These NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.2

Question 1.

Give first step you will use to separate the variable and then solve the equation:

(a) x – 1 = 0

(b) x + 1 = 0

(c) x – 1 = 5

(d) x + 6 = 2

(e) y – 4 = -7

(f) y – 4 = 4

(g) y + 4 = 4

(h) y + 4 = – 4

Answer:

(a) x – 1 = 0

Adding 1 on both sides, we get.

x – 1 + 1 = 0 + 1

x = 1

The solution is x = 1.

(b) x + 1 = 0

Subtracting 1 from both sides, we get.

x + 1 – 1 = 0 – 1

x = -1.

The solution is x = – 1

(c) x – 1 = 5

Adding 1 to both sides, we get.

x – 1 + 1 = 5 + 1

x = 6

The solution is x = 6.

(d) x + 6 = 2

Subtracting 6 from both sides, we get

x + 6 – 6 = 2 – 6

x = – 4

The solution is x = – 4.

(e) y – 4 = – 7

Adding 4 on both sides, we get

y – 4 + 4 = -7 + 4

y = – 3

The solution is y = – 3.

(f) y – 4 = 4

Adding + 4 on both sides, we get

y – 4 + 4 = 4 + 4

y = 8

The solution is y = 8.

(g) y + 4 = 4

Subtracting 4 from both sides, we get

y + 4 – 4 = 4 – 4

y = 0

The solution is y = 0.

(h) y + 4 = – 4

Adding – 4 on both sides, we get

y + 4 – 4 = -4 -4

y = – 8

The solution is y = – 8.

Question 2.

Give first step you will use to separate the variable and then solve the equation:

(a) 31 = 42

(b) \(\frac{\mathrm{b}}{2}\) = 6

(c) \(\frac{\mathrm{p}}{7}\) = 4

(d) 4x = 25

(e) 8y = 36

(f) \(\frac{z}{3}=\frac{5}{4}\)

(g) \(\frac{a}{5}=\frac{7}{15}\)

(h) 20t = -10

Answer:

(a) 31 = 42

Dividing both sides by 3, we get

\(\frac{31}{3}=\frac{42}{3}\)

1 = 14

The solution is 1 = 14.

(b) \(\frac{\mathrm{b}}{2}\) = 6

Multiplying both sides by 2, we get. b

\(\frac{\mathrm{b}}{2}\) × 2 = 6 × 2

b = 12

The solution is b = 12.

(c) \(\frac{\mathrm{p}}{7}\) = 4

Multiplying both sides by 7, we get

\(\frac{\mathrm{p}}{7}\) × 7 = 4 × 7

p = 28

The solution is p = 28.

(d) 4x = 25

Dividing both sides by 4, we get

\(\frac{4 x}{4}=\frac{25}{4}\)

x = \(\frac{25}{4}\)

The solution is x = \(\frac{25}{4}\) or x = 6 \(\frac{1}{4}\) .

(e) 8y = 36

Dividing both sides by 8, we get

\(\frac{8 y}{8}=\frac{36}{8}\)

y = \(\frac{9}{2}\)

The solution is y = \(\frac{9}{2}\) or y = 4 \(\frac{1}{2}\)

(f) (f) \(\frac{z}{3}=\frac{5}{4}\)

Multiplying both sides by 3, we get

\(\frac{z}{3}\) × 3 = \(\frac{5}{4}\) × 3

z = \(\frac{25}{4}\)

The solution is z = \(\frac{15}{4}\) or y = 3 \(\frac{3}{4}\)

(g) \(\frac{a}{5}=\frac{7}{15}\)

Multiplying both sides by 5, we get

\(\frac{a}{5}\) × 5 = \(\frac{7}{15}\) × 5

a = \(\frac{7}{3}\)

The solution is a = \(\frac{7}{3}\) or a = 2 \(\frac{1}{3}\).

(h) 20t = – 10

Dividing both sides by 20, we get

\(\frac{20 \mathrm{t}}{20}=\frac{-10}{20}\)

t = \(\frac{-1}{2}\)

The solution is t = \(\frac{-1}{2}\).

Question 3.

Give the steps you will use to separate the variable and then solve the equation:

(a) 3n – 2 = 46

(b) 5m +7 = 17

(c) \(\frac{20 \mathrm{p}}{3}\) = 40

(d) \(\frac{3 \mathrm{p}}{10}\) = 6

Answer:

(a) 3n – 2 = 46

Adding 2 on both sides, we get

3n – 2 + 2 =46 + 2

3n = 48

Dividing both sides by 3, we get

\(\frac{3 n}{3}=\frac{48}{3}\)

n = 16

The solution is n = 16.

