NCERT Solutions Class 7 Maths Chapter 4 Simple Equations Ex 4.2

NCERT Solutions Class 7 Mathematics 
Chapter – 4 (Simple Equations)

The NCERT Solutions in English Language for Class 7 Mathematics Chapter – 4 Simple Equations Exercise 4.2 has been provided here to help the students in solving the questions from this exercise. 

Chapter : 4 Simple Equations

Exercise – 4.2 

1. Give first the 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

Solution –
(a) x – 1 = 0
We have to add 1 to both the side of given equation,
Then we get,
⇒ x – 1 + 1 = 0 + 1
⇒ x = 1

(b) x + 1 = 0
We have to subtract 1 to both the side of given equation,
Then we get,
⇒ x + 1 – 1 = 0 – 1
⇒ x = – 1

(c) x – 1 = 5
We have to add 1 to both the side of given equation,
⇒ x – 1 + 1 = 5 + 1
⇒ x = 6

(d) x + 6 = 2
We have to subtract 6 to both the side of given equation,
⇒ x + 6 – 6 = 2 – 6
⇒ x = – 4

(e) y – 4 = – 7
We have to add 4 to both the side of given equation,
⇒ y – 4 + 4 = – 7 + 4
⇒ y = – 3

(f) y – 4 = 4
We have to add 4 to both the side of given equation,
⇒ y – 4 + 4 = 4 + 4
⇒ y = 8

(g) y + 4 = 4
We have to subtract 4 to both the side of given equation,
⇒ y + 4 – 4 = 4 – 4
⇒ y = 0

(h) y + 4 = – 4
We have to subtract 4 to both the side of given equation,
⇒ y + 4 – 4 = – 4 – 4
⇒ y = – 8

2. Give first the step you will use to separate the variable and then solve the equation:
(a) 3l = 42
(b) \mathbf{\frac{b}{2}} = 6

(c) \mathbf{\frac{p}{7}} = 4
(d) 4x = 25
(e) 8y = 36

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

(g) \mathbf{\frac{a}{5}} = \mathbf{\frac{7}{15}}
(h) 20t = – 10

Solution –
(a) 3l = 42
Now we have to divide both sides of the equation by 3,
⇒ \frac{3l}{3} = \frac{42}{3}
⇒ l = 14

(b) \mathbf{\frac{b}{2}} = 6
Now we have to multiply both sides of the equation by 2,
⇒ \frac{b}{2} × 2= 6 × 2
⇒ b = 12

(c) \mathbf{\frac{p}{7}} = 4
Now we have to multiply both sides of the equation by 7,
⇒ \frac{p}{7} × 7 = 4 × 7
⇒ p = 28

(d) 4x = 25
Now we have to divide both sides of the equation by 4,
⇒ \frac{4x}{4} = \frac{25}{4}

⇒ x = \frac{25}{4} = 6\frac{1}{4}

(e) 8y = 36
Now we have to divide both sides of the equation by 8,
⇒ \frac{8y}{8} = \frac{36}{6}

⇒ x = \frac{9}{2} = 4\frac{1}{2}

(f) \mathbf{\frac{z}{3}} = \mathbf{\frac{5}{4}}
Now we have to multiply both sides of the equation by 3,
⇒ \frac{z}{3} × 3 = \frac{5}{4} × 3

⇒ x = \frac{15}{4} = 3\frac{3}{4}

(g) \mathbf{\frac{a}{5}} = \mathbf{\frac{7}{15}}
Now we have to multiply both sides of the equation by 5,
⇒ \frac{a}{5} × 5 = \frac{7}{15} × 5

⇒ a = \frac{7}{3} = 2\frac{1}{3}

(h) 20t = – 10
Now we have to divide both sides of the equation by 20,
⇒ \frac{20t}{20} = \frac{-10}{20}

⇒ x = – ½

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) \mathbf{\frac{20p}{3}} = 40

(d) \mathbf{\frac{3p}{10}} = 6

Solution –
(a) 3n – 2 = 46
First we have to add 2 to the both sides of the equation,
⇒ 3n – 2 + 2 = 46 + 2
⇒ 3n = 48
Now,
We have to divide both sides of the equation by 3,
⇒ \frac{3n}{3} = \frac{48}{3}
⇒ n = 16

