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# NCERT Solutions for Class 7 Maths Exercise 4.2

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NCERT solutions for Maths Simple Equations

## NCERT Solutions for Class 7 Maths Simple Equations

Class –VII Mathematics (Ex. 4.2)

###### Question 1.Give first the step you will use to separate the variable and then solve the equations:

(a) {tex}x – 1 = 0{/tex}

(b) {tex}x + 1 = 0{/tex}

(c) {tex}x – 1 = 5{/tex}

(d) {tex}x + 6 = 2{/tex}

(e) {tex}y – 4 = – 7{/tex}

(f) {tex}y – 4 = 4{/tex}

(g) {tex}y + 4 = 4{/tex}

(h) {tex}y + 4 = – 4{/tex}

(a) {tex}x – 1 = 0{/tex} {tex} \Rightarrow {/tex} {tex}x – 1 + 1 = 0 + 1{/tex} [Adding 1 both sides]

{tex} \Rightarrow {/tex} {tex}x = 1{/tex}

(b) {tex}x + 1 = 0{/tex} {tex} \Rightarrow {/tex} {tex}x + 1 – 1 = 0 – 1{/tex} [Subtracting 1 both sides]

{tex} \Rightarrow {/tex} {tex}x = – 1{/tex}

(c) {tex}x – 1 = 5{/tex} {tex} \Rightarrow {/tex} {tex}x – 1 + 1 = 5 + 1{/tex} [Adding 1 both sides]

{tex} \Rightarrow {/tex} {tex}x = 6{/tex}

(d) {tex}x + 6 = 2{/tex} {tex} \Rightarrow {/tex} {tex}x + 6 – 6 = 2 – 6{/tex} [Subtracting 6 both sides]

{tex} \Rightarrow {/tex} {tex}x = – 4{/tex}

(e) {tex}y – 4 = – 7{/tex} {tex} \Rightarrow {/tex} {tex}y – 4 + 4 = – 7 + 4{/tex} [Adding 4 both sides]

{tex} \Rightarrow {/tex} {tex}y = – 3{/tex}

(f) {tex}y – 4 = 4{/tex} {tex} \Rightarrow {/tex} {tex}y – 4 + 4 = 4 + 4{/tex} [Adding 4 both sides]

{tex} \Rightarrow {/tex} {tex}y = 8{/tex}

(g) {tex}y + 4 = 4{/tex} {tex} \Rightarrow {/tex} {tex}y + 4 – 4 = 4 – 4{/tex} [Subtracting 4 both sides]

{tex} \Rightarrow {/tex} {tex}y = 0{/tex}

(h) {tex}y + 4 = – 4{/tex} {tex} \Rightarrow {/tex} {tex}y + 4 – 4 = – 4 – 4{/tex} [Subtracting 4 both sides]

{tex} \Rightarrow {/tex} {tex}y = – 8{/tex}

NCERT Solutions for Class 7 Maths Exercise 4.2

###### Question 2.Give first the step you will use to separate the variable and then solve the equations

(a) {tex}3l = 42{/tex}

(b) {tex}\frac{b}{2} = 6{/tex}

(c) {tex}\frac{p}{7} = 4{/tex}

(d) {tex}4x = 25{/tex}

(e) {tex}8y = 36{/tex}

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

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

(h) {tex}20t = – 10{/tex}

(a) {tex}3l = 42{/tex} {tex} \Rightarrow {/tex} {tex}\frac{{3l}}{3} = \frac{{42}}{3}{/tex} [Dividing both sides by 3]

{tex} \Rightarrow {/tex} {tex}l = 14{/tex}

(b) {tex}\frac{b}{2} = 6{/tex} {tex} \Rightarrow {/tex} {tex}\frac{b}{2} \times 2 = 6 \times 2{/tex} [Multiplying both sides by 2]

{tex} \Rightarrow {/tex} {tex}b = 12{/tex}

(c) {tex}\frac{p}{7} = 4{/tex} {tex} \Rightarrow {/tex} {tex}\frac{p}{7} \times 7 = 4 \times 7{/tex} [Multiplying both sides by 7]

{tex} \Rightarrow {/tex} {tex}p = 28{/tex}

(d) {tex}4x = 25{/tex} {tex} \Rightarrow {/tex} {tex}\frac{{4x}}{4} = \frac{{25}}{4}{/tex} [Dividing both sides by 4]

{tex} \Rightarrow {/tex} {tex}x = \frac{{25}}{4}{/tex}

(e) {tex}8y = 36{/tex} {tex} \Rightarrow {/tex} {tex}\frac{{8y}}{8} = \frac{{36}}{8}{/tex} [Dividing both sides by 8]

