HBSE 7th Class Maths Solutions Chapter 4 Simple Equations Ex 4.2

Haryana State Board HBSE 7th Class Maths Solutions Chapter 4 Simple Equations Ex 4.2 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 4 Simple Equations Exercise 4.2

Question 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
L.H.S. = x- R.H.S.
1 + 1 = x = 0 + 1 = 1
∴ x = 1
Which is the required solution.

HBSE 7th Class Maths Solutions Chapter 4 Simple Equations Ex 4.2

(b) x + 1 = 0
L.H.S. = x + 1 – 1 = x
R.H.S. = 0 – 1 = -1
∴ x = -1
Which is the required solution.

(c) x – 1 = 5
L.H.S. = x – 1 + 1 = x
R.H.S. = 5 + 1 = 6
∴ x = 6
Which is the required solution.

(d) x + 6 =2
L.H.S. = x + 6 – 6 = x
R.H.S. = 2 – 6 = – 4
∴ x = -4
Which is the required solution.

(e) y- 4 = -7
L.H.S. = y – 4 + 4 = y
R.H.S. = – 7 + 4 = – 3
∴ y = -3
Which is the required solution.

(f) y – 4 = 4
L.H.S. = y – 4 + 4 = y
R.H.S. = 4 + 4 = 8
∴ y = 8
Which is the required solution.

(g) y + 4 = 4
L.H.S. = y + 4 + 4= y
R.H.S. = 4 – 4 = 0
∴ y = 0
Which is the required solution.

HBSE 7th Class Maths Solutions Chapter 4 Simple Equations Ex 4.2

(h) y + 4 = – 4
L.H.S. = y + 4 — 4 = y
R.H.S. = – 4 – 4 = – S
∴ y = -8
Which is the required solution.

Question 2.
Give first the step you will use to separate the variable and then solve the equation.
(a) 3l = 42
(b) \(\frac{b}{2}\) = 6
(c) \(\frac{p}{7}\) = 4
(d) 4x = 25
(e) 8y = 36
(f) \(\frac{z}{3}=\frac{5}{4}\)
(g) \(\frac{a}{5}=\frac{7}{15}\)
(h) 20t = – 10.
Solution:
(a) 3l = 42
L.H.S. = 3l = \(\frac{3}{3}\) x l = l
R.H.S. = 42 = \(\frac{42}{3}\) = 14
∴ l = 14
Which is the required solution.

b) \(\frac{b}{2}\) = 6
L.H.S = \(\frac{b}{2}=\frac{b}{2}\) x 2 = b
R.H.S = 6 = 6 x 2 = 12
∴ b = 12.
Which is the required solution.

(c) \(\frac{p}{7}\) = 4
L.H.S. = \(\frac{p}{7}=\frac{p}{7}\) x 7 = p
R.H.S. = 4 = 4 x 7 = 28
∴ p = 28
Which is the required solution.

(d) 4 x = 25
L.H.S = 8y = \(\frac{8y}{8}\) = y
R.H.S = 36 = \(\frac{36}{8}=\frac{9}{2}\)
∴ y = \(\frac{9}{2}\)
Which is the required solution.

(f) \(\frac{z}{3}=\frac{5}{4}\)
L.H.S = \(\frac{z}{3}=\frac{z}{3}\) x 3 = z
R.H.S = \(\frac{5}{4}=\frac{5}{4} \times 3=\frac{15}{4}\)
∴ y = \(\frac{15}{4}\)
Which is the required solution.

HBSE 7th Class Maths Solutions Chapter 4 Simple Equations Ex 4.2

(f) \(\frac{a}{5}=\frac{7}{15}\)
L.H.S. = \(\frac{a}{5}=\frac{a}{5}\) x 5 = a
R.H.S = \(\frac{7}{15}=\frac{7}{15}\) x 5 = \(\frac{7}{3}\)
∴ a = \(\frac{7}{3}\)
Which is the required solution.

