Math Ncert Exemplar 2019 Solutions for Class 7 Maths Chapter 4 Simple Equations are provided here with simple step-by-step explanations. These solutions for Simple Equations are extremely popular among class 7 students for Maths Simple Equations Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Math Ncert Exemplar 2019 Book of class 7 Maths Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Math Ncert Exemplar 2019 Solutions. All Math Ncert Exemplar 2019 Solutions for class 7 Maths are prepared by experts and are 100% accurate.

#### Question 1:

In the given question, there are four options out of which, one is correct. Choose the correct one.
The solution of the equation ax + b = 0 is

(a) $\frac{\mathit{a}}{\mathit{b}}$

(b) –b

(c) $\mathit{-}\frac{b}{a}$

(d) $\frac{b}{a}$

ax + b = 0
ax = –b
x = $\frac{-b}{a}$

Hence, the correct answer is option (c).

#### Question 2:

In the given question, there are four options out of which, one is correct. Choose the correct one.
If a and b are positive integers, then the solution of the equation ax = b will always be a
(a) positive number
(b) negative number
(c) 1
(d) 0

a and b are positive integers.
ax = b
⇒ $x=\frac{b}{a}$
Since, a, b > 0
⇒ $\frac{b}{a}>0$

Hence, the correct answer is option (a).

#### Question 3:

In the given question, there are four options out of which, one is correct. Choose the correct one.
Which of the following is not allowed in a given equation?
(a) Adding the same number to both sides of the equation.
(b) Subtracting the same number from both sides of the equation.
(c) Multiplying both sides of the equation by the same non-zero number.
(d) Dividing both sides of the equation by the same number.

Dividing both sides of a equation by the same non-zero number is possible only.

Hence, the correct answer is option (d).

#### Question 4:

In the given question, there are four options out of which, one is correct. Choose the correct one.
The solution of which of the following equations is neither a fraction nor an integer?
(a) 2x + 6 = 0
(b) 3x – 5 = 0
(c) 5x – 8 = x + 4
(d) 4x + 7 = x + 2

(a) 2x + 6 = 0
⇒ 2x = –6
= –3

(b) 3x – 5 = 0
⇒ 3x = 5
x$\frac{5}{3}$

(c) 5– 8 = x + 4
⇒ 5x – 8 = 4 + 8
⇒ 4x = 12
x = 3

(d) 4x + 7 = x + 2
⇒ 4x – x = 2 – 7
⇒ 3x = –5
x = $-\frac{5}{3}$
Thus, only $\frac{-5}{3}$ is neither a fraction nor an integer.

Hence, the correct answer is option (d).

#### Question 5:

In the given question, there are four options out of which, one is correct. Choose the correct one.
The equation which cannot be solved in integers is
(a) 5y – 3 = – 18
(b) 3x – 9 = 0
(c) 3z + 8 = 3 + z
(d) 9y + 8 = 4y – 7

(a) 5y – 3 = –18
⇒ 5y = –18 + 3
y = $\frac{–15}{5}=–3$

(b) 3x – 9 = 0
⇒ 3x = 9
x = 3

(c) 3z + 8 = 3 + z
⇒ 3z z = 3 – 8
⇒ 2z = –5
z = $\frac{-5}{2}$

(d) 9y + 8 = 4y – 7
⇒ 9y – 4y = –78
⇒ 5y = –15
y = –3
Thus, only z = $\frac{-5}{2}$ is not an integer.

Hence, the correct answer is option (c).

#### Question 6:

In the given question, there are four options out of which, one is correct. Choose the correct one.
If 7x + 4 = 25, then x is equal to
(a) $\frac{29}{7}$
(b) $\frac{100}{7}$
(c) 2
(d) 3

7x + 4 = 25
⇒ 7x = 25 – 4
⇒ 7x = 21
x = 3

Hence, the correct answer is option (d).

#### Question 7:

In the given question, there are four options out of which, one is correct. Choose the correct one.
The solution of the equation 3x + 7 = – 20 is
(a) $\frac{17}{7}$
(b) –9
(c) 9
(d)

3x + 7 = –20
⇒ 3x = –20 – 7
⇒ 3x = –27
⇒  x = –9

Hence, the correct answer is option (b).

#### Question 8:

In the given question, there are four options out of which, one is correct. Choose the correct one.
The value of y for which the expressions (y – 15) and (2y + 1) become equal is
(a) 0
(b) 16
(c) 8
(d) – 16

y – 15 = 2y + 1
– 2y = 1 + 15
⇒ –y = 16
y = –16

Hence, the correct answer is option (d).

#### Question 9:

In the given question, there are four options out of which, one is correct. Choose the correct one.
If k + 7 = 16, then the value of 8k – 72 is
(a) 0
(b) 1
(c) 112
(d) 56

k + 7 = 16
k = 16 – 7
k = 9
∴ 8k – 72 = 8 × 9 – 72
= 72 – 72
= 0

â€‹Hence, the correct answer is option (a).

#### Question 10:

In the given question, there are four options out of which, one is correct. Choose the correct one.
If 43m = 0.086, then the value of m is
(a) 0.002
(b) 0.02
(c) 0.2
(d) 2

43m = 0.086
m$\frac{0.086}{43}$
m = 0.002

Hence, the correct answer is option (a).

#### Question 11:

In the given question, there are four options out of which, one is correct. Choose the correct one.
x exceeds 3 by 7, can be represented as
(a) x + 3 = 2
(b) x + 7 = 3
(c) x – 3 = 7
(d) x – 7 = 3

exceeds 3 can be represented as x = 3 + 7
x – 3 = 7

Hence, the correct answer is option (c).

