Exercise 4.2 (Revised) - Chapter 4 - Simple Equations - Ncert Solutions class 7 - Maths

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Chapter 4 - Simple Equations - NCERT Solutions for Class 7 Maths | Comprehensive Guide

Ex 4.2 Question 1.

Give first the step you will use to separate the variable and then solve the equations:
(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) $x-1=0 \Rightarrow x-1+1=0+1$ [Adding 1 both sides]
$
\Rightarrow x=1
$
(b) $x+1=0 \Rightarrow x+1-1=0-1$ [Subtracting 1 both sides]
$
\Rightarrow x=-1
$

(c) $x-1=5 \Rightarrow x-1+1=5+1$ [Adding 1 both sides]
$
\Rightarrow x=6
$
(d) $x+6=2 \Rightarrow x+6-6=2-6$ [Subtracting 6 both sides]
$
\Rightarrow x=-4
$
(e) $y-4=-7 \Rightarrow y-4+4=-7+4$ [Adding 4 both sides]

$
\Rightarrow y=-3
$
(f) $y-4=4 \Rightarrow y-4+4=4+4$ [Adding 4 both sides]
$
\Rightarrow y=8
$
(g) $y+4=4 \Rightarrow y+4-4=4-4$ [Subtracting 4 both sides]
$
\Rightarrow y=0
$
(h) $y+4=-4 \Rightarrow y+4-4=-4-4$ [Subtracting 4 both sides]
$
\Rightarrow y=-8
$

Ex 4.2 Question 2.

Give first the step you will use to separate the variable and then solve the equations
(a) $3 l=42$
(b) $\frac{b}{2}=6$
(c) $\frac{p}{7}=4$
(d) $4 x=25$
(e) $8 y=36$
(f) $\frac{z}{3}=\frac{5}{4}$
(g) $\frac{a}{5}=\frac{7}{15}$

(h) $20 t=-10$

Answer:

(a) $3 l=42 \Rightarrow \frac{3 l}{3}=\frac{42}{3}$ [Dividing both sides by 3] $\Rightarrow l=14$
(b) $\frac{b}{2}=6 \Rightarrow \frac{b}{2} \times 2=6 \times 2$ [Multiplying both sides by 2] $\Rightarrow b=12$

(c) $\frac{p}{7}=4 \Rightarrow \frac{p}{7} \times 7=4 \times 7$ [Multiplying both sides by 7]
$
\Rightarrow p=28
$
(d) $4 x=25 \Rightarrow \frac{4 x}{4}=\frac{25}{4}$ [Dividing both sides by 4 ]
$
\Rightarrow x=\frac{25}{4}
$
(e) $8 y=36 \Rightarrow \frac{8 y}{8}=\frac{36}{8}$ [Dividing both sides by 8 ]
$
\Rightarrow y=\frac{9}{2}
$
(f) $\frac{z}{3}=\frac{5}{4} \Rightarrow \frac{z}{3} \times 3=\frac{5}{4} \times 3$ [Multiplying both sides by 3]
$
\Rightarrow z=\frac{15}{4}
$
(g) $\frac{a}{5}=\frac{7}{15} \Rightarrow \frac{a}{5} \times 5=\frac{7}{15} \times 5$ [Multiplying both sides by 5]
$
\Rightarrow a=\frac{7}{3}
$
(h) $20 t=-10 \Rightarrow \frac{20 t}{20}=\frac{-10}{20}$ [Dividing both sides by 20]
$
\Rightarrow t=\frac{-1}{2}
$

Ex 4.2 Question 3.

Give first the step you will use to separate the variable and then solve the equations
(a) $3 n-2=46$
(b) $5 m+7=17$
(c) $\frac{20 p}{3}=40$
(d) $\frac{3 p}{10}=6$

Answer:

(a) $3 n-2=46$
Step I: $3 n-2+2=46+2 \Rightarrow 3 n=48$
[Adding 2 both sides]

Step II: $\frac{3 n}{3}=\frac{48}{3} \Rightarrow n=16$ [Dividing both sides by 3 ]
(b) $5 m+7=17$

