An AP containing a finite number of terms is called finite AP. A finite
AP has a last term.
For example: 3,5,7,9,11,13,15,17,19,21
Infinite AP: An AP which does not have a finite number of terms is called infinite
AP. Such APs do not have a last term.
For example: 5,10,15,20,25,30, 35,40,45………………

Sum of N Terms of AP

For an AP, the sum of the first n terms can be calculated if the first term, common
difference and the total terms are known. The formula for the arithmetic
progression sum is explained below:
Consider an AP consisting “n” terms.

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Sn = n/2[2a + (n − 1) × d]

Example: Let us take the example of adding natural numbers up to 15 numbers.

AP = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15
Given, a = 1, d = 2-1 = 1 and an = 15
Now, by the formula we know;
Sn = n/2[2a + (n − 1) × d]
S15 = 15/2[2.1+(15-1).1]
= 15/2[2+14]
= 15/2 [16]
= 15 x 8
= 120
Hence, the sum of the first 15 natural numbers is 120.

Sum of AP when the Last Term is Given

Formula to find the sum of AP when first and last terms are given as follows:
S = n/2 (first term + last term)

List of Arithmetic Progression Formulas

The list of formulas is given in a tabular form used in AP. These formulas are
useful to solve problems based on the series and sequence concept.
General Form of AP a, a + d, a + 2d, a + 3d, . . .

The nth term of AP an = a + (n – 1) × d

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Sum of n terms in AP S = n/2[2a + (n − 1) × d]

Sum of all terms in a finite AP with the last term as ‘l’ n/2(a + l)

Arithmetic Progressions Solved Examples

Below are the problems to find the nth term and the sum of the sequence, which
are solved using AP sum formulas in detail. Go through them once and solve the
practice problems to excel in your skills.

1. Find the value of n, if a = 10, d = 5, an = 95.

Solution:
Given, a = 10, d = 5, an = 95
From the formula of general term, we have:
an = a + (n − 1) × d
95 = 10 + (n − 1) × 5
(n − 1) × 5 = 95 – 10 = 85
(n − 1) = 85/ 5
(n − 1) = 17
n = 17 + 1
n = 18
2. Find the 20th term for the given AP:3, 5, 7, 9, ……
Solution: Given,
3, 5, 7, 9, ……
a = 3, d = 5 – 3 = 2, n = 20
an = a + (n − 1) × d
a20 = 3 + (20 − 1) × 2

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a20 = 3 + 38
⇒a20 = 41
3. Find the sum of the first 30 multiples of 4.
Solution:
The first 30 multiples of 4 are: 4, 8, 12, ….., 120
Here, a = 4, n = 30, d = 4
We know,
S30 = n/2 [2a + (n − 1) × d]
S30 = 30/2[2 (4) + (30 − 1) × 4]
S30 = 15[8 + 116]
S30 = 1860

Geometric Progression

A geometric progression (GP) is a sequence of numbers in which each

successive term is the product of its preceding term and a fixed number. This fixed
number is called the common ratio. For example, 4, 16, 64, 256, ... is a GP as every
number is obtained by multiplying a fixed number 4 to its previous term.

General form:

a, ar, ar 2 , … … , ar n−1 , … … is a G.P. with the first term ‘a’ and common ratio ‘r’.

Common ratio:
ak+1
r= ,k ≥ 1
ak

𝐧𝐭𝐡 term of G.P. : 𝐚𝐫 𝐧−𝟏

a(rn−1 )
Sum of n terms of G.P Sn =
(r−1)

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1. Find the nth term for the AP: 11, 17, 23, 29, …
Solution:
Here, a = 11,
d = 17 – 11 = 23 – 17 = 29 – 23 = 6
We know that nth term of an AP is a + (n – 1) d
nth term for the given AP = 11 + (n – 1) 6
nth term for the given AP = 5 + 6 n
We can verify the answer by putting values of ‘n’.
n = 1 -> First term = 5 + 6 = 11
n = 2 -> Second term = 5 + 12 = 17
n = 3 -> Third term = 5 + 18 = 23and so on …
2. Find the sum of the AP in the above question till the first 10 terms.
Solution :
From the above question,
nth term for the given AP = 5 + 6 n
First term = 5 + 6 = 11
Tenth term = 5 + 60 = 65
Sum of 10 terms of the AP = 0.5 n
(first term + last term) = 0.5 x 10 (11 + 65)
Sum of 10 terms of the AP = 5 x 76 = 380
3. Find the sum of the series 32, 16, 8, 4, … upto infinity.
Solution :
First term, a = 32
Common ratio, r = 16 / 32 = 8 / 16 = 4 / 8 = 1 / 2 = 0.5
We know that for an infinite GP,
Sum of terms = a / (1 – r)
Sum of terms of the GP = 32 / (1 – 0.5) = 32 / 0.5 = 64

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4. Find the middle term of the A.P. 6, 13, 20, … , 216.

