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NCERT Solutions for Class 7 Maths Exercise 4.1 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. 4.1)**

**Question 1.**Complete the last column of the table:

S. No. | Equation | Value | Say, whether the Equation is satisfied. (Yes / No) |

1 | {tex}x + 3 = 0{/tex} | {tex}x = 3{/tex} | |

2 | {tex}x + 3 = 0{/tex} | {tex}x = 0{/tex} | |

3 | {tex}x + 3 = 0{/tex} | {tex}x =- 3{/tex} | |

4 | {tex}x – 7 = 1{/tex} | {tex}x = 7{/tex} | |

5 | {tex}x – 7 = 1{/tex} | {tex}x = 8{/tex} | |

6 | {tex}5x = 25{/tex} | {tex}x = 0{/tex} | |

7 | {tex}5x = 25{/tex} | {tex}x = 5{/tex} | |

8 | {tex}5x = 25{/tex} | {tex}x =- 5{/tex} | |

9 | {tex}\frac{m}{3} = 2{/tex} | {tex}m =- 6{/tex} | |

10 | {tex}\frac{m}{3} = 2{/tex} | {tex}m = 0{/tex} | |

11 | {tex}\frac{m}{3} = 2{/tex} | {tex}m = 6{/tex} |

**Answer:**

S. No. | Equation | Value | Say, whether the Equation is satisfied. (Yes / No) |

1 | {tex}x + 3 = 0{/tex} | {tex}x = 3{/tex} | No |

2 | {tex}x + 3 = 0{/tex} | {tex}x = 0{/tex} | No |

3 | {tex}x + 3 = 0{/tex} | {tex}x =- 3{/tex} | Yes |

4 | {tex}x – 7 = 1{/tex} | {tex}x = 7{/tex} | No |

5 | {tex}x – 7 = 1{/tex} | {tex}x = 8{/tex} | Yes |

6 | {tex}5x = 25{/tex} | {tex}x = 0{/tex} | No |

7 | {tex}5x = 25{/tex} | {tex}x = 5{/tex} | Yes |

8 | {tex}5x = 25{/tex} | {tex}x =- 5{/tex} | No |

9 | {tex}\frac{m}{3} = 2{/tex} | {tex}m =- 6{/tex} | No |

10 | {tex}\frac{m}{3} = 2{/tex} | {tex}m = 0{/tex} | No |

11 | {tex}\frac{m}{3} = 2{/tex} | {tex}m = 6{/tex} | Yes |

NCERT Solutions for Class 7 Maths Exercise 4.1

**Question 2.**Check whether the value given in the brackets is a solution to the given equation or not:

(a) {tex}n + 5 = 19\left( {n = 1} \right){/tex}

(b) {tex}7n + 5 = 19\left( {n = – 2} \right){/tex}

(c) {tex}7n + 5 = 19\left( {n = 2} \right){/tex}

(d) {tex}4p – 3 = 13\left( {p = 1} \right){/tex}

(e) {tex}4p – 3 = 13\left( {p = – 4} \right){/tex}

(f) {tex}4p – 3 = 13\left( {p = 0} \right){/tex}

**Answer:**

(a) {tex}n + 5 = 19\left( {n = 1} \right){/tex}

Putting {tex}n = 1{/tex} in L.H.S.,

1 + 5 = 6

{tex}\because {/tex} L.H.S. {tex} \ne {/tex} R.H.S.,

{tex}\therefore {/tex} {tex}n = 1{/tex} is not the solution of given equation.

(b) {tex}7n + 5 = 19\left( {n = – 2} \right){/tex}

Putting {tex}n = – 2{/tex} in L.H.S.,

{tex}7\left( { – 2} \right) + 5 = – 14 + 5 = – 9{/tex}

{tex}\because {/tex} L.H.S. {tex} \ne {/tex} R.H.S.,

{tex}\therefore {/tex} {tex}n = – 2{/tex} is not the solution of given equation.

(c) {tex}7n + 5 = 19\left( {n = 2} \right){/tex}

Putting {tex}n = 2{/tex} in L.H.S.,

{tex}7\left( 2 \right) + 5 = 14 + 5 = 19{/tex}

{tex}\because {/tex} L.H.S. {tex} = {/tex} R.H.S.,

{tex}\therefore {/tex} {tex}n = 2{/tex} is the solution of given equation.

(d) {tex}4p – 3 = 13\left( {p = 1} \right){/tex}

Putting {tex}p = 1{/tex} in L.H.S.,

{tex}4\left( 1 \right) – 3 = 4 – 3 = 1{/tex}

{tex}\because {/tex} L.H.S. {tex} \ne {/tex} R.H.S.,

{tex}\therefore {/tex} {tex}p = 1{/tex} is not the solution of given equation.

