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.

(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.

(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.

(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.

(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

(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

(l) 2q + 6 = 12

or 2q = 12 – 6

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

∴ q = 3.