These Class 7 Maths Extra Questions Chapter 4 Simple Equations will help students prepare well for the exams.

## Class 7 Maths Chapter 4 Extra Questions Simple Equations

### Class 7 Maths Simple Equations Extra Questions – 2 Marks

Question 1.

p + 3 = 10, p = –1

Solution:

p + 3 = 10 ∵ p = –1

– 1 + 3 = 2 ≠ 10

∴ p = –1 is Not the solution.

Question 2.

7p – 3 = 1, p = 5

Solution:

7p – 3 = 1 ∵ p = 5

7(5) – 3 = 35 – 3 = 32 ≠ 1

∴ p = 51 is Not the solution.

Question 3.

1 – 3p = 7, p = 0

Solution:

1 – 3p = 7 ∵ p = 0

1 – 3(0) = 1 – 0 = 1 ≠ 7

∴ p = 0 is not the solution.

Question 4.

7 – x = –1, x = 8

Solution:

7 – x = –1 ∵ x = 8

7 – 8 = –1

∴ x = 8 is the solution.

Question 5.

3p = –12, p = –4

Solution:

3p = –12,

∵ p = –4

3(–4) = –12

–12 = –12

∴ p = –4 is the solution.

Question 6.

\(\frac{3 n}{2}\) + 1 = 0, n = \(\frac{-2}{3}\)

Solution:

\(\frac{3 n}{2}\) + 1 = 0

∵ n = \(\frac{-2}{3}\) = \(\frac{3 \times-2}{2 \times 3}\) + 1 = – 1 + 1 = 0

∴ n = \(\frac{-2}{3}\) is the solution.

Question 7.

4l – 3 = 8l, l = 1.

Solution:

4l – 3 = 8l ∵ l = 1

4(1) – 3 = 4 – 3 = 1

8l = 8 × 1 = 8

l ≠ 8

∴ l = 1 is not the solution.

Question 8.

18x = 0, x = 0

Solution:

18x = 0 ∵ x = 0

18x = 18 × 0 = 0

∴ x = 0 is the solution.

Express the following statements in the form of simple equations.

Question 9.

i) 3 is added to twice of x, result is – 1.

ii) 8 times p is –54.

Solution:

i) 2x + 3 = –1

ii) 8p = –54

Question 10.

i) 1 is subtracted from b gives 10

ii) Three fourth of a number z is 9.

Solution:

i) b – 1 = 10

ii) \(\frac { 3 }{ 4 }\) z = 9

Question 11.

i) \(\frac { 7 }{ 2 }\) is added to thrice of p is equal to 7.

ii) 8 times of 8x is equal to 64.

Solution:

i) 3p + \(\frac { 7 }{ 2 }\) = 7

ii) 8 × 8x = 64 ⇒ 64x = 64

Question 12.

i) Twice of a is added to – 7, the result is 10.5.

ii) 8 times of ‘x plus 2 ‘ is 91 .

Solution:

i) 2a – 7 = 10.5

ii) 8(x+2) = 91

Question 13.

i) Ten times of n is equal to 1000.

ii) p is divided by 6 gives the quotient 8.

Solution:

i) 10n = 1000

ii) \(\frac { p }{ 6 }\) = 8

Question 14.

x – 3 = 5

Solution:

x – 3 = 5

Adding ‘ 3 ‘ on both sides

x – 3 + 3 = 5 + 3

x = 8

Question 15.

1 + x = –9

Solution:

1 + x = –9

Adding – 1 on both sides

1 + x – 1 = – 9 – 1

x = –10

Question 16.

l – 3 = – 1

Solution:

l – 3 = – 1

Adding 3 on both sides

l – 3 + 3 = – 1 + 3 ⇒ l = 2

Question 17.

y – 7 = – 4

Solution:

y – 7 = – 4

Adding 7 on both sides

y – 7 + 7 = – 4 + 7 ⇒ y = 3

Question 18.

3p = 1

Solution:

3p = 1

Divide both sides by 3

\(\frac{3 \mathrm{p}}{3}\) = \(\frac { 1 }{ 3 }\) ⇒ p = \(\frac { 1 }{ 3 }\)

Question 19.

4x = 20

Solution:

4x = 20

Divide both sides by 4

\(\frac { 4x }{ 4 }\) = \(\frac { 20 }{ 4 }\) – x – 5

Question 20.