{tex} \Rightarrow {/tex} {tex}y = \frac{9}{2}{/tex}

(f) {tex}\frac{z}{3} = \frac{5}{4}{/tex} {tex} \Rightarrow {/tex} {tex}\frac{z}{3} \times 3 = \frac{5}{4} \times 3{/tex} [Multiplying both sides by 3]

{tex} \Rightarrow {/tex} {tex}z = \frac{{15}}{4}{/tex}

(g) {tex}\frac{a}{5} = \frac{7}{{15}}{/tex} {tex} \Rightarrow {/tex} {tex}\frac{a}{5} \times 5 = \frac{7}{{15}} \times 5{/tex} [Multiplying both sides by 5]

{tex} \Rightarrow {/tex} {tex}a = \frac{7}{3}{/tex}

(h) {tex}20t = – 10{/tex} {tex} \Rightarrow {/tex} {tex}\frac{{20t}}{{20}} = \frac{{ – 10}}{{20}}{/tex} [Dividing both sides by 20]

{tex} \Rightarrow {/tex} {tex}t = \frac{{ – 1}}{2}{/tex}

NCERT Solutions for Class 7 Maths Exercise 4.2

###### Question 3.Give first the step you will use to separate the variable and then solve the equations

(a) {tex}3n – 2 = 46{/tex}

(b) {tex}5m + 7 = 17{/tex}

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

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

(a) {tex}3n – 2 = 46{/tex}

Step I: {tex}3n – 2 + 2 = 46 + 2{/tex} {tex} \Rightarrow {/tex} {tex}3n = 48{/tex}

Step II: {tex}\frac{{3n}}{3} = \frac{{48}}{3}{/tex} {tex} \Rightarrow {/tex} {tex}n = 16{/tex} [Dividing both sides by 3]

(b) {tex}5m + 7 = 17{/tex}

Step I: {tex}5m + 7 – 7 = 17 – 7{/tex} {tex} \Rightarrow {/tex} {tex}5m = 10{/tex} [Subtracting 7 both sides]

Step II: {tex}\frac{{5m}}{5} = \frac{{10}}{5}{/tex} {tex} \Rightarrow {/tex}{tex}m = 2{/tex} [Dividing both sides by 5]

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

Step I: {tex}\frac{{20p}}{3} \times 3 = 40 \times 3{/tex} {tex} \Rightarrow {/tex} {tex}20p = 120{/tex} [Multiplying both sides by 3]

Step II: {tex}\frac{{20p}}{{20}} = \frac{{120}}{{20}}{/tex} {tex} \Rightarrow {/tex} {tex}p = 6{/tex} [Dividing both sides by 20]

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

Step I: {tex}\frac{{3p}}{{10}} \times 10 = 6 \times 10{/tex} {tex} \Rightarrow {/tex} {tex}3p = 60{/tex} [Multiplying both sides by 10]

Step II: {tex}\frac{{3p}}{3} = \frac{{60}}{3}{/tex} {tex} \Rightarrow {/tex} {tex}p = 20{/tex} [Dividing both sides by 3]

NCERT Solutions for Class 7 Maths Exercise 4.2

###### Question 4.Solve the following equation:

(a) {tex}10p = 100{/tex}

(b) {tex}10p + 10 = 100{/tex}

(c) {tex}\frac{p}{4} = 5{/tex}

(d) {tex}\frac{{ – p}}{3} = 5{/tex}

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

(f) {tex}3s = – 9{/tex}

(g) {tex}3s + 12 = 0{/tex}

(h) {tex}3s = 0{/tex}

(i) {tex}2q = 6{/tex}

(j) {tex}2q – 6 = 0{/tex}

(k) {tex}2q + 6 = 0{/tex}

(l) {tex}2q + 6 = 12{/tex}

(a) {tex}10p = 100{/tex} {tex} \Rightarrow {/tex} {tex}\frac{{10p}}{{10}} = \frac{{100}}{{10}}{/tex} [Dividing both sides by 10]

{tex} \Rightarrow {/tex} {tex}p = 10{/tex}

(b) {tex}10p + 10 = 100{/tex} {tex} \Rightarrow {/tex} {tex}10p + 10 – 10 = 100 – 10{/tex} [Subtracting both sides 10]

{tex} \Rightarrow {/tex} {tex}10p = 90{/tex} {tex} \Rightarrow {/tex} {tex}\frac{{10p}}{{10}} = \frac{{90}}{{10}}{/tex} [Dividing both sides by 10]

