NCERT Solutions for Class 7 Maths Exercise 4.1

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NCERT solutions for Maths Simple Equations Download as PDF

NCERT Solutions for Class 7 Maths Exercise 4.1

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.EquationValueSay, 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.EquationValueSay, 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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