\(\frac { b }{ 3 }\) = 4

Solution:

\(\frac { b }{ 3 }\) = 4

Multiply both sides by 3

\(\frac { b }{ 3 }\) × 3 = 4 × 3 ⇒ b = 12

Question 21.

Solve 2p – 5 = 5

Solution:

2p – 5 = 5

Adding 5 on both sides

2p – 5 + 5 = 5 + 5

2p = 10

p = \(\frac { 10 }{ 2 }\) (divide both sides by 2)

p = 5

Question 22.

Solve 1 – 2x = 3

Solution:

1 – 2x = 3

Subtracting 1 from both sides

1 – 2x – 1 = 3 – 1

– 2x = 2

Divide both sides by ‘–2’

\(\frac{-2 x}{-2}\) = \(\frac{2}{-2}\)

⇒ x = –1

Question 23.

2x + 5 = 1, x = – 2,1

Solution:

2x + 5 = 1, x = –2

Case (i) :

LHS = 2x + 5 = 2(– 2) + 5 = – 4 + 5 = 1

RHS = 1

LHS = RHS

∴ x = –2 is the solution.

Case (ii): x = 1

LHS = 2x + 5 = 2(1) + 5 = 2 + 5 = 7

RHS = 1

LHS ≠ RHS

x = 2 is not the solution.

Question 24.

9p = 18, p = 2,–2

Solution:

9p = 18

Let p = 2

9p = 9 × 2 = 18

Let p = –2

9p = 9 × – 2 = –18

∴ p = 2 is the solution.

Question 25.

Solve “Add 8 to eight times of a number you get 24” What is that number?

Solution:

Let the number be x

8 times of the number = 8x

According to the problem

8 + 8x = 24

8x = 24 – 8 (Transposing 8 to RHS)

8x = 16

x = \(\frac { 16 }{ 8 }\) (Divide both sides by 8 )

x = 2

∴ The required number = 2

### Simple Equations Extra Questions Class 7 – 3 Marks

Question 1.

Anwar’s father’s age is 5 years more than three times Anwar’s age. Find Anwar’s age, if his father is 47 years.

Solution:

Let the age of Anwar = x years

Three times of x = 3x

According to the problem

3x + 5 = 47

3x = 47 – 5 (Transposing 5 to RHS)

3x = 42

x = \(\frac { 42 }{ 3 }\) (Divide both sides by 3)

x = 14 years

∴ Age of Anwar = 14 years.

Question 2.

Find a number such that one fifth of the number is 4 more than 10.

Solution:

Let the required number = p

One fifth of the number = \(\frac { p }{ 5 }\)

According to the problem

\(\frac { p }{ 5 }\) – 10 = 4

\(\frac { p }{ 5 }\) = 10 + 4 (Transposing 10 to RHS)

\(\frac { p }{ 5 }\) = 14

p = 14 × 5 (Multiply both sides by 5)

p = 70

∴ Required number = 70

Question 3.

Solve 4(x – 1) = 16

Solution:

4(x – 1) = 16

x – 1 = \(\frac { 16 }{ 4 }\) (divide by 4 on both sides)

x – 1 = 4

x = 4 + 1 (adding 1 on both sides)

x = 5

Question 4.

Solve 0 = – 8 + (p – 3) 2

Solution:

0 = –8 + (p – 3)2

8 = (p – 3) 2 (adding 8 on both sides)

\(\frac { 8 }{ 2 }\) = p – 3 (divide by 2 on both sides)

4 = p – 3

p = 4 + 3 (adding 3 on both sides)

p = 7

Question 5.

Solve \(\frac { a }{ 2 }\) – 3 = 2

Solution:

\(\frac { a }{ 2 }\) – 3 = 2

\(\frac { a }{ 2 }\) = 2 + 3 (adding 3 on both sides)

\(\frac { a }{ 2 }\) = 5

a = 5 × 2 (Multiply by 2 on both sides)

a = 10

Question 6.

Solve 3(5 – x) = –30

Solution:

3(5–x) = – 30

5 – x = \(\frac{-30}{5}\) (Divide by 3 on both sides)

5 – x = – 6

–x = – 6 – 5 (Transposing 5 to RHS)

–x = – 11

x = 11 (Divide both sides by –1)

Question 7.

Solve \(\frac{8x}{3}\) = \(\frac{3}{8}\)

Solution:

\(\frac{8x}{3}\) = \(\frac{3}{8}\)

Multiply both sides by 3

8x = \(\frac{3}{8}\) × 3

8x = \(\frac{9}{8}\)

x = \(\frac{9}{8 \times 8}\) (Divide both sides by 8)

x = \(\frac{9}{64}\)

Question 8.