Step I: $5 m+7-7=17-7 \Rightarrow 5 m=10$ [Subtracting 7 both sides]
Step II: $\frac{5 m}{5}=\frac{10}{5} \Rightarrow m=2$ [Dividing both sides by 5]
(c) $\frac{20 p}{3}=40$

Step I: $\frac{20 p}{3} \times 3=40 \times 3 \Rightarrow 20 p=120$ [Multiplying both sides by 3]
Step II: $\frac{20 p}{20}=\frac{120}{20} \Rightarrow p=6$ [Dividing both sides by 20 ]
(d) $\frac{3 p}{10}=6$

Step I: $\frac{3 p}{10} \times 10=6 \times 10 \Rightarrow 3 p=60$ [Multiplying both sides by 10 ]
Step II: $\frac{3 p}{3}=\frac{60}{3} \Rightarrow p=20$ [Dividing both sides by 3]

Ex 4.2 Question 4.

Solve the following equation:
(a) $10 p=100$
(b) $10 p+10=100$
(c) $\frac{p}{4}=5$
(d) $\frac{-p}{3}=5$
(e) $\frac{3 p}{4}=6$
(f) $3 s=-9$
(g) $3 s+12=0$
(h) $3 s=0$
(i) $2 q=6$

(j) $2 q-6=0$
(k) $2 q+6=0$
(1) $2 q+6=12$

Answer:

(a) $10 p=100 \Rightarrow \frac{10 p}{10}=\frac{100}{10}$ [Dividing both sides by 10 ]
$
\Rightarrow p=10
$
(b) $10 p+10=100 \Rightarrow 10 p+10-10=100-10$ [Subtracting both sides 10]
$\Rightarrow 10 p=90 \Rightarrow \frac{10 p}{10}=\frac{90}{10}$ [Dividing both sides by 10 ]
$
\Rightarrow p=9
$
(c) $\frac{p}{4}=5 \Rightarrow \frac{p}{4} \times 4=5 \times 4$ [Multiplying both sides by 4]
$
\Rightarrow p=20
$
(d) $\frac{-p}{3}=5 \Rightarrow \frac{-p}{3} \times(-3)=5 \times(-3)$ [Multiplying both sides by -3 ]
$
\Rightarrow p=-15
$
(e) $\frac{3 p}{4}=6 \Rightarrow \frac{3 p}{4} \times 4=6 \times 4$ [Multiplying both sides by 4 ]
$\Rightarrow 3 p=24 \Rightarrow \frac{3 p}{3}=\frac{24}{3}$ [Dividing both sides by 3]

$
\Rightarrow p=8
$
(f) $3 s=-9 \Rightarrow \frac{3 s}{3}=\frac{-9}{3}$ [Dividing both sides by 3]
$
\Rightarrow s=-3
$
(g) $3 s+12=0 \Rightarrow 3 s+12-12=0-12$ [Subtracting both sides 10]
$\Rightarrow 3 s=-12 \Rightarrow \frac{3 s}{3}=\frac{-12}{3}$ [Dividing both sides by 3]
$
\Rightarrow s=-4
$
(h) $3 s=0 \Rightarrow \frac{3 s}{3}=\frac{0}{3}$ [Dividing both sides by 3]

$
\Rightarrow s=0
$
(i) $2 q=6 \Rightarrow \frac{2 q}{2}=\frac{6}{2}$ [Dividing both sides by 2 ]
$
\Rightarrow q=3
$
(j) $2 q-6=0 \Rightarrow 2 q-6+6=0+6$ [Adding both sides 6 ]
$\Rightarrow 2 q=6 \Rightarrow \frac{2 q}{2}=\frac{6}{2}$ [Dividing both sides by 2 ]
$
\Rightarrow q=3
$
(k) $2 q+6=0 \Rightarrow 2 q+6-6=0-6$ [Subtracting both sides 6]
$\Rightarrow 2 q=-6 \Rightarrow \frac{2 q}{2}=\frac{-6}{2}$ [Dividing both sides by 2]
$
\Rightarrow q=-3
$
(l) $2 q+6=12 \Rightarrow 2 q+6-6=12-6$ [Subtracting both sides 6]
$\Rightarrow 2 q=6 \Rightarrow \frac{2 q}{2}=\frac{6}{2}$ [Dividing both sides by 2]
$
\Rightarrow q=3
$