(e) {tex}4p – 3 = 13\left( {p = – 4} \right){/tex}

Putting {tex}p = – 4{/tex} in L.H.S.,

{tex}4\left( { – 4} \right) – 3 = – 16 – 3 = – 19{/tex}

{tex}\because {/tex} L.H.S. {tex} \ne {/tex} R.H.S.,

{tex}\therefore {/tex} {tex}p = – 4{/tex} is not the solution of given equation.

(f) {tex}4p – 3 = 13\left( {p = 0} \right){/tex}

Putting {tex}p = 0{/tex} in L.H.S.,

{tex}4\left( 0 \right) – 3 = 0 – 3 = – 3{/tex}

{tex}\because {/tex} L.H.S. {tex} \ne {/tex} R.H.S.,

{tex}\therefore {/tex} {tex}p = 0{/tex} is not the solution of given equation.

NCERT Solutions for Class 7 Maths Exercise 4.1

**Question 3.**Solve the following equations by trial and error method:

(i) {tex}5p + 2 = 17{/tex}

(ii) {tex}3m – 14 = 4{/tex}

**Answer:**

(i) {tex}5p + 2 = 17{/tex}

Putting {tex}p = – 3{/tex} in L.H.S. {tex}5\left( { – 3} \right) + 2{/tex} = {tex} – 15 + 2 = – 13{/tex}

{tex}\because {/tex}{tex} – 13 \ne 17{/tex} Therefore, {tex}p = – 3{/tex} is not the solution.

Putting {tex}p = – 2{/tex} in L.H.S. {tex}5\left( { – 2} \right) + 2 = {/tex}{tex} – 10 + 2 = – 8{/tex}

{tex}\because {/tex}{tex} – 8 \ne 17{/tex} Therefore, {tex}p = – 2{/tex} is not the solution.

Putting {tex}p = – 1{/tex} in L.H.S. {tex}5\left( { – 1} \right) + 2 = {/tex}{tex} – 5 + 2 = – 3{/tex}

{tex}\because {/tex}{tex} – 3 \ne 17{/tex} Therefore, {tex}p = – 1{/tex} is not the solution.

Putting {tex}p = 0{/tex} in L.H.S. {tex}5\left( 0 \right) + 2 = {/tex}{tex}0 + 2 = 2{/tex}

{tex}\because {/tex}{tex}2 \ne 17{/tex} Therefore, {tex}p = 0{/tex} is not the solution.

Putting {tex}p = 1{/tex} in L.H.S. {tex}5\left( 1 \right) + 2 = {/tex}{tex}5 + 2 = 7{/tex}

{tex}\because {/tex}{tex}7 \ne 17{/tex} Therefore, {tex}p = 1{/tex} is not the solution.

Putting {tex}p = 2{/tex} in L.H.S. {tex}5\left( 2 \right) + 2 = {/tex}{tex}10 + 2 = 12{/tex}

{tex}\because {/tex}{tex}12 \ne 17{/tex} Therefore, {tex}p = 2{/tex} is not the solution.

Putting {tex}p = 3{/tex} in L.H.S. {tex}5\left( 3 \right) + 2 = {/tex}{tex}15 + 2 = 17{/tex}

{tex}\because {/tex}{tex}17 = 17{/tex} Therefore, {tex}p = 3{/tex} is the solution.

(ii) {tex}3m – 14 = 4{/tex}

Putting {tex}m = – 2{/tex} in L.H.S. {tex}3\left( { – 2} \right) – 14 = – 6 – 14 = – 20{/tex}

{tex}\because {/tex}{tex} – 20 \ne 4{/tex} Therefore, {tex}m = – 2{/tex} is not the solution.

Putting {tex}m = – 1{/tex} in L.H.S. {tex}3\left( { – 1} \right) – 14 = – 3 – 14 = – 17{/tex}

{tex}\because {/tex}{tex} – 17 \ne 4{/tex} Therefore, {tex}m = – 1{/tex} is not the solution.

Putting {tex}m = 0{/tex} in L.H.S. {tex}3\left( 0 \right) – 14 = 0 – 14 = – 14{/tex}

{tex}\because {/tex}{tex} – 14 \ne 4{/tex} Therefore, {tex}m = 0{/tex} is not the solution.

Putting {tex}m = 1{/tex} in L.H.S. {tex}3\left( 1 \right) – 14 = 3 – 14 = – 11{/tex}

{tex}\because {/tex}{tex} – 11 \ne 4{/tex} Therefore, {tex}m = 1{/tex} is not the solution.

Putting {tex}m = 2{/tex} in L.H.S. {tex}3\left( 2 \right) – 14 = 6 – 14 = – 8{/tex}

{tex}\because {/tex}{tex} – 8 \ne 4{/tex} Therefore, {tex}m = 2{/tex} is not the solution.

Putting {tex}m = 3{/tex} in L.H.S. {tex}3\left( 3 \right) – 14 = 9 – 14 = – 5{/tex}

{tex}\because {/tex}{tex} – 5 \ne 4{/tex} Therefore, {tex}m = 3{/tex} is not the solution.