(b) 5m + 7 = 17

Adding – 7 on both sides, we get

5m + 7 – 7 = 17 – 7

5m = 10

Dividing both sides by 5, we get

\(\frac{5 m}{5}=\frac{10}{5}\)

m = 2

The solution is m = 2

(c) \(\frac{20 \mathrm{p}}{3}\) = 40

Multiplying both sides by 3, we get.

\(\frac{20 \mathrm{p}}{3}\) × 3 = 40 × 3

20 p = 120

Dividing both sides by 20 we get.

\(\frac{20 p}{20}=\frac{120}{20}\)

p = 6

The solution is p = 6.

(d) \(\frac{3 \mathrm{p}}{10}\) = 6

Multiplying both sides by 10, we get 3p

\(\frac{3 \mathrm{p}}{10}\) × 10 = 6 × 10

3p = 60

Dividing both sides by 3, we get

\(\frac{3 p}{3}=\frac{60}{3}\)

p = 20

The solution is p = 20.

Question 4.

Solve the following equations:

(a) 10p = 100

(b) 10p + 10 = 100

(c) \(\frac{\mathrm{p}}{4}\)

(d) \(\frac{-p}{3}\) = 5

(e) \(\frac{3 \mathrm{p}}{4}\) = 6

(f) 3s = – 9

(g) 3s + 12 = 0

(h) 3s = 0

(i) 2q = 6

(j) 2q – 6 = 0

(k) 2q + 6 = 0

(l) 2q + 6 = 12

Answer:

(a) 10p = 100

Dividing both sides by 10, we get

\(\frac{10 \mathrm{p}}{10}=\frac{100}{10}\)

p = 10 10p = 90

The solution is p = 10.

(b) 10p + 10 = 100

10p + 10 – 10= 100 – 10

Dividing both sides by 10, we get

\(\frac{10 p}{10}=\frac{90}{10}\)

p = 9

The solution is p = 9.

(c) \(\frac{\mathrm{p}}{4}\) = 5

Multiplying both sides by 4, we get

\(\frac{\mathrm{p}}{4}\) × 4 = 5 × 4

p = 20

The solution is p = 20.

(d) \(\frac{-p}{3}\) = 5

Multiplying both sides by 3, we get

-p = 15

\(\frac{-p}{3}\) × 3 = 5 × 3

Multiplying both sides by (- 1)

– p × (- 1)= 15 × (- 1)

p = – 15

The solution is p = -15.

(e) \(\frac{3 \mathrm{p}}{4}\) = 6

Multiplying both sides by 4, we get

\(\frac{3 \mathrm{p}}{4}\) × 4 = 6 × 4

3p = 24

Dividing both sides by 3, we get

\(\frac{3 p}{3}=\frac{24}{3}\)

p = 8

The solution is p = 8.

(f) 3s = – 9

Dividing both sides by 3, we get

\(\frac{3 s}{3}=\frac{-9}{3}\)

s = – 3

The solution is s = – 3.

(g) 3s + 12 = 0

Subtracting 12 from both sides, we get

3s + 12- 12 = 0 – 12

3s = – 12

Dividing both sides by 3, we get

\(\frac{3 s}{3}=\frac{-12}{3}\)

s = – 4

The solution is s = – 4.

(h) 3s =0

Dividing both sides by 3, we get

\(\frac{3 s}{3}=\frac{0}{3}\)

s = 0

The solution is s = 0.

(i) 2q = 6

Dividing both sides by 2, we get

\(\frac{2 \mathrm{q}}{2}=\frac{6}{2}\)

q = 3

The solution is q = 3.

(j) 2q – 6 = 0

Adding 6 on both sides, we get

2q – 6 + 6= 0 + 6

2q = 6

Dividing both sides by 2, we get

\(\frac{2 \mathrm{q}}{2}=\frac{6}{2}\)

q = 3

The solution is q = 3.

(k) 2q + 6 =0

Subtracting 6 from both sides, we get

2q + 6 – 6 = 0 – 6

2q = – 6

Dividing both sides by 2, we get

\(\frac{2 q}{2}=\frac{-6}{2}\)

q = -3

The solution is q = – 3.

(l) 2q + 6 =12

Subtracting 6 from both sides, we get

2q + 6 – 6 = 12 – 6

2q = 6

Dividing both sides by 2, we get

\(\frac{2 \mathrm{q}}{2}=\frac{6}{2}\)

q = 3

The solution is q = 3.