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**NCERT solutions for Maths Simple Equations ****Download as PDF**

## NCERT Solutions for Class 7 Maths Simple Equations

**Class –VII Mathematics (Ex. 4.4)**

**Question 1.**Set up equations and solve them to find the unknown numbers in the following cases:

- Add 4 to eight times a number; you get 60.
- One-fifth of a number minus 4 gives 3.
- If I take three-fourth of a number and add 3 to it, I get 21.
- When I subtracted 11 from twice a number, the result was 15.
- Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
- Ibenhal thinks of a number. If she adds 19 to it divides the sum by 5, she will get 8.
- Answer thinks of a number. If he takes away 7 from {tex}\frac{5}{2}{/tex} of the number, the result is {tex}\frac{{11}}{2}.{/tex}

**Answer:**

(a) Let the number be {tex}x.{/tex}

According to the question, {tex}8x + 4 = 60{/tex}

{tex} \Rightarrow {/tex} {tex}8x = 60 – 4{/tex} {tex} \Rightarrow {/tex} {tex}8x = 56{/tex}

{tex} \Rightarrow {/tex} {tex}x = \frac{{56}}{8}{/tex} {tex} \Rightarrow {/tex} {tex}x = 7{/tex}

(b) Let the number be {tex}y.{/tex}

According to the question, {tex}\frac{y}{5} – 4 = 3{/tex}

{tex} \Rightarrow {/tex} {tex}\frac{y}{5} = 3 + 4{/tex} {tex} \Rightarrow {/tex} {tex}\frac{y}{5} = 7{/tex}

{tex} \Rightarrow {/tex} {tex}y = 7 \times 5{/tex} {tex} \Rightarrow {/tex} {tex}y = 35{/tex}

(c) Let the number be {tex}z.{/tex}

According to the question, {tex}\frac{3}{4}z + 3 = 21{/tex}

{tex} \Rightarrow {/tex} {tex}\frac{3}{4}z = 21 – 3{/tex} {tex} \Rightarrow {/tex} {tex}\frac{3}{4}z = 18{/tex} {tex} \Rightarrow {/tex} {tex}3z = 18 \times 4{/tex}

{tex} \Rightarrow {/tex} {tex}3z = 72{/tex} {tex} \Rightarrow {/tex} {tex}z = \frac{{72}}{3}{/tex} {tex} \Rightarrow {/tex} {tex}z = 24{/tex}

(d) Let the number be {tex}x.{/tex}

According to the question, {tex}2x – 11 = 15{/tex}

{tex} \Rightarrow {/tex} {tex}2x = 15 + 11{/tex} {tex} \Rightarrow {/tex} {tex}2x = 26{/tex}

{tex} \Rightarrow {/tex} {tex}x = \frac{{26}}{2}{/tex} {tex} \Rightarrow {/tex} {tex}x = 13{/tex}

(e) Let the number be {tex}m.{/tex}

According to the question, {tex}50 – 3m = 8{/tex}

{tex} \Rightarrow {/tex} {tex} – 3m = 8 – 50{/tex} {tex} \Rightarrow {/tex} {tex} – 3m = – 42{/tex}

{tex} \Rightarrow {/tex} {tex}m = \frac{{ – 42}}{{ – 3}}{/tex} {tex} \Rightarrow {/tex} {tex}m = 14{/tex}

(f) Let the number be {tex}n.{/tex}

According to the question, {tex}\frac{{n + 19}}{5} = 8{/tex}

{tex} \Rightarrow {/tex} {tex}n + 19 = 8 \times 5{/tex} {tex} \Rightarrow {/tex} {tex}n + 19 = 40{/tex}

{tex} \Rightarrow {/tex} {tex}n = 40 – 19{/tex} {tex} \Rightarrow {/tex} {tex}n = 21{/tex}

(g) Let the number be {tex}x.{/tex}

According to the question, {tex}\frac{5}{2}x – 7 = \frac{{11}}{2}{/tex}

{tex} \Rightarrow {/tex} {tex}\frac{5}{2}x = \frac{{11}}{2} + 7{/tex} {tex} \Rightarrow {/tex} {tex}\frac{5}{2}x = \frac{{11 + 14}}{2}{/tex}

{tex} \Rightarrow {/tex} {tex}\frac{5}{2}x = \frac{{25}}{2}{/tex} {tex} \Rightarrow {/tex} {tex}5x = \frac{{25 \times 2}}{2}{/tex} {tex} \Rightarrow {/tex} {tex}5x = 25{/tex}

{tex} \Rightarrow {/tex} {tex}x = \frac{{25}}{5}{/tex} {tex} \Rightarrow {/tex} {tex}x = 5{/tex}

NCERT Solutions for Class 7 Maths Exercise 4.4

**Question 2.**Solve the following:

- 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. What is the lowest score?
- In an isosceles triangle, the base angles are equal. The vertex angle is {tex}40^\circ .{/tex} What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is {tex}180^\circ .{/tex})
- Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?