#### Question 12:

In the given question, there are four options out of which, one is correct. Choose the correct one.
The equation having 5 as a solution is:
(a) 4x + 1 = 2
(b) 3 – x = 8
(c) x – 5 = 3
(d) 3 + x = 8

(a) 4x + 1 = 2
⇒ 4x = 2 – 1 = 1
⇒ $x=\frac{1}{4}$

(b) 3 – x = 8
⇒ –x = 8 – 3
⇒ –x = 5
x = –5

(c) x – 5 = 3
x = 5 + 3
x = 8

(d) 3 + x = 8
x = 8 – 3
x = 5
Thus, equation 3 + x = 8 has 5 as a solution.

Hence, the correct answer is option (d).

#### Question 13:

In the given question, there are four options out of which, one is correct. Choose the correct one.
The equation having – 3 as a solution is:
(a) x + 3 =1
(b) 8 + 2x = 3
(c) 10 + 3x = 1
(d) 2x + 1 = 3

(a) x + 3 = 1
x = 1 – 3
x = –2

(b) 8 + 2x = 3
⇒ 2x = –5
x = $\frac{-5}{2}$

(c) 10 + 3=1
⇒ 3x = –9
x = –3

(d) 2x + 1 = 3
⇒ 2x = 2
x = 1
Thus, equation 10 + 3x = 1 has –3 as a solution.

Hence, the correct answer is option (c).

#### Question 14:

In the given question, there are four options out of which, one is correct. Choose the correct one.
Which of the following equations can be formed starting with x = 0 ?
(a) 2x + 1 = – 1
(b)$\frac{x}{2}$ + 5 = 7
(c) 3x – 1 = – 1
(d) 3x – 1 = 1

(a) 2+ 1 = –1
⇒ 2x = –1 –1
x = –1

(b) $\frac{x}{2}+5=7$
$⇒\frac{x}{2}=2\phantom{\rule{0ex}{0ex}}⇒x=2×2=4$

(c) 3x – 1 = –1
⇒ 3x = 0
x = 0

(d) 3x – 1 = 1
⇒ 3x = 2
$x=\frac{2}{3}$
Thus, equation 3x – = –1 has 0 as a solution.

Hence, the correct answer is option (c).

#### Question 15:

In the given question, there are four options out of which, one is correct. Choose the correct one.
Which of the following equations cannot be formed using the equation x = 7 ?
(a) 2x + 1 =15
(b) 7x – 1 = 50
(c) x – 3 = 4
(d)$\frac{x}{7}$ – 1 = 0

(a) 2x + 1 = 15
⇒ 2x = 15 – 1 = 14
⇒ $x=\frac{14}{2}=7$

(b) 7x – 1 = 50
⇒ 7x = 50 + 1 = 51
x = $\frac{51}{7}$

(c) x – 3 = 4
x = 3 + 4 = 7

(d) $\frac{x}{7}-1=0$
$⇒\frac{x}{7}=1\phantom{\rule{0ex}{0ex}}⇒x=7×1=7$

Thus, equation 7x – 1 = 50 can not be formed with x = 7.

Hence, the correct answer is option (b).

#### Question 16:

In the given question, there are four options out of which, one is correct. Choose the correct one.
If  $\frac{x}{2}$= 3, then the value of 3x + 2 is
(a) 20
(b) 11
(c) $\frac{13}{2}$
(d) 8

$\frac{x}{2}=3$
x = 3 × 2
x = 6
∴ 3x + 2 = 3 × 6 + 2
= 18 + 2
= 20

Hence, the correct answer is option (a).

#### Question 17:

In the given question, there are four options out of which, one is correct. Choose the correct one.
Which of the following numbers satisfy the equation –6 + x = –12 ?
(a) 2
(b) 6
(c) – 6
(d) – 2

–6 + x = –12
x = –12 – (–6)
x = –12 + 6
x = –6

Hence, the correct answer is option (c).

#### Question 18:

In the given question, there are four options out of which, one is correct. Choose the correct one.
Shifting one term from one side of an equation to another side with a change of sign is known as
(a) commutativity
(b) transposition
(c) distributivity
(d) associativity

Shifting one term from one side of an equation to another side with a change of sign is known as transposition.

Hence, the correct answer is option (b).

#### Question 19:

Fill in the blanks to make the statements true.
The sum of two numbers is 60 and their difference is 30.
(a) If smaller number is x, the other number is ___________.(use sum)
(b) The difference of numbers in term of x is _________.
(c) The equation formed is ___________.
(d) The solution of the equation is ____________.
(e) The numbers are _________ and _________.

(a) 60 – x
(b) 60 – 2x
(c) 60 – 2x = 30
(d) 60 – 2x = 30
⇒ – 2x = 30 – 60
⇒ 2x = 30
⇒ x = 15
(e) 60 – x = 60 – 15 = 45
Numbers are 15 and 45.

#### Question 20:

Fill in the blanks to make the statements true.
Sum of two numbers is 81. One is twice the other___________.
(a) If smaller number is x, the other number is___________ .
(b) The equation formed is ___________.
(c) The solution of the equation is ____________.
(d) The numbers are ___________ and __________.

(a) 2x
(b) x + 2x = 81
⇒ 3x = 81
(c) We have,
3x = 81
x = 27
Thus, the solution of the equation is x = 27.
(d) The two numbers are x and 2x.
Here, x = 27
⇒ 2x = 2 × 27
= 54
â€‹Thus, the two numbers are 27 and 54.

#### Question 21:

Fill in the blanks to make the statements true.
In a test Abha gets twice the marks as that of Palak. Two times Abha's marks and three times Palak's marks make 280.
(a) If Palak gets x marks, Abha gets ___________ marks .
(b) The equation formed is ____________.
(c) The solution of the equation is___________ .
(d) Marks obtained by Abha are ____________.