Solution:
In the given AP,
 First term a1 = 6
 Last term an = 216
 Common difference = 7
Now, to find the number of terms, n = (an – a1)/d + 1
n = (216 – 6)/7 + 1
n = 210/7 + 1
n = 30 + 1
n = 31
So, middle term is (n + 1) / 2
= (31 + 1)/2
= 16
Now to calculate the middle term
a16 = a1 + 15×d
a16 = 6 + 15 × 7
a16 = 6 + 105 = 111
So the middle term of the given AP is 111
5. In an AP, if ? = −2, ? = 5 and ?? = 0, then find the value of ? ?
Solution:
Given values are:
 d = -2
 n=5
 an = 0
So, to calculate follow these steps:
an = a + (n-1)d
0 = a + (5-1)(-2)
0=a–8
a=8
6. Number of bacteria in a dish are 100, and they are increasing by double the
previous value every hour. Find the number of bacteria in the dish after 6
hours.
Solution:
Here, every year the number becomes 2 times. A constant number is being
multiplied to the previous term to get the new term. This is a geometric
progression.
100,200, 400 … and so on.
Here a = 100 and r= 2
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Using the formula for sum till nth term of the GP

Sn = a(1–rn)1–r1–ra(1–rn)
n = 6. Plugging in the values in the formula
Sn = a(1–rn)1–r1–ra(1–rn)
⇒Sn = a(1–rn)1–r1–ra(1–rn)
⇒ S6 = 100(1–26)1–21–2100(1–26)
⇒ S6 = 100(26–1)2–12–1100(26–1)
⇒ S6 = 100(64–1)2–12–1100(64–1)
⇒ S6 = 6300
There must be 63,00 bacteria in the dish now.

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15.GEOMETRY
Point

A point is an exact location

Line Segment

The straight path between two points A and B is called a line segment AB. A
line segment has two end points.

Ray

On extending a line segment AB indefinitely in one direction we get the ray

AB. Ray AB has one end point, namely A.

LINE

A line segment AB extended indefinitely in both directions is called line AB.

1. A line contains infinitely many points.

2. Through a given points , infinitely many lines can be drawn.
3. One and only one line can be drawn to pass through two given points A and
B.
4. Two line meet in a point.
5. Two planes meet in a line.

Collinear

In the given figure, the points A,B,C are collinear.

Concurrent Lines

Three or more lines intersecting at the same points are called concurrent
lines.

Angle

Two rays OA and OB having a common end points O form angle AOB,
written as ∠AOB

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Measure of an Angle

The amount of turning from OA to OB is called the measure of ∠AOB

written as m(∠AOB).

An angle of 360°

If a ray OA starting from its original position OA , rotates about O in

anticlockwise direction and after a complete rotation comes back to its original
position , then we say that it has rotated through 360. This complete rotation is
divided into 360° equal parts. Then, each part is called 1 degree , written as 1°

1° = 60 minutes, written as 60'

1 minute = 60 seconds, written as 60"

Types of Angle

1. Right angle - An angle whose measure is 90° is called a right angle.

2. Acute angle - An angle whose measure is less than 90° is called an acute
angle.
3. Obtuse angle - An angle whose measure is more than 90° but less than
180°, is called an obtues angle.
4. Straight angle - An angle whose measure is 180° is called a Straight angle.
5. Reflex angle - An angle whose measure is more than 180° but less than
360°, is called a Reflex angle.
6. Complete angle - An angle whose measure is 360°, is called a complete
angle.
7. Equal angle - Two angles are said to be equal , if they have the same
measure.
8. Complementary angleTwo angles are said to be complementary if the sum
of their measures is 90. For example, angles measuring 65° and 25° are
complementary angle.