**Answer:**

(a) Let the lowest marks be {tex}y.{/tex}

According to the question, {tex}2y + 7 = 87{/tex}

{tex} \Rightarrow {/tex} {tex}2y = 87 – 7{/tex} {tex} \Rightarrow {/tex} {tex}2y = 80{/tex} {tex} \Rightarrow {/tex} {tex}y = \frac{{80}}{2}{/tex}

{tex} \Rightarrow {/tex} {tex}y = 40{/tex}

Thus, the lowest score is 40.

(b) Let the base angle of the triangle be {tex}b.{/tex}

Given, {tex}a = 40^\circ ,b = c{/tex}

Since, {tex}a + b + c = 180^\circ {/tex} [Angle sum property of a triangle]

{tex} \Rightarrow {/tex} {tex}40^\circ + b + b = 180^\circ {/tex}

{tex} \Rightarrow {/tex} {tex}40^\circ + 2b = 180^\circ {/tex}

{tex} \Rightarrow {/tex} {tex}2b = 180^\circ – 40^\circ {/tex} {tex} \Rightarrow {/tex} {tex}2b = 140^\circ {/tex}

{tex} \Rightarrow {/tex} {tex}b = \frac{{140^\circ }}{2}{/tex} {tex} \Rightarrow {/tex} {tex}b = 70^\circ {/tex}

Thus, the base angles of the isosceles triangle are {tex}70^\circ {/tex} each.

(c) Let the score of Rahul be {tex}x{/tex} runs and Sachin’s score is {tex}2x.{/tex}

According to the question, {tex}x + 2x = 198{/tex}

{tex} \Rightarrow {/tex} {tex}3x = 198{/tex} {tex} \Rightarrow {/tex} {tex}x = \frac{{198}}{3}{/tex}

{tex} \Rightarrow {/tex} {tex}x = 66{/tex}

Thus, Rahul’s score = 66 runs

And Sachin’s score = 2 x 66 = 132 runs.

NCERT Solutions for Class 7 Maths Exercise 4.4

**Question 3.**Solve the following:

- Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?
- Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age?
- People of Sundergram planted a total of 102 trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruit trees planted?

**Answer:**

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

According to the question, {tex}5m + 7 = 37{/tex}

{tex} \Rightarrow {/tex} {tex}5m = 37 – 7{/tex} {tex} \Rightarrow {/tex} {tex}5m = 30{/tex}

{tex} \Rightarrow {/tex} {tex}m = \frac{{30}}{5}{/tex} {tex} \Rightarrow {/tex} {tex}m = 6{/tex}

Thus, Parmit has 6 marbles.

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

Then her father’s age = {tex}\left( {3y + 4} \right){/tex} years

According to question, {tex}3y + 4 = 49{/tex}

{tex} \Rightarrow {/tex} {tex}3y = 49 – 4{/tex} {tex} \Rightarrow {/tex} {tex}3y = 45{/tex}

{tex} \Rightarrow {/tex} {tex}y = \frac{{45}}{3}{/tex} {tex} \Rightarrow {/tex} {tex}y = 15{/tex}

Thus, the age of Laxmi is 15 years.

(iii) Let the number of fruit trees be {tex}t.{/tex}

Then the number of non-fruits tree = {tex}3t + 2{/tex}

According to the question, {tex}t + 3t + 2 = 102{/tex}

{tex} \Rightarrow {/tex} {tex}4t + 2 = 102{/tex} {tex} \Rightarrow {/tex} {tex}4t = 102 – 2{/tex}

{tex} \Rightarrow {/tex} {tex}4t = 100{/tex} {tex} \Rightarrow {/tex} {tex}t = \frac{{100}}{4}{/tex}

{tex} \Rightarrow {/tex} {tex}t = 25{/tex}

Thus, the number of fruit trees are 25.

NCERT Solutions for Class 7 Maths Exercise 4.4

**Question 4.**Solve the following riddle:

I am a number, Tell my identity!

Take me seven times over, And add a fifty!

To reach a triple century, You still need forty!

**Answer:**

Let the number be {tex}n.{/tex}

According to the question, {tex}7n + 50 + 40 = 300{/tex}

{tex} \Rightarrow {/tex} {tex}7n + 90 = 300{/tex} {tex} \Rightarrow {/tex} {tex}7n = 300 – 90{/tex}

{tex} \Rightarrow {/tex} {tex}7n = 210{/tex} {tex} \Rightarrow {/tex} {tex}n = \frac{{210}}{7}{/tex}

{tex} \Rightarrow {/tex} {tex}n = 30{/tex}

Thus, the required number is 30.

## NCERT Solutions for Class 7 Maths Exercise 4.4

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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