{tex} \Rightarrow {/tex} {tex}p = 9{/tex}

(c) {tex}\frac{p}{4} = 5{/tex} {tex} \Rightarrow {/tex} {tex}\frac{p}{4} \times 4 = 5 \times 4{/tex} [Multiplying both sides by 4]

{tex} \Rightarrow {/tex} {tex}p = 20{/tex}

(d) {tex}\frac{{ – p}}{3} = 5{/tex} {tex} \Rightarrow {/tex} {tex}\frac{{ – p}}{3} \times \left( { – 3} \right) = 5 \times \left( { – 3} \right){/tex} [Multiplying both sides by – 3]

{tex} \Rightarrow {/tex} {tex}p = – 15{/tex}

(e) {tex}\frac{{3p}}{4} = 6{/tex} {tex} \Rightarrow {/tex} {tex}\frac{{3p}}{4} \times 4 = 6 \times 4{/tex} [Multiplying both sides by 4]

{tex} \Rightarrow {/tex} {tex}3p = 24{/tex} {tex} \Rightarrow {/tex} {tex}\frac{{3p}}{3} = \frac{{24}}{3}{/tex} [Dividing both sides by 3]

{tex} \Rightarrow {/tex} {tex}p = 8{/tex}

(f) {tex}3s = – 9{/tex} {tex} \Rightarrow {/tex} {tex}\frac{{3s}}{3} = \frac{{ – 9}}{3}{/tex} [Dividing both sides by 3]

{tex} \Rightarrow {/tex} {tex}s = – 3{/tex}

(g) {tex}3s + 12 = 0{/tex} {tex} \Rightarrow {/tex} {tex}3s + 12 – 12 = 0 – 12{/tex} [Subtracting both sides 10]

{tex} \Rightarrow {/tex} {tex}3s = – 12{/tex} {tex} \Rightarrow {/tex} {tex}\frac{{3s}}{3} = \frac{{ – 12}}{3}{/tex} [Dividing both sides by 3]

{tex} \Rightarrow {/tex} {tex}s = – 4{/tex}

(h) {tex}3s = 0{/tex} {tex} \Rightarrow {/tex} {tex}\frac{{3s}}{3} = \frac{0}{3}{/tex} [Dividing both sides by 3]

{tex} \Rightarrow {/tex} {tex}s = 0{/tex}

(i) {tex}2q = 6{/tex} {tex} \Rightarrow {/tex} {tex}\frac{{2q}}{2} = \frac{6}{2}{/tex} [Dividing both sides by 2]

{tex} \Rightarrow {/tex} {tex}q = 3{/tex}

(j) {tex}2q – 6 = 0{/tex} {tex} \Rightarrow {/tex} {tex}2q – 6 + 6 = 0 + 6{/tex} [Adding both sides 6]

{tex} \Rightarrow {/tex} {tex}2q = 6{/tex} {tex} \Rightarrow {/tex} {tex}\frac{{2q}}{2} = \frac{6}{2}{/tex} [Dividing both sides by 2]

{tex} \Rightarrow {/tex} {tex}q = 3{/tex}

(k) {tex}2q + 6 = 0{/tex} {tex} \Rightarrow {/tex} {tex}2q + 6 – 6 = 0 – 6{/tex} [Subtracting both sides 6]

{tex} \Rightarrow {/tex} {tex}2q = – 6{/tex} {tex} \Rightarrow {/tex} {tex}\frac{{2q}}{2} = \frac{{ – 6}}{2}{/tex} [Dividing both sides by 2]

{tex} \Rightarrow {/tex} {tex}q = – 3{/tex}

(l) {tex}2q + 6 = 12{/tex} {tex} \Rightarrow {/tex} {tex}2q + 6 – 6 = 12 – 6{/tex} [Subtracting both sides 6]

{tex} \Rightarrow {/tex} {tex}2q = 6{/tex} {tex} \Rightarrow {/tex} {tex}\frac{{2q}}{2} = \frac{6}{2}{/tex} [Dividing both sides by 2]

{tex} \Rightarrow {/tex} {tex}q = 3{/tex}

## NCERT Solutions for Class 7 Maths Exercise 4.2

NCERT Solutions Class 7 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 7 Maths includes text book solutions from Class 7 Maths Book . NCERT Solutions for CBSE Class 7 Maths have total 15 chapters. 7 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 7 solutions PDF and Maths ncert class 7 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.

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