(a) 2x
(b) 2(2x) + 3(x) = 280
⇒ 4x + 3x = 280
⇒ 7x = 280
(c) We have,
7x = 280
x = $\frac{280}{7}$ = 40
Thus, the solution of the equation is x = 40.
(d) Marks obtained by Abha are 80.

#### Question 22:

Fill in the blanks to make the statements true.
The length of a rectangle is two times its breadth. Its perimeter is 60 cm.
(a) If the breadth of rectangle is x cm, the length of the rectangle is_____________.
(b) Perimeter in terms of x is ____________.
(c) The equation formed is ____________.
(d) The solution of the equation is _____________.

(a) 2 × x = 2x cm
(b) Perimeter of rectangle = 2 × (length + breadth)
= 2 × (x + 2x)
= 2 × 3x
=
6x
Thus, perimeter of rectangle in terms of x is 6x cm.
(c) The equation formed is 6x = 60.
(d) We have,
6x = 60
⇒ x = 10
Thus, the solution of the equation is x = 10.

#### Question 23:

Fill in the blanks to make the statements true.
In a bag there are 5 and 2 rupee coins. If they are equal in number and their worth is â‚¹ 70, then
(a) The worth of x coins of â‚¹ 5 each ___________.
(b) The worth of x coins of â‚¹ 2 each ___________.
(c) The equation formed is_____________ .
(d) There are__________ 5 rupee coins and___________ 2 rupee coins.

(a) 5x
(b) 2x
(c) There are equal number of â‚¹5 and â‚¹2 coins in the bag. So,
5x + 2x = 70
⇒ 7x = 70
Thus, the equation formed is 7x = 70.
(d) We have,
7x = 70
x = 10
Thus, there are 10 â‚¹5 coins and 10 â‚¹2 coins.

Hence, the correct answer is option (d).

#### Question 24:

Fill in the blanks to make the statements true.
In a Mathematics quiz, 30 prizes consisting of 1st and 2nd prizes only are to be given. 1st and 2nd prizes are worth â‚¹ 2000 and â‚¹ 1000, respectively. If the total prize money is â‚¹ 52,000 then show that:
(a) If 1st prizes are x in number the number of 2nd prizes are ___________.
(b) The total value of prizes in terms of x are ____________.
(c) The equation formed is ____________.
(d) The solution of the equation is ______________.
(e) The number of 1st prizes are______________ and the number of 2nd prizes are _______________.

(a) 30 – x
(b) 2000x + 1000(30 – x)
(c) We have,
2000x + 1000(30 – x) = 52000
⇒ 2000x + 30000 – 1000x = 52000
⇒ 1000x = 22000
(d) We have,
1000x = 2000
x = 22
â€‹(e) Thus, the number of 1st prizes are 22 and the number of 2nd prizes are 8.

#### Question 25:

Fill in the blanks to make the statements true.
If z + 3 = 5, then z = _____________ .

We have,
z + 3 = 5
Subtracting 3 from both sides, (z + 3) – 3 = 5 – 3
z = 2

#### Question 26:

Fill in the blanks to make the statements true.
___________is the solution of the equation 3x – 2 =7.

We have,
3x – 2 = 7
⇒ 3x = 7 + 2
⇒ 3x = 9
x = 3
Thus, x = 3 is the solution of equation 3x – 2 = 7.

#### Question 27:

Fill in the blanks to make the statements true.
____________is the solution of 3x + 10 = 7.

We have,
3x + 10 = 7
⇒ 3x = 7 – 10
⇒ 3x = –3
x = –1
Thus, x = –1 is the solution of 3x + 10 = 7.

#### Question 28:

Fill in the blanks to make the statements true.
If 2x + 3 = 5, then value of 3x + 2 is ______________.

We have,
2x + 3 = 5
⇒ 2x = 5 – 3
⇒ 2x = 2
x = 1
Now,
3x + 2 = 3(1) + 2
= 3 + 2
= 5
Thus, the value of 3x + 2 is 5.

#### Question 29:

Fill in the blanks to make the statements true.
In integers, 4x – 1 = 8 has ___________solution.

We have,
4x – 1 = 8
⇒ 4x = 8 + 1
⇒ 4x = 9
x =
Thus. the equation 4x – 1 = 8 has no solution in integers.

#### Question 30:

Fill in the blanks to make the statements true.
In natural numbers, 4x + 5 = – 7 has ___________solution.

We have.
4x + 5 = –7
⇒ 4x = –7 – 5
⇒ 4x = –12
x = –3, which is not a natural number
Thus, the equation 4x + 5 = –7 has no solution in natural numbers.

#### Question 31:

Fill in the blanks to make the statements true.
In natural numbers, x – 5 = – 5 has ____________solution.

We have.
x – 5 = –5
= –5 + 5
⇒ = 0, which is not a natural number
Thus, the equation x – 5 = –5 has no solution in natural numbers.

#### Question 32:

Fill in the blanks to make the statements true.
In whole numbers, x + 8 = 12 – 4 has___________ solution.

We have.
x + 8 = 12 – 4
x + 8 = 8
= 8 – 8
⇒ = 0, which is a whole number
Thus, the equation x + 8 = 12 – 4 has one solution in natural numbers.

#### Question 33:

Fill in the blanks to make the statements true.
If 5 is added to three times a number, it becomes the same as 7 is subtracted from four times the same number. This fact can be represented as _______.

Let the number be x.
When 5 is added to three times a number, it becomes (3x + 5)
When 7 is subtracted from four times of the number, it becomes (4x – 7).
Now,
3x + 5 = 4x – 7
This fact can be represented as 3x + 5 = 4x – 7.