(b) 5m + 7 = 17
First we have to subtract 7 to the both sides of the equation,
⇒ 5m + 7 – 7 = 17 – 7
⇒ 5m = 10
Now,
We have to divide both sides of the equation by 5,
⇒ \frac{5m}{5} = \frac{10}{5}
⇒ m = 2

(c) \mathbf{\frac{20p}{3}} = 40
First we have to multiply both sides of the equation by 3,
⇒ \frac{20p}{3} × 3 = 40 × 3
⇒ 20p = 120
Now,
We have to divide both sides of the equation by 20,
⇒ \frac{20p}{20} = \frac{120}{20}
⇒ p = 6

(d) \mathbf{\frac{3p}{10}} = 6
First we have to multiply both sides of the equation by 10,
⇒ \frac{3p}{10} × 10 = 6 × 10
⇒ 3p = 60
Now,
We have to divide both sides of the equation by 3,
⇒ \frac{3p}{3} = \frac{60}{3}
⇒ p = 20

4. Solve the following equations:
(a) 10p = 100
(b) 10p + 10 = 100
(c) \mathbf{\frac{p}{4}} = 5

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

(e) \mathbf{\frac{3p}{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

Solution –

(a) 10p = 100
We have to divide both sides of the equation by 10,
⇒ \frac{10p}{10} = \frac{100}{10}
⇒ p = 10

(b) 10p + 10 = 100
First we have to subtract 10 to the both sides of the equation,
⇒ 10p + 10 – 10 = 100 – 10
⇒ 10p = 90
Now,
We have to divide both sides of the equation by 10,
⇒ \frac{10p}{10} = \frac{90}{10}
⇒ p = 9

(c) \mathbf{\frac{p}{4}} = 5
We have to multiply both sides of the equation by 4,
⇒ \frac{p}{4} × 4 = 5 × 4
⇒ p = 20

(d) \mathbf{\frac{-p}{3}} = 5
We have to multiply both sides of the equation by – 3,
⇒ \frac{-p}{3} × (- 3) = 5 × (- 3)
⇒ p = – 15

(e) \mathbf{\frac{3p}{4}} = 6
First we have to multiply both sides of the equation by 4,
⇒ \frac{3p}{4} × 4 = 6 × 4
⇒ 3p = 24
Now,
We have to divide both sides of the equation by 3,
⇒ \frac{3p}{3} = \frac{24}{3}
⇒ p = 8

(f) 3s = – 9
We have to divide both sides of the equation by 3,
⇒ \frac{3s}{3} = \frac{-9}{3}
⇒ s = -3

(g) 3s + 12 = 0
First we have to subtract 12 to the both sides of the equation,
⇒ 3s + 12 – 12 = 0 – 12
⇒ 3s = -12
Now,
We have to divide both sides of the equation by 3,
⇒ \frac{3s}{3} = \frac{-12}{3}
⇒ s = – 4

(h) 3s = 0
We have to divide both sides of the equation by 3,
Then, we get,
⇒ \frac{3s}{3} = \frac{0}{3}
⇒ s = 0

(i) 2q = 6
We have to divide both sides of the equation by 2,
⇒ \frac{2q}{2} = \frac{6}{2}
⇒ q = 3

(j) 2q – 6 = 0
First we have to add 6 to the both sides of the equation,
⇒ 2q – 6 + 6 = 0 + 6
⇒ 2q = 6
Now,
We have to divide both sides of the equation by 2,
⇒ \frac{2q}{2} = \frac{6}{2}
⇒ q = 3

(k) 2q + 6 = 0
First we have to subtract 6 to the both sides of the equation,
⇒ 2q + 6 – 6 = 0 – 6
⇒ 2q = – 6
Now,
We have to divide both sides of the equation by 2,
⇒ \frac{2q}{2} = \frac{-6}{2}
⇒ q = – 3

(l) 2q + 6 = 12
First we have to subtract 6 to the both sides of the equation,
⇒ 2q + 6 – 6 = 12 – 6
⇒ 2q = 6
Now,
We have to divide both sides of the equation by 2,
⇒ \frac{2q}{2} = \frac{6}{2}
⇒ q = 3

 

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