These NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.4 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.4

Question 1.

Set up equations and solve them to find the unknown numbers in the following cases:

(a) Add 4 to eight times a number; you get 60.

(b) One-fifth of a number minus 4 gives 3.

(c) If I take three-fourths of a number and add 3 to it, I get 21.

(d) When I subtracted 11 from twice a number, the result was 15.

(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.

(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.

(g) Anwar thinks of a number. If he takes away 7 from \(\frac{5}{2}\) of the number, the result is 23.

Answer:

(a) Let the required number be x

∴ Eight times of the number = 8x

According to the given question, we get

8x + 4 =60

Transposing 4 to R.H.S

8x = 60 – 4

8x = 56

Dividing both sides by 8, we get

\(\frac{8 x}{8}=\frac{56}{8}\)

x = 7

∴ The required number = 7

(b) Let the number be x

one-fifth of the number = \(\frac{1}{5}\) x

According to the given question, we get

\(\frac{1}{5}\) x – 4 = 3

Transposing -4 to R.H.S

\(\frac{1}{5}\)x = 3 + 4

\(\frac{1}{5}\)x = 7

Multiplying both sides by 5, we get 1

\(\frac{1}{5}\)x × 5 = 7 × 5

x = 35

∴ The required number = 35

(c) Let the number be x

Three-fourths of the number = \(\frac{3}{4}\) x

According to the given question, we get

\(\frac{3}{4}\) x + 3 = 21

Transposing 3 to R.H.S

\(\frac{3}{4}\) x = 21 – 3

\(\frac{3}{4}\) x = 18

Multiplying both sides by 4, we get

\(\frac{3}{4}\) x × 4 = 18 × 4

3x = 72

Dividing both sides by 3, we get

\(\frac{3 x}{3}=\frac{72}{3}\)

x = 24

∴ Thus, the required number = 24

(d) Let the number be x

Twice a number = 2x

According to the given question

2x – 11 = 15

Transposing -11 to R.H.S

2x = 15 + 11

2x = 26

Dividing both sides by 2, we get

\(\frac{2 \mathrm{x}}{2}=\frac{26}{2}\)

x = 13

∴ The required number 13.

(e) Let the number of notebooks with Munna be ‘x’

Thrice the number of notebooks = 3x

According to the given question, we get

50 – 3x = 8

Transposing 50 to R.H.S

-3x = 8 – 50

-3x = -42

Dividing both sides by (-3), we get

\(\frac{-3 x}{-3}=\frac{-42}{-3}\)

x = 14

∴ The required number of notebooks is 14.

(f) Let the number be x

According to the given question, we get

\(\frac{x+19}{5}\) = 8

Multiplying both sides by 5, we get

\(\frac{x+19}{5}\) × 5 = 8 × 5

x + 19 = 40

Transposing 19 to R.H.S

x = 40 – 19

x = 21

∴ The required number = 21

(g) Let the number be x

\(\frac{5}{2}\) of the number = \(\frac{5}{