#### Question 34:

Fill in the blanks to make the statements true.
x + 7 = 10 has the solution _____________.

We have,
x + 7 = 10
x = 10 – 7
x = 3
Thus, x + 7 = 10 has the solution x = 3.

#### Question 35:

Fill in the blanks to make the statements true.
x – 0 =_____________ ; when 3x = 12.

We have,
3x = 12
x = 4
x – 0 = 4 – 0
x – 0 = 4
Thus, x – 0 = when 3x = 12.

#### Question 36:

Fill in the blanks to make the statements true.
x – 1=__________ ; when 2x = 2.

We have,
2x = 2
x = 1
x – 1 = 1 – 1
– 1 = 0
Thus, x – 1 =  when 2x = 2.

#### Question 37:

Fill in the blanks to make the statements true.
x – _________= 15; when $\frac{x}{2}$ = 6.

$\frac{x}{2}=6$
x = 2 × 6
x = 12
Since, 12 + 3 = 15
⇒ 12 – (–3) = 15
x –  (–3)  = 15

#### Question 38:

Fill in the blanks to make the statements true.
The solution of the equation x + 15 = 19 is ___________.

x + 15 = 19
x = 19 – 15
x = 4

#### Question 39:

Fill in the blanks to make the statements true.
Finding the value of a variable in a linear equation that__________ the equation is called a_________ of the equation.

Finding the value of a variable in a linear equation that satisfies the equation is called a root of the equation.

#### Question 40:

Fill in the blanks to make the statements true.
Any term of an equation may be transposed from one side of the equation to the other side of the equation by changing the ___________of the term.

Any term of an equation may be transposed from one side of the equation to the other side of the equation by charging the sign of term.

#### Question 41:

Fill in the blanks to make the statements true.

#### Question 42:

Fill in the blanks to make the statements true.
If 3 – x = – 4, then x =__________.

3 – x = –4
⇒ –x = –4 – 3
⇒ –x = –7
x = 7

#### Question 43:

Fill in the blanks to make the statements true.
If x $-\frac{1}{2}$ = $-\frac{1}{2}$, then x = ____________.

$x-\frac{1}{2}=\frac{-1}{2}\phantom{\rule{0ex}{0ex}}⇒x=\frac{-1}{2}+\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒x=0$

#### Question 44:

Fill in the blanks to make the statements true.
If $\frac{1}{6}-$ x = $\frac{1}{6}$, then x = ___________.

$\frac{1}{6}-x=\frac{1}{6}\phantom{\rule{0ex}{0ex}}⇒-x=\frac{1}{6}-\frac{1}{6}\phantom{\rule{0ex}{0ex}}⇒-x=0\phantom{\rule{0ex}{0ex}}⇒x=0$

#### Question 45:

Fill in the blanks to make the statements true.
If 10 less than a number is 65, then the number is _____________.

Let the number be x.
x – 10 = 65
x = 65 + 10
x = 75
Thus, the number is 75.

#### Question 46:

Fill in the blanks to make the statements true.
If a number is increased by 20, it becomes 45. Then the number is ________.

Let the number be p.
p + 20 = 45
p = 45 – 20
p = 25
Thus, the number is 25.

#### Question 47:

Fill in the blanks to make the statements true.
If 84 exceeds another number by 12, then the other number is _________.

Let the number is 25.
∴ 84 – x = 12
⇒ –x = 12 – 84
⇒ –x = –72
x = 72
Thus, the number is 72.

#### Question 48:

Fill in the blanks to make the statements true.
If $\frac{7}{8}=\frac{7}{8}$, then x = __________.

$x-\frac{7}{8}=\frac{7}{8}$
⇒ $x=\frac{7}{8}+\frac{7}{8}$
⇒ $x=\frac{7+7}{8}$
⇒ $x=\frac{14}{8}$
⇒ $\frac{7}{4}$

#### Question 49:

State whether the statements are True or False.
5 is the solution of the equation 3x + 2 = 17.

3x + 2 = 17
⇒ 3x = 17 – 2 = 15
⇒ $\frac{3x}{3}=\frac{15}{3}$
x = 5
Thus, 5 is the solution of 3x + 2 = 17.

#### Question 50:

State whether the statements are True or False.
$\frac{9}{5}$is the solution of the equation 4x – 1 = 8.

4x – 1 = 8
⇒ 4x = 8 + 1 = 9
⇒ $\frac{4x}{4}=\frac{9}{4}$
⇒ $\frac{9}{4}$
Thus, the statement is false.

#### Question 51:

State whether the statements are True or False.
4x – 5 = 7 does not have an integer as its solution.

4x – 5 = 7
⇒ 4= 7 + 5 = 12
⇒ $\frac{4x}{4}=\frac{12}{4}$
x = 3, is an integer
Thus, the statement is false.

#### Question 52:

State whether the statements are True or False.
One third of a number added to itself gives 10, can be represented as $\frac{x}{3}+10$ x.

Let the number be x.
∴ One-third of the number = $\frac{x}{3}$
As per the question,
$\frac{x}{3}+x=10$
Thus, the statement is false.

#### Question 53:

State whether the statements are True or False.
$\frac{3}{2}$is the solution of the equation 8x – 5 = 7.

8x – 5 = 7
⇒ 8x = 7 + 5 = 12
⇒ $\frac{8x}{8}=\frac{12}{8}$
⇒ $x=\frac{3}{2}$
Thus, the statement is true.

#### Question 54:

State whether the statements are True or False.
If 4x – 7 = 11, then x = 4.

4x – 7 = 11
⇒ 4x = 11 + 7 = 18
⇒ $\frac{4x}{4}=\frac{18}{4}$
⇒ $x=\frac{9}{2}$
Thus, the statement is false.

