NCERT Solutions for Class 7 Maths Exercise 4.2

NCERT Solutions for Class 7 Maths Exercise 4.2 book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 7 Maths chapter wise NCERT solution for Maths Book all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

NCERT solutions for Maths Simple Equations Download as PDF

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}

Answer:

(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}

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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}

Answer:

(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}

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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}

Answer:

(a) {tex}3n – 2 = 46{/tex}

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

[Adding 2 both sides]

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]

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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}

Answer:

(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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