(h) 20t = -10
L.H.S. = 20t = \(\frac{20 t}{20}\) = t
R.H.S = -10 = \(\frac{-10}{20}=-\frac{1}{2}\)
∴ t = \(-\frac{1}{2}\)
Which is the required solution.

Question 3.
Give the step you will use to separate the variable and then solve the equation:
(a) 3n – 2 = 46
(b) 5m + 7 = 17
(c) \(\frac{20 p}{3}\) = 40
(d) \(\frac{3 p}{10}\) = 6
Solution:
3n – 2 = 46
L.H.S. = 3n – 2 + 2
= 3n = \(\frac{3n}{3}\) = n
R.H.S. = 46 + 2 = 48 = \(\frac{48}{3}\) = 16
∴ n = 16
Which is the required solution.

HBSE 7th Class Maths Solutions Chapter 4 Simple Equations Ex 4.2

(b) 5m + 7 = 17
L.H.S. = 5 m + 7 = 5m + 7-7
= 5m = \(\frac{5m}{5}\) = m
R.H.S = 17 = 17 – 7 = 10
= \(\frac{10}{5}\) = 2
∴ m = 2
Which is the required solution.

(c) \(\frac{20 p}{3}\) = 40
L.H.S = \(\frac{20 p}{3}=\frac{20 p}{3} \times 3\)
= 20 p = \(\frac{20 p}{20}\) = p
R.H.S. = 40 = 40 x 3
= 120 = \(\frac{120}{20}\) = 6
∴ P = 6
Which is the required solution.

(d) \(\frac{3 p}{10}\) = 6
L.H.S.
= \(\frac{3 p}{10}=\frac{3 p}{10} \times 10=3 p=\frac{3 p}{3}=p\)
R.H.S. = 6 x 10 = 60 = \(\frac{60}{3}\) = 20
∴ p = 20
Which is the required solution.

Question 4.
Solve the following equations
(a) 10p = 100
(b) 10p + 10 = 100
(c) \(\frac{p}{4}\) = 5
(d) \(\frac{-p}{3}\) = 5
(e) \(\frac{3 p}{4}\) = 6
(f) 3s = – 9
(g) 3s + 12 = 0
(h) 3s = 0
(j) 2q – 6 = 0
(k) 2q + 6 = 0
(l) 2q + 6 = 12
Solution:
(a) 10p = 100
\(\frac{10 p}{10}=\frac{100}{10}\)
∴ p = 10

(b) 10p + 10 = 100
10p = 100 – 10
10p = 90
\(\frac{10 p}{10}=\frac{90}{10}\)
∴ p = 9

HBSE 7th Class Maths Solutions Chapter 4 Simple Equations Ex 4.2

(c) \(\frac{p}{4}\) = 5
\(\frac{p}{4}\) x 4 = 5 x 4
∴ p = 20

(d) \(\frac{-p}{3}\) x 3 = 5 x 3
-p = 15
∴ p = -15

(e) \(\frac{3 p}{4}\) = 6
\(\frac{3 p}{4}\) x 4 = 6 x 4
3p = 24
\(\frac{3 p}{3}=\frac{24}{3}\)
∴ p = 8

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

(g) 3s+ 12 = 0
3s = -12
\(\frac{3 s}{3}=\frac{-12}{3}\)
∴ s = -4

(h) 3s = 0
or \(\frac{3 s}{3}=\frac{0}{3}\)
∴ s = 0

(i) 2q = 6
or \(\frac{2 q}{2}=\frac{6}{2}\)
∴ q = 3

(j) 2q – 6 = 6
or, 2q = +6
\(\frac{2 q}{2}=\frac{6}{2}\)
∴ q = 3

(k) 2q + 6 = 0
or 2q = -6
\(\frac{2 q}{2}=\frac{-6}{2}\)
∴ q = -3

HBSE 7th Class Maths Solutions Chapter 4 Simple Equations Ex 4.2

(l) 2q + 6 = 12
or 2q = 12 – 6
\(\frac{2 q}{2}=\frac{6}{2}\)
∴ q = 3.

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