#### Question 55:

State whether the statements are True or False.
If 9 is the solution of variable x in the equation $\frac{5x-7}{2}=y$ , then the value of y is 28

$\frac{5x-7}{2}=y\phantom{\rule{0ex}{0ex}}⇒\frac{5×9–7}{2}=y\phantom{\rule{0ex}{0ex}}⇒y=\frac{45–7}{2}\phantom{\rule{0ex}{0ex}}⇒y=\frac{38}{2}\phantom{\rule{0ex}{0ex}}⇒y=19$
Thus, the statement is false.

#### Question 56:

Match each of the entries in Column I with the appropriate entries in Column II.
Column I                                                            Column II
(i) x + 5 = 9
(ii) x – 7 = 4
(C) 4
(iv) 5x = 30                                                        (D) 6
(v) The value of y which satisfies 3y = 5          (E) 11
(vi) If p = 2, then the value of $\frac{1}{3}$(1 – 3p)         (F) – 60
(G) 3

(i) x + 5 = 9
x = 9 – 5
x = 4     (C)
(ii) x – 7 = 4
⇒ x = 7 + 4
x = 11   (E)
(iii) $\frac{x}{12}=–5$
x = I2 × (–5)
x = –60    (F)
(iv) 5x = 30
⇒ $\frac{5x}{5}=\frac{30}{5}$
x = 6    (D)
(v) 3y = 5
⇒ $\frac{3y}{3}=\frac{5}{3}$
⇒ $y=\frac{5}{3}$    (B)
(vi) $\frac{1}{3}\left(1-3p\right)=\frac{1}{3}\left(1-3×2\right)$

#### Question 57:

Express the given statement as an equation.
13 subtracted from twice of a number gives 3 .

Let the number be x.
∴ 2x – 13 = 3
Thus, the given statement can be written in equation as 2x – 13 = 3.

#### Question 58:

Express the given statement as an equation.
One-fifth of a number is 5 less than that number.

Let the number be x.
So, one-fifth of the number = $\frac{x}{5}$x5
Therefore, $\frac{x}{5}=x-5$ is the required equation.

#### Question 59:

Express the given statement as an equation.
A number is 7 more than one-third of itself.

Let the number be x.
So, one-third of number = $\frac{x}{3}$
7 more than $\frac{x}{3}=\frac{x}{3}+7$
Therefore, $x=\frac{x}{3}+7$ is the required equation

#### Question 60:

Express the given statement as an equation.
Six times a number is 10 more than the number.

Let the number be x.
So, six times of the number = 6x
Therefore, 6xx + 10 is the required equation.

#### Question 61:

Express the given statement as an equation.
If 10 is subtracted from half of a number, the result is 4.

Let the number be x.
So, half of the number $=\frac{x}{2}$ x2
On subtracting 10 from it, we get $\frac{x}{2}-10$x210
Therefore , $\frac{x}{2}-10=4$ is the required equation.

#### Question 62:

Express the given statement as an equation.
Subtracting 5 from p, the result is 2.

The number is p.
On subtracting 5 from it, we get p – 5.
p – 5 = 2 is the required equation.

#### Question 63:

Express the given statement as an equation.
Five times a number increased by 7 is 27.

Let the number be x.
So, five times of the number = 5x
When increased by 7, it gives the expression 5x +7
∴ 5x + 7 = 27 is the required equation.

#### Question 64:

Express the given statement as an equation.
Mohan is 3 years older than Sohan. The sum of their ages is 43 years.

Let Sohan is x years old.
So, Mohan is x + 3 years old.
∴ Sum of their ages be x + (x + 3).
x + (x + 3) = 43 is the required equation.

#### Question 65:

Express the given statement as an equation.
If 1 is subtracted from a number and the difference is multiplied by $\frac{1}{2}$, the result is 7.

Let the number be x.
On subtracting 1 from it, we get x – 1.
Multiplying it by $\frac{1}{2}$ we get $\frac{1}{2}\left(x-1\right)$
$\frac{1}{2}\left(x-1\right)=7$ is the required equation.

#### Question 66:

Express the given statement as an equation.
A number divided by 2 and then increased by 5 is 9.

Let the number be x.
Dividing the number by 2, we get $\frac{x}{2}$.
When, increased by 5, it gives the expression $\frac{x}{2}+5$
$\frac{x}{2}+5=9$ is the required equation.

#### Question 67:

Express the given statement as an equation.
The sum of twice a number and 4 is 18.

Let the number be x.
So, twice of the number = 2x
On adding 4 to it, we get 2x + 4
∴ 2x + 4 = 18 is the required equation.

#### Question 68:

The age of Sohan Lal is four times that of his son Amit. If the difference of their ages is 27 years, find the age of Amit.

Let the age of Amit be x years.
So, age of Sohan Lal = 4x years
According to question, 4xx = 27
$⇒3x=27\phantom{\rule{0ex}{0ex}}⇒x=9$
Thus, Amit is 9 years old.

#### Question 69:

A number exceeds the other number by 12. If their sum is 72, find the numbers.

Let the number be x.
∴ Other number = x +12
According to question,
x + x + 12 = 72
⇒ 2x + 12 = 72
⇒ 2x = 72 – 12 = 60
x = 30
x + 12 = 30 + 12 = 42
Hence, the numbers are 30 and 42

#### Question 70:

Seven times a number is 12 less than thirteen times the same number. Find the number.

Let the number be x.
So, seven times of the number = 7x
Thirteen times of the number = 13x
According to question,
7x = 13x – 12
⇒ 12 = 13x – 7x
⇒ 6x = 12
x = 2x=126=2
Thus, 2 is the required number.