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9. Supplementary angle - Two angle are said to be supplementary if the sum

of their measures is 180°. For example, angles measures 70° and 110° are
supplementary.
10.Adjacent angle - Two angles are called adjacent angle if they have the same
vertex and a common arm such that non-common arms are on either side of
the comman arm. In the given figure , ∠AOC and ∠BOC are adjacent angle.

Important Results

If a ray stands on a line , than the sum of two adjacent angle so formed is 180° In
the given figure , ray CP stands on line AB.

∴ ∠ACD + ∠BCD = 180°.

The sum of all angle formed on the same side of a line at a given point on the line
is 180°. In the given figure four angle are formed on the same side of AOB.

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∴ ∠AOE + ∠EOD + ∠DOC + ∠COD = 180°.

The sum of all angle around a point is 360° In the given figure five angle are
formed around a point O.

∴∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA=360°.

Vertically Opposite Angles

If two lines A Band CD intersect at a point O, then AOC , BOD and BOC , AOD
are two pair of vertically opposites angle Vertically opposite angle are always
equal.

∴ ∠AOC = ∠BOD and ∠AOD = ∠BOC

Parallel Lines

If two lines lie in the same plane and do not intersect when produced on either side
then such lines are said to be paralleled and we write , l||m.

Traversal line cutting parallel lines

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Let two parallel lines AB and CD be cut by a transversal EF. Then

Corresponding angle are equal

(∠1 = ∠5), (∠4= ∠8 ), (∠2 = ∠6) , (∠3 = ∠7)

Alternate interior angles are equal.

(∠3 =∠5 ) and (∠4 =∠6 )

Consecutive interior angles are supplementary

∠4+∠5 = 180° and ∠3 +∠6 = 180°.

Triangle

A figure bounded by three straight lines is called a triangle. In the given figure , we
have ∆ABC; ∆ABC having three vertices A,B,C. In has three angles, namely
∠A,∠B and ∠C. It has three sides , namely AB, AC and BC.

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Types of Triangle

1. A triangle having all sides equal is called an equilateral triangle.

2. A triangle having two sides equal, is called an isosceles triangle.
3. A triangle having all sides of different lengths,is called a scalene triangle.
4. A triangle one of whose angles measures 90°,is called a right triangle.
5. A triangle one of whose angle lies between 90° and 180° is called an obtuse
triangle.
6. A triangle each of whose angle is acute, is called an acute triangle.
7. The sum of all sides of a triangle is called the perimeter of the triangle.
8. The sum of two sides of a triangle is greater than the third side.
9. In a right angled ABC in which ∠B = 90°, we have AC2 =AB2+BC2. This is
called Pythagoras Theorem.

Quadrilateral

A figure bounded by four straight line is called a quadrilateral. The sum of all
angles of a quadrilateral is 360°.

1. Rectangle - A quadrilateral is called a rectangle, if its opposite side are

equal and each of its angle is 90°. In given fig. ABCD is a rectangle.

2. Square - A quadrilateral is called a square, if all of its sides are equal and
each of its angles measures 90°. In given fig. ABCD is square in which AB
= BC = CD = DA.

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3. Parallelogram - A quadrilateral is called a parallelogram, if its opposite

sides are parallel. In given fig. ABCD is a parallelogram in which AB = DC
& AD = BC.

4. Rhombus - A parallelogram having all sides equal is called a rhombus. In

given fig. ABCD is a rhombus in which AB =BC =CD=DA, AB || DC and
AD || BC.

Important Facts

1. A quadrilateral is a rectangle if opposite sides are equal and its diagonals are
equal.
2. A quadrilateral is a Square if all sides are equal and the diagonal are equal.
3. A quadrilateral is a parallelogram, if opposite sides are equal.
4. A quadrilateral is a parallelogram but not a rectangle, if opposite sides are
equal but the diagonals are not equal.
5. A quadrilateral is a rhombus but not a square if all their sides are equal and
the diagonals are not equal.

Results on Quadrilateral

1. In a parallelogram, we have
1. Opposite sides are equal.
2. Opposite angles are equal.
3. Each diagonal bisects the parallelogram.

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4. Diagonals of a parallelogram bisect each other.