Putting {tex}m = 4{/tex} in L.H.S. {tex}3\left( 4 \right) – 14 = 12 – 14 = – 2{/tex}

{tex}\because {/tex}{tex} – 2 \ne 4{/tex} Therefore, {tex}m = 4{/tex} is not the solution.

Putting {tex}m = 5{/tex} in L.H.S. {tex}3\left( 5 \right) – 14 = 15 – 14 = 1{/tex}

{tex}\because {/tex}{tex}1 \ne 4{/tex} Therefore, {tex}m = 5{/tex} is not the solution.

Putting {tex}m = 6{/tex} in L.H.S. {tex}3\left( 6 \right) – 14 = 18 – 14 = 4{/tex}

{tex}\because {/tex}{tex}4 = 4{/tex} Therefore, {tex}m = 6{/tex} is the solution.

NCERT Solutions for Class 7 Maths Exercise 4.1

**Question 4.**Write equations for the following statements:

(i) The sum of numbers {tex}x{/tex} and 4 is 9.

(ii) 2 subtracted from {tex}y{/tex} is 8.

(iii) Ten times {tex}a{/tex} is 70.

(iv) The number {tex}b{/tex} divided by 5 gives 6.

(v) Three-fourth of {tex}t{/tex} is 15.

(vi) Seven times {tex}m{/tex} plus 7 gets you 77.

(vii) One-fourth of a number {tex}x{/tex} minus 4 gives 4.

(viii) If you take away 6 from 6 times {tex}y,{/tex} you get 60.

(ix) If you add 3 to one-third of {tex}z,{/tex} you get 30.

**Answer:**

(i) {tex}x + 4 = 9{/tex}

(ii) {tex}y – 2 = 8{/tex}

(iii) {tex}10a = 70{/tex}

(iv) {tex}\frac{b}{5} = 6{/tex}

(v) {tex}\frac{3}{4}t = 15{/tex}

(vi) {tex}7m + 7 = 77{/tex}

(vii) {tex}\frac{x}{4} – 4 = 4{/tex}

(viii) {tex}6y – 6 = 60{/tex}

(ix) {tex}\frac{z}{3} + 3 = 30{/tex}

NCERT Solutions for Class 7 Maths Exercise 4.1

**Question 5. **Write the following equations in statement form:

(i) {tex}p + 4 = 15{/tex}

(ii) {tex}m – 7 = 3{/tex}

(iii) {tex}2m = 7{/tex}

(iv) {tex}\frac{m}{5} = 3{/tex}

(v) {tex}\frac{{3m}}{5} = 6{/tex}

(vi) {tex}3p + 4 = 25{/tex}

(vii) {tex}4p – 2 = 18{/tex}

(viii) {tex}\frac{p}{2} + 2 = 8{/tex}

**Answer:**

(i) The sum of numbers {tex}p{/tex} and 4 is 15.

(ii) 7 subtracted from {tex}m{/tex} is 3.

(iii) Two times {tex}m{/tex} is 7.

(iv) The number {tex}m{/tex} is divided by 5 gives 3.

(v) Three-fifth of the number {tex}m{/tex} is 6.

(vi) Three times {tex}p{/tex} plus 4 gets 25.

(vii) If you take away 2 from 4 times {tex}p,{/tex} you get 18.

(viii) If you added 2 to half is {tex}p,{/tex} you get 8.

NCERT Solutions for Class 7 Maths Exercise 4.1

**Question 6.**Set up an equation in the following cases:

(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. (Tale {tex}m{/tex} to be the number of Parmit’s marbles.)

(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. (Take Laxmi’s age to be {tex}y{/tex} years.)

(iii) The teacher tells the class that the highest marks obtained by a student in her class are twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to be {tex}l.{/tex} )

(iv) In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be {tex}b{/tex} in degrees. Remember that the sum of angles of a triangle is {tex}180^\circ .{/tex})

**Answer:**

(i) Let {tex}m{/tex} be the number of Parmit’s marbles.

{tex}\therefore {/tex} {tex}5m + 7 = 37{/tex}

(ii) Let the age of Laxmi be {tex}y{/tex} years.

{tex}\therefore {/tex} {tex}3y + 4 = 49{/tex}

(iii) Let the lowest score be {tex}l.{/tex}

{tex}\therefore {/tex} {tex}2l + 7 = 87{/tex}

(iv) Let the base angle of the isosceles triangle be {tex}b,{/tex} so vertex angle = {tex}2b.{/tex}

{tex}\therefore {/tex} {tex}2b + b + b = 180^\circ {/tex}{tex} \Rightarrow {/tex} {tex}4b = 180^\circ {/tex} [Angle sum property of a {tex}\Delta {/tex}]

## NCERT Solutions for Class 7 Maths Exercise 4.1

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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