#### Question 71:

The interest received by Karim is â‚¹ 30 more than that of Ramesh. If the total interest received by them is â‚¹ 70, find the interest received by Ramesh.

Let the interest received by Ramesh be x.
So, interest received by Karim = â‚¹ (30+ x)
According to question,
x + x + 30 = 70
⇒ 2x = 70 – 30 = 40
x = 20
Thus, â‚¹ 20 is the interest received by Ramesh.

#### Question 72:

Subramaniam and Naidu donate some money in a Relief Fund. The amount paid by Naidu is â‚¹ 125 more than that of Subramaniam. If
the total money paid by them is â‚¹ 975, find the amount of money donated by Subramaniam.

Let the amount of money donated by Subramaniam be x.
So, the amount paid by Naidu is (x + 125).
According to question,
x + x + 125 = 975
⇒ 2x = 975 – 125 = 850
x = 425
Thus, â‚¹425 is donated by Subramaniam.

#### Question 73:

In a school, the number of girls is 50 more than the number of boys. The total number of students is 1070. Find the number of girls.

Let the number of girls be x.
So, the number of boys = x – 50
According to question,
x + x – 50 = 1070
⇒ 2x = 1070 + 50 = 1120
x = 560
Thus, 560 are girls.

#### Question 74:

Two times a number increased by 5 equals 9. Find the number.

Let the number be x.
So, two times of the number = 2x
When, increased by 5, it gives the expression 2x + 5
∴ 2x + 5 = 9
⇒ 2x = 9 – 5 = 4
x = 2
Thus, x = 2 is the required number.

#### Question 75:

9 added to twice a number gives 13. Find the number.

Let the number be x.
So, twice of the number = 2x
On adding 9 to it, we get 2x + 9
∴ 2x + 9 = 13
⇒ 2x = 13 – 9 = 4
x = 2
Thus, x = 2 is the required number.

#### Question 76:

1 subtracted from one-third of a number gives 1. Find the number.

Let the number be x.
So, one third of the number $\frac{x}{3}$.
On subtracting 1 from it, we get $\frac{x}{3}-1$x31
$\frac{x}{3}-1=1\phantom{\rule{0ex}{0ex}}⇒\frac{x}{3}=1+1=2\phantom{\rule{0ex}{0ex}}⇒x=6$
Thus, x = 6 is the required number.

#### Question 77:

After 25 years, Rama will be 5 times as old as he is now. Find his present age.

Let present age of Rama be x years.
So, five times of his age = 5x.
According to question,
5x = x + 25
⇒ 5xx = 25
⇒ 4x = 25
x = $\frac{25}{4}$

Thus, at present Rama is $\frac{25}{4}$ years old.

#### Question 78:

After 20 years, Manoj will be 5 times as old as he is now. Find his present age.

Let present age of Manoj be x years.
So, five times of his age = 5x
According to question,
5 = x + 20
⇒ 5xx = 20
⇒ 4x = 20
x = 5
Thus, at present Manoj is 5 years old.

#### Question 79:

My younger sister's age today is 3 times, what it will be 3 years from now minus 3 times what her age was 3 years ago. Find her present age.

Let present age of my younger sister be x years.
After 3 years, her age will be = (x + 3) years
Before 3 years, her age was = (x – 3) years
According to question,
x = 3 (x + 3) – 3 (x – 3)
x = 3 (x + 3 – x + 3)
x = 3(6) = 18
Thus, at present my younger sister is 18 years old.

#### Question 80:

If 45 is added to half a number, the result is triple the number. Find the number.

Let the number be x.
So, half of the number $\frac{x}{2}$.
On adding 45 to it, we get $\frac{x}{2}+45$.

$\frac{x}{2}+45=3x\phantom{\rule{0ex}{0ex}}3x-\frac{x}{2}=45\phantom{\rule{0ex}{0ex}}\frac{6x-x}{2}=45\phantom{\rule{0ex}{0ex}}\frac{5x}{2}=45\phantom{\rule{0ex}{0ex}}x=18$

Thus, the number is 18.

#### Question 81:

In a family, the consumption of wheat is 4 times that of rice. The total consumption of the two cereals is 80 kg. Find the quantities of rice and wheat consumed in the family.

Let the quantity of rice consumed in the family be x kg.
So, quantity of wheat consumed = 4x kg
According to question, x + 4x = 80
⇒ 5x = 80
x = 16

Thus, 16 kg rice consumed in the family and 4 × 16 = 64 kg wheat consumed in the family.

#### Question 82:

In a bag, the number of one rupee coins is three times the number of two rupees coins. If the worth of the coins is â‚¹ 120, find the number of 1 rupee coins.

Let two rupees coins are x in numbers.
So, one rupee coins are 3x in numbers.
According to question, 1(3x) + 2(x) = 120
⇒ 3x + 2x = 120
⇒ 5x = 120
x = 24
Thus, 3 × 24 = 72 coins are of one rupee coins.

#### Question 83:

Anamika thought of a number. She multiplied it by 2, added 5 to the product and obtained 17 as the result. What is the number she had thought of ?

Let the number Anamika thought be x.
Multiplying it by 2, we get 2x
On adding 5 to it, we get 2x + 5
∴ 2x + 5 = 17
⇒ 2x = 17 – 5 = 12
x = 6
Thus, she had thought of number 6.

#### Question 84:

One of the two numbers is twice the other. The sum of the numbers is 12. Find the numbers.

Let the one number be x.
So, other number = 2x
According to question,
x + 2x = 12
⇒ 3x = 12
x = 4

∴ Other number is 2x = 2 × 4 = 8.

#### Question 85:

The sum of three consecutive integers is 5 more than the smallest of the integers. Find the integers.