2. Diagonals of a rectangle are equal.
3. Diagonals of a rhombus bisect each other at right angles.

Results on Circle

1. The perpendicular from the center to a chord bisects the chord.

2. There is one and only one circle passing through three non collinear points.
3. Angle in a semi circle is a right angle.
4. Opposite angles of a cyclic quadrilateral are supplementary.
5. Angle in the same segment of a circle is equal.
6. The tangent at any point of a circle is perpendicular to the radius through the
point of contact.
7. Two tangent to a circle from a point outside it are equal.
8. If PT is a tangent to a circle and PAB is a secant, Then PA x PB= PT2

1. A, B, C are the three angles of a Δ. If A − B = 15° and B − C = 30°. Then

∠A is equal to :

A. 65° b)80° c) 75° 85°

Answer: B

Explanation:

Since A, B and C are the angles of a Δ,

∴ A + B + C = 180° ...... (1)

According to question,

A – B = 15°

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A = B + 15°...... (2)

B – C = 30°;

B = C + 30° ....... (3)

ut the value of B from equation (2) in Equation (1), we will get

∴ A = B + 15°

A = C + 30° + 15°

A = C + 45° ......... (4)

From the equation,

∴ A + B + C = 180°

(C + 45°) + (C + 30°) + C = 180°

3C + 45° + 30° = 180°

3C = 180° – 75° = 105°

C = 35° ........ (5)

From equation (4)

A = C + 45°

Put the value of C from equation (5), we will get

∠A = 35° + 45° = 80°.

2. In a right angled Δ ABC, Right angled at A, AD ⊥ BC. Then:

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A. AD2 = BD * CD B. AD2 = AB* AC C. AD2 = BD * AB D. AD2 =

CD* AC

Explanation:

From above given right angle triangle, we can see that

Since ∠1 + ∠2 = ∠2 + ∠3 (Each 90°)
∴ ∠1 = ∠3 [Each 90°]
Also ∠5 = ∠6 [Each 90°]
∴ ΔADB ∼ ΔCDA [AA Similarity]

4. If the sides of a right triangle are x, x + 1 and x – 1, then the hypotenuse:

A. 5 b) 4 c) 1 d)0
Explanation:

From given figure, we have

Let, (x + 1) be the hypotenuse and perpendicular = x - 1, base = x.
By Pythagoras theorem,
∴ (x + 1)2 = x2 + (x + 1)2 ⇒ x = 4
∴ Hypotenuse = 4 + 1 = 5.

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5. In Δ ABC, the median BE intersects AC at E, if BG = 6 cm, where G is

the centroid, then BE is equal to:

A. 7 cm b)9 cm c) 8 cm d)10 cm
Explanation:

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16.TRIGNOMETRY

In a right angled OAB, where BOA = ,

𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟
sin 𝜃 =
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑏𝑎𝑠𝑒
𝑐𝑜𝑠𝜃 =
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟
tan 𝜃 =
𝑏𝑎𝑠𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑐𝑜𝑠𝑒𝑐 𝜃 =
𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
sec 𝜃 =
𝑏𝑎𝑠𝑒
𝑏𝑎𝑠𝑒
cot 𝜃 =
𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟
Relation between trigonometric ratio:
1
𝑐𝑜𝑠𝑒𝑐 𝜃 =
sin 𝜃
1
sec 𝜃 =
cos 𝜃

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cos 𝜃
cot 𝜃 =
sin 𝜃
1
cot 𝜃 =
tan 𝜃
sin 𝜃
tan 𝜃 =
cos 𝜃
tan 𝜃. cot 𝜃 = 1

Trigonometric identities:

𝑠𝑖𝑛2 𝜃 + 𝑐𝑜𝑠 2 𝜃 = 1

1 + 𝑡𝑎𝑛2 𝜃 = 𝑠𝑒𝑐 2 𝜃

1 + 𝑐𝑜𝑡 2 𝜃 = 𝑐𝑜𝑠𝑒𝑐 2 𝜃

Trigonometric ratios of some special cases:

Angle 𝜽 𝟎° 𝟑𝟎° 𝟒𝟓° 𝟔𝟎° 𝟗𝟎°

sin 𝜽 0 1 1 √3 1
2 √2 2
cos 𝜽 1 √3 1 1 0
2 √2 2
tan 𝜽 0 1 1 √3 Not defined
√3
Cosec 𝜽 Not defined 2 √2 2 1
√3
Sec 𝜽 1 2 √2 2 Not defined
√3
Cot 𝜽 Not defined √3 1 1 0
√3

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Trigonometric ratio of complementary angle:

sin (90°- A) = cos A ; cos (90°- A) = sin A

tan (90°- A) = cot A; cot (90°- A) = tan A

sec (90°- A) = csc A; csc (90°- A) = sec A

1. Angle of Elevation:

Suppose a man from a point O looks up at an object P, placed above the level of
his eye. Then, the angle which the line of sight makes with the horizontal through
O, is called the angle of elevation of P as seen from O.
Angle of elevation of P from O = AOP.