Solve \(\frac{2 x}{5}\) =– 4

Solution:

\(\frac{2 x}{5}\) =– 4

2x = –4 × 5(Multiply both sides by 5)

2x = –20

x = \(\frac{-20}{2}\) (Divide both sides by 2)

x = –10

Question 9.

Solve 3k – 12 = – 6

Solution:

3k – 12 = – 6

3k = – 6 + 12 (adding 12 on both sides)

3k = 6

k = \(\frac{6}{3}\) (divide both sides by 3)

k = 2

Question 10.

Solve \(\frac{2}{3}\)(x + 1) = 1

Solution:

\(\frac{2}{3}\)(x + 1) = 1

x – 1 = 1 × \(\frac{3}{2}\)

x – 1 = \(\frac{3}{2}\)

x = \(\frac{3}{2}\) + 1 ⇒ x = \(\frac{3+2}{2}\) ⇒ x = \(\frac{5}{2}\)

### Extra Questions of Simple Equations Class 7 – 5 Marks

Question 1.

Solve \(\frac{2 x+1}{2}\) + 1 = \(\frac{x-2}{3}\) – 1

Solution:

Question 2.

Solve 2(2x + 5) – 3(2x – 1) = 7

Solution:

2(2x + 5) – 3(2x – 1) = 7

4x + 10– 6x + 3 = 7

–2x + 13 = 7

–2x = 7 – 13

–2x = – 6

\(\frac{-2 x}{-2}\) = \(\frac{-6}{-2}\) ⇒ x = 3

Question 3.

Solve \(\frac{p+3}{10}\) – 2 = p

Solution:

\(\frac{p+3}{10}\) – 2 = p

\(\frac{p+3-20}{10}\) = p

p – 17 = 10p

p – 10p = 17

–9p = 17 ⇒ p = \(\frac{17}{-9}\)

p = \(\frac{-17}{9}\)

Question 4.

Solve \(\frac{x}{3}\) + 4 = \(\frac{x}{2}\) – 1

Solution:

\(\frac{x}{3}\) + 4 = \(\frac{x}{2}\) – 1

\(\frac{x}{3}\) – \(\frac{x}{2}\) = – 1 – 4

\(\frac{2 x-3 x}{6}\) = –5

\(\frac{-x}{6}\) = –5 ⇒ \(\frac{x}{6}\) = 5

⇒ x = 5 × 6 ⇒ x = 30

Question 5.

3(x – 1) = 12

Solution:

3(x – 1) = 12

x – 1 = \(\frac{12}{3}\)

x – 1 = 4

x = 4 + 1 ⇒ x = 5

Verification :

LHS = 3(x – 1) = 3(5 – 1) = 3 × 4 = 12

RHS = 12

∴ LHS = RHS

Question 6.

3x – \(\frac{1}{3}\) = 3

Solution:

3x – \(\frac{1}{3}\) = 3

3x = 3 + \(\frac{1}{3}\) ⇒ 3x = \(\frac{9+1}{3}\)

3x = \(\frac{10}{3}\) ⇒ x = \(\frac{10}{3}\) × \(\frac{1}{3}\) ⇒ x = \(\frac{10}{9}\)

Verification :

LHS = 3x – \(\frac{1}{3}\) = \(3\left(\frac{10}{9}\right)\) – \(\frac{1}{3}\)

= \(\frac{10}{3}\) – \(\frac{1}{3}\) = \(\frac{10-1}{3}\) = \(\frac{9}{3}\) = 3

RHS = 3

∴ LHS = RHS

Question 7.

\(\frac{2}{8x}\) = 16

Solution:

Question 8.

Find the three consecutive natural numbers so that the sum of the first and second is 15 more than the third.

Solution:

Let the first number = x

Second number = x + 1

Third number = x + 1 + 1 = x + 2

According to the problem

x + x + 1 = 15 + x + 2

2x + 1 = 17 + x

2x – x = 17 – 1 ⇒ x = 16

First number = 16

Second number = 16 + 1 = 17

Third number = 17 + 1 = 18

∴ The three consecutive number are 16,17 and 18.

Question 9.

The sum of two consecutive odd number is 68. Find the number.

Solution:

Let the first odd number = x

Second odd number = x + 2

According to the problem

x + x + 2 = 68

2x + 2 = 68