Let the smallest integer be x
So, next two consecutive integer would be x + 1 and x + 2 respectively.
According to question,
x + x + 1 + x + 2 = x + 5
⇒ 3x + 3 = x + 5
⇒ 3xx = 5 – 3
⇒ 2x = 2
x = 1
Thus, next two consecutive integers are 2 and 3 respectively.

#### Question 86:

A number when divided by 6 gives the quotient 6. What is the number?

Let the number be x.
Dividing it by 6, we get $\frac{x}{6}$.
$\frac{x}{6}=6\phantom{\rule{0ex}{0ex}}x=6×6\phantom{\rule{0ex}{0ex}}x=36$
Thus, x = 36 is the required number.

#### Question 87:

The perimeter of a rectangle is 40m. The length of the rectangle is 4 m less than 5 times its breadth. Find the length of the rectangle.

Let the breadth of rectangle be x m.
So, five times of breadth = 5x
Therefore, length of rectangle = (5x – 4) m
Perimeter of rectangle = 40 m
⇒ 2(x + 5x – 4) = 40
⇒ 2(6x – 4) = 40
⇒ 6x – 4 = 20
⇒ 6x = 20 + 4 = 24
x = 4
Thus, length of rectangle = (5 × 4 – 4) = 16 m

#### Question 88:

Each of the 2 equal sides of an isosceles triangle is twice as large as the third side. If the perimeter of the triangle is 30 cm, find the length of each side of the triangle.

Let length of equal sides of an isosceles triangle be 2x cm.
So, the third side of triangle x cm.
Perimeter of triangle = 30 cm
x + 2x + 2x = 30
⇒ 5x = 30
x = 6
Thus, 6 cm, 12 cm and 12 cm are the required sides of the triangle.

#### Question 89:

The sum of two consecutive multiples of 2 is 18. Find the numbers.

Let first multiple of 2 be x.
So, next multiple of 2 = x + 2
According to question,
x + x + 2 = 18
⇒ 2x = 18 – 2 = 16
x = 8
Thus, 8 and 10 are required numbers.

#### Question 90:

Two complementary angles differ by 20°. Find the angles.

Let one angle be x.
So, complement of x = 90°– x
According to question,
x – (90°– x) = 20°
x – 90° + x = 20°
⇒ 2x = 20° + 90° = 110°
x = 55°
∴ Complement of x = 90°– x = 90°– 55° = 35°
Thus, 55° and 35° are required complementary angles.

#### Question 91:

150 has been divided into two parts such that twice the first part is equal to the second part. Find the parts.

Let first part be x.
So, other part = 150 – x
According to question, 2x = 150 – x
⇒ 2x + x = 150
⇒ 3x = 150
x = 50

Thus, other part = 150 – x = 150 – 50 = 100
Hence, 150 has been divided into 50 and 100.

#### Question 92:

In a class of 60 students, the number of girls is one third the number of boys. Find the number of girls and boys in the class.

Let the number of boys in class = x
So, the number of girls in class = 60 – x
According to question,
$60-x=\frac{x}{3}$
⇒ 180 – 3x = x
⇒ 180 = 3x + x
⇒ 4x = 180
x = 45
Thus, number of boys in the class = 45
And number of girls in the class = 60 – 45 = 15

#### Question 93:

Two-third of a number is greater than one-third of the number by 3. Find the number.

Let the number be x.
So, two-third of numbers $\frac{2}{3}x$.
According to question,
$\frac{2}{3}x=\frac{1}{3}x+3\phantom{\rule{0ex}{0ex}}\frac{2x}{3}-\frac{x}{3}=3\phantom{\rule{0ex}{0ex}}\frac{2x-x}{3}=3\phantom{\rule{0ex}{0ex}}\frac{x}{3}=3\phantom{\rule{0ex}{0ex}}x=9$
Thus, x = 9 is the required number.

#### Question 94:

A number is as much greater than 27 as it is less than 73. Find the number.

Let the number be x.
According to question,
x – 27 = 73 – x
x + x = 73 + 27
⇒ 2x = 100
x = 50

Thus, x = 50 is the required number.

#### Question 95:

A man travelled two fifth of his journey by train, one-third by bus, one-fourth by car and the remaining 3 km on foot. What is the length of his total journey?

Let the length of total journey be x km.
∴ Journey by train = x km
∴ Joumey by bus =
∴ Journey by car = $\frac{x}{3}$ km
And journey on foot = $\frac{x}{4}$ km
$x=\frac{2x}{5}+\frac{x}{3}+\frac{x}{4}+3\phantom{\rule{0ex}{0ex}}x=\frac{24x+20x+15x}{60}+3\phantom{\rule{0ex}{0ex}}x=\frac{59x}{60}+3\phantom{\rule{0ex}{0ex}}x-\frac{59x}{60}=3\phantom{\rule{0ex}{0ex}}\frac{x}{60}=3\phantom{\rule{0ex}{0ex}}x=3×60\phantom{\rule{0ex}{0ex}}x=180$

Thus, x = 180 km is the length of total journey.

#### Question 96:

Twice a number added to half of itself equals 24. Find the number.

Let the number be x.
According to question,
$2x+\frac{x}{2}=24\phantom{\rule{0ex}{0ex}}\frac{4x+x}{2}=24\phantom{\rule{0ex}{0ex}}5x=2×24=48\phantom{\rule{0ex}{0ex}}x=\frac{48}{5}\phantom{\rule{0ex}{0ex}}x=9.6$

Thus, x = 9.6 is the required number.

#### Question 97:

Thrice a number decreased by 5 exceeds twice the number by 1. Find the number.

Let the number be x.
So, thrice of the number = 3x
When it decreased by 5, we get 3x – 5
According to question,
3x – 5 = 2x + 1
⇒ 3x – 2x = 1 + 5
x = 6
Thus, x = 6 is the required number.