2. Angle of Depression:

Suppose a man from a point O looks down at an object P, placed below the level of
his eye, then the angle which the line of sight makes with the horizontal through O,
is called the angle of depression of P as seen from O.

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1. Prove that (sin4θ – cos4θ +1) cosec2θ = 2

Solution:

L.H.S. = (sin4θ – cos4θ +1) cosec2θ

= [(sin2θ – cos2θ) (sin2θ + cos2θ) + 1] cosec2θ

Using the identity sin2A + cos2A = 1,

= (sin2θ – cos2θ + 1) cosec2θ

= [sin2θ – (1 – sin2θ) + 1] cosec2θ

= 2 sin2θ cosec2θ

= 2 sin2θ (1/sin2θ)

=2

= RHS

2. Prove that (√3 + 1) (3 – cot 30°) = tan360° – 2 sin 60°.

Solution:

LHS = (√3 + 1)(3 – cot 30°)

= (√3 + 1)(3 – √3)

= 3√3 – √3.√3 + 3 – √3

= 2√3 – 3 + 3

= 2√3

RHS = tan360° – 2 sin 60°

= (√3)3 – 2(√3/2)

= 3√3 – √3
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= 2√3

Therefore, (√3 + 1) (3 – cot 30°) = tan360° – 2 sin 60°.

Hence proved.

3. If tan(A + B) = √3 and tan(A – B) = 1/√3 ; 0° < A + B ≤ 90°; A > B, find

A and B.

Solution:

tan(A + B) = √3

tan(A + B) = tan 60°

A + B = 60°….(i)

And

tan(A – B) = 1/√3

tan(A – B) = tan 30°

A – B = 30°….(ii)

Adding (i) and (ii),

A + B + A – B = 60° + 30°

2A = 90°

A = 45°

Substituting A = 45° in (i),

45° + B = 60°

B = 60° – 45° = 15°

Therefore, A = 45° and B = 15°.

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4. If sin 3A = cos (A – 26°), where 3A is an acute angle, find the value of

A.

Solution:

Given,

sin 3A = cos(A – 26°); 3A is an acute angle

cos(90° – 3A) = cos(A – 26°) {since cos(90° – A) = sin A}

⇒ 90° – 3A = A – 26

⇒ 3A + A = 90° + 26°

⇒ 4A = 116°

⇒ A = 116°/4

⇒ A = 29°

5. If A, B and C are interior angles of a triangle ABC, show that sin (B +

C/2) = cos A/2.

Solution:

We know that, for a given triangle, the sum of all the interior angles of a
triangle is equal to 180°

A + B + C = 180° ….(1)

B + C = 180° – A

Dividing both sides of this equation by 2, we get;

⇒ (B + C)/2 = (180° – A)/2

⇒ (B + C)/2 = 90° – A/2

Take sin on both sides,

sin (B + C)/2 = sin (90° – A/2)

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⇒ sin (B + C)/2 = cos A/2 {since sin(90° – x) = cos x}

7. If tan θ + sec θ = l, prove that sec θ = (l2 + 1)/2l.

Solution:

Given,

tan θ + sec θ = l….(i)

We know that,

sec2θ – tan2θ = 1

(sec θ – tan θ)(sec θ + tan θ) = 1

(sec θ – tan θ) l = 1 {from (i)}

sec θ – tan θ = 1/l….(ii)

Adding (i) and (ii),

tan θ + sec θ + sec θ – tan θ = l + (1/l)

2 sec θ = (l2 + 1)l

sec θ = (l2 + 1)/2l

Hence proved.