#### Question 98:

A girl is 28 years younger than her father. The sum of their ages is 50 years. Find the ages of the girl and her father.

Let the age of father be x years.
So, age of his girl = (x – 28) years
According to question,
x + x – 28 = 50
⇒ 2x = 50 + 28 = 78
x = 39
Thus, age of father = 39 years
And age of his girl = 11 years

#### Question 99:

The length of a rectangle is two times its width. The perimeter of the rectangle is 180 cm. Find the dimensions of the rectangle.

Let the width of rectangle be x cm.
So, the length of rectangle = 2x
Perimeter of rectangle = 180 cm
⇒ 2(2x + x) = 180
⇒ 6x = 180
x = 30
Thus, width of rectangle = 30 cm
And length of rectangle = 2 × 30 = 60 cm

#### Question 100:

Look at this riddle?
If she answers the riddle correctly how ever will she pay for the pencils?

Let the cost of one pencil be â‚¹ x
So, the cost of 5 pencils = â‚¹ 5x
Cost of 7 pencils = â‚¹ 7x
According to question,
7x = 5x + 6
⇒ 7x – 5x = 6
⇒ 2x = 6
x = 3
Cost of 10 pencils = â‚¹ (3 × 10) = â‚¹ 30

#### Question 101:

In a certain examination, a total of 3768 students secured first division in the years 2006 and 2007. The number of first division in 2007
exceeded those in 2006 by 34. How many students got first division in 2006?

Let number of students got first division in 2006 be x.
So, the number of students got first division in 2007 = 3768 – x.
According to question,
3768 – x = x + 34
⇒ 3768 – 34 = x + x
⇒ 2x = 3734
x = 1867
Thus, 1867 students got first division in 2006.

#### Question 102:

Radha got â‚¹ 17,480 as her monthly salary and over-time. Her salary exceeds the over-time by â‚¹ 10,000. What is her monthly salary ?

Let Radha’s monthly salary = â‚¹ x
So, money got by her in over time = â‚¹ (17480 – x)
According to question,
x = 17480 – x + 10000
x + x = 17480 + 10000
⇒ 2x = 27480
x = 13740
Thus, â‚¹13740 is her monthly salary.

#### Question 103:

If one side of a square is represented by 18x – 20 and the adjacent side is represented by 42 – 13x, find the length of the side of the square.

Since, square has all sides equal
∴ 18x – 20 = 42 – 13x
⇒ 18x + 13x = 42 + 20
⇒ 31x = 62
x = 2
∴ Side of square = 18 × 2 – 20 = 36 – 20 = 16
Thus, length of the side of the square is 16 units.

#### Question 104:

Follow the directions and correct the given incorrect equation, written in Roman numerals:
(a) Remove two of these matchsticks to make a valid equation:

(b) Move one matchstick to make the equation valid. Find two different solutions.
graphics needs to change

(a) Incorrect equation is

After removing two matchsticks the correct equation is
GRAPHIC

(b) Incorrect equation is
GRAPHIC

After removing one matchstick the correct equation is
GRAPHIC

#### Question 105:

What does a duck do when it flies upside down? The answer to this riddle is hidden in the equation given below:
If i + 69 = 70, then i = ? If 8u = 6u + 8, then u =?
If 4a = –5a + 45, then a = ? if 4q + 5 = 17, then q =?
If –5t – 60 = – 70, then t = $\frac{1}{4}s$ + 98 = 100, then s =?
If $\frac{5}{3}$p + 9 = 24, then p =_____?
If 3c = c +12, then c =_____?
If 3 (k + 1) = 24, then k =_____?
For riddle answer : substitute the number for the letter it equals

i + 69 = 70
⇒ i = 70 – 69 = 1

8u = 6u +8
⇒ 8u – 6u = 8
⇒ 2u = 8
⇒ u = 4

4a = –5a + 45
⇒ 4a + 5a = 45
⇒ 9a – 45
⇒ a = 5

4q + 5 = 17
⇒ 4q = 17 – 5
⇒ 4q = 12
⇒ q = 3

–5t – 60 = – 70
⇒ –5t = –70 + 60
⇒ –5t = –10
⇒ t = 2

$\frac{1}{4}s+98=100\phantom{\rule{0ex}{0ex}}\frac{1}{4}s=2\phantom{\rule{0ex}{0ex}}s=8\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{5}{3}p+9=24\phantom{\rule{0ex}{0ex}}\frac{5}{3}p=15\phantom{\rule{0ex}{0ex}}p=9\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}3c=c+12\phantom{\rule{0ex}{0ex}}3c-c=12\phantom{\rule{0ex}{0ex}}2c=12\phantom{\rule{0ex}{0ex}}c=6\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}3\left(k+1\right)=24\phantom{\rule{0ex}{0ex}}k+1=8\phantom{\rule{0ex}{0ex}}k=7$

Thus, 1 = i, 2 = t, 3 = q, 4 = u, 5 = a, 6 = c, 7 = k, 8 = s, 9 = p.

#### Question 106:

The three scales below are perfectly balanced if • = 3. What are the values of âˆ† and * ?

Given that • = 3.

Solving equations, we get
$5*=2*+2•+2•\phantom{\rule{0ex}{0ex}}5*-2*=4•\phantom{\rule{0ex}{0ex}}3*=4×3\phantom{\rule{0ex}{0ex}}*=4$
$\begin{array}{rcl}3∆& =& 3×4+3×3\\ & =& 12+9\\ & =& 21\\ ∆& =& 7\end{array}$

#### Question 107:

The given figure represents a weighing balance. The weights of some objects in the balance are given. Find the weight of each square and
the circle.