8. Prove that (cos A – sin A + 1)/ (cos A + sin A – 1) = cosec A + cot A, using
the identity cosec2A = 1 + cot2A.

Solution:

LHS = (cos A – sin A + 1)/ (cos A + sin A – 1)

Dividing the numerator and denominator by sin A, we get;

= (cot A – 1 + cosec A)/(cot A + 1 – cosec A)

Using the identity cosec2A = 1 + cot2A ⇒ cosec2A – cot2A = 1,

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= [cot A – (cosec2A – cot2A) + cosec A]/ (cot A + 1 – cosec A)

= [(cosec A + cot A) – (cosec A – cot A)(cosec A + cot A)] / (cot A + 1 –

cosec A)

= [(cosec A + cot A) (1 – cosec A + cot A)]/ (1 – cosec A + cot A)]

= cosec A + cot A

= RHS

Hence proved.

9. Prove that: (cosec A – sin A)(sec A – cos A) = 1/(tan A + cot A)

Solution:

LHS = (cosec A – sin A)(sec A – cos A)

= [(1/sin A) – sin A) [(1/cos A) – cos A]

= [(1 – sin2A)/ sin A] [(1 – cos2A)/ cos A]

Using the identity sin2A + cos2A = 1,

= (cos2A/sin A) (sin2A/cos A)

= cos A sin A….(i)

RHS = 1/(tan A + cot A)

= 1/[(sin A/cos A) + (cos A/sin A)]

= (sin A cos A)/ (sin2A + cos2A)

= (sin A cos A)/1

= sin A cos A….(ii)

From (i) and (ii),

LHS = RHS

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i.e. (cosec A – sin A)(sec A – cos A) = 1/(tan A + cot A)

Hence proved.

10. If a sin θ + b cos θ = c, prove that a cosθ – b sinθ = √(a2 + b2 – c2).

Solution:

Given,

a sin θ + b cos θ = c

Squaring on both sides,

(a sin θ + b cos θ)2 = c2

a2 sin2θ + b2 cos2θ + 2ab sin θ cos θ = c2

Using the identity sin2A + cos2A = 1,

a2(1 – cos2θ) + b2(1 – sin2θ) + 2ab sin θ cos θ = c2

a2 – a2 cos2θ + b2 – b2 sin2θ + 2ab sin θ cos θ = c2

a2 + b2 – c2 = a2 cos2θ + b2 sin2θ – 2ab sin θ cos θ

a2 + b2 – c2 = (a cos θ – b sin θ )2

⇒ a cos θ – b sin θ = √(a2 + b2 – c2)

Hence proved.

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17.STATISTICS
MEAN

Mean is the average of the given numbers and is calculated by dividing the
sum of given numbers by the total number of numbers.
sum of all observations
Mean =
total number of observation
Mode:

Mode of a list of observations is the value among the observations which

occurs the highest number of times.

Median:

Median of a list of observations is the middle-most value of the

observations.

Let there is a list of n observations, then

𝑛+1 𝑡ℎ
If n is odd, then Median = ( ) observation
2

𝑛 𝑡ℎ 𝑛 𝑡ℎ
( ) 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛+ ( +1) 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛
2 2
If n is even, then Median =
2

Relation between Mean, Median and Mode = 3 Median – 2 Mean

PROBLEMS:

1. In a week, temperature of a certain place is measured during winter are as

follows 26°C, 24°C, 28°C, 31°C, 30°C, 26°C, 24°C. Find the mean
temperature of the week.
Solution:
Mean temperature of the week

(26 + 24 + 28 + 31 + 30 + 26 + 24)
=
7
= 27°C
Mean temperature of the week 27° C

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2. The mean weight of 4 members of a family is 60 kg. Three of them have the
weight 56 kg, 68 kg and 72 kg respectively. Find the weight of the fourth
member.
Solution:
Weight of 4 members = 4 × 60 kg
= 240 kg
Weight of three members = 56 kg + 68 kg + 72 kg
= 196 kg
Weight of the fourth member = 240 kg – 196 kg
= 44 kg
3. In a class test in mathematics, 10 students scored 75 marks, 12 students
scored 60 marks, 8 students scored 40 marks and 3 students scored 30
marks. Find the mean of their score.
Solution:
Total marks of 10 students = 10 × 75 = 750
Total marks of 12 students = 12 × 60 = 720
Total marks of 8 students = 8 × 40 = 320
Total marks of 3 students = 3 × 30 = 90
Total marks of (10 + 12 + 8 + 3) 33 students
= 750 + 720 + 320 + 90
= 1880
Mean of marks = 188033
= 56.97 (or) 57 approximately
4. Find the median of the given data: 36, 44, 86, 31, 37, 44, 86, 35, 60, 51
Solution:
Arrange the values in ascending order we get
31, 35, 36, 37, 44, 44, 51, 60, 86, 86
The number of values = 10 which is even
Median = Average of 5th and 6th value
= 44+442
= 882
= 44
∴ Median = 44
5. The median of observation 11, 12, 14, 18, x + 2, x + 4, 30, 32, 35, 41
arranged in ascending order is 24. Find the values of x.
Solution:
The given observation is 11, 12, 14, 18, x + 2, x + 4, 30, 32, 35, 41 (is
ascending order)
The number of values =10
Median = Average of 5th and 6th value
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24 = x+2.+x+42
24 = 2x+62
2x + 6 = 48
2x = 48 – 6
2x = 42
x = 422
= 21
The value of x = 21
6. Find the mode of the given data: 3.1, 3.2, 3.3, 2.1, 1.3, 3.3, 3.1
Solution:
3.1 occuring two times
3.3 occuring two times
∴ 3.1 and 3.3 are the mode (bimodal)
7. For the data 11, 15, 17, x + 1, 19, x – 2, 3 if the mean is 14, find the value of
x. Also find the mode of the data.
Solution:
Arithmetic mean

11 + 15 + 17 + 𝑥 + 1 + 19 + 𝑥 − 2 + 3
14 =
7
2𝑥 + 64
14 =
7

∴ 2x + 64 = 14 × 7
2x = 98 – 64
2x = 34
x = 342
= 17
The given numbers are 11, 15, 17, 18, 19, 15 and 3
15 occuring two times
∴ Mode = 15
The value of x = 17 and mode = 15

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18.DATA INTERPRETATION
DATA INTERPRETATION

It is nothing but drawing conclusions and inference from a comprehensive

data presented numerically in tabular form or pictorial form by means of an
illustration graphs, pie charts etc. Thus the act of organizing and interpreting data
to get meaningful information is data interpretation.

Bar Charts:
Study the bar chart and answer the question based on it.
Production of Fertilizers by a Company (in 1000 tones) Over the Years

1. What was the percentage decline in the production of fertilizers

from 1997 to 1998?

A. 33(1/3)% B. 20%

C. 25% D. 21%

EXPLANATION

2.
The average production of 1996 and 1997 was exactly equal to the average
production of which of the following pairs of years?

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A. 2000 and 2001 B. 1999 and 2000

C. 1998 and 2000 D. 1995 and 2001

EXPLANATION

3. What was the percentage increase in production of fertilizers in 2002 compared to that in 1995?

A. 320% B. 300%

C. 220% D. 200%

EXPLANATION

4.
In which year was the percentage increase in production as compared to
the previous year the maximum?

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A. 2002 B. 2001

C. 1997 D. 1996

EXPLANATION

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Pie Charts:
Study the pie chart below and answer the questions that follow :

The above pie chart shows the sales of four different types of articles in a
shop.
a) What is the central angle of type A ?
b) If the total sale is 1200, what is the sale of B ?
c) What is the difference between the central angle of C and D ?
Solution :
a) Central angle of A = Percentage of A x 360 degrees = (35 / 100) x 360 = 126
degrees
b) Sales of B = 20 % of 1200 = 240
c) Difference between the central angle of C and D = 40 % of 360 – 5 % of 360 = 35
% of 360 = 126 degrees

Line Charts:
Directions: ( 1–5):Read the graph and answer questions .
Income and Expenditure of a company over the year (in lakhs of rupees).

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1. The ratio of the average income of all the years to the average profit is :
(a) 24 : 13 (b) 48 : 17 (c) 12 : 7 (d) 6 : 5

2. Percentage increase in profit in 1986 over 1982 is:

(a) 150 % (b) 120 % (c) 100% (d) 80%

3. The total income exceeds the total expenditure over the year 1982 to 1986 by:
(a) 85 lakhs (b) 105 lakhs (c) 115 lakhs (d) 120 lakhs

4. What is the difference in profit between 1983 and 1984 (in lakhs of rupees) :
(a) No profit (b) 5 (c) 10 (d) 15

5. The number of years in which the income is more than the average income of the
given year is:
(a) One (b) Two (c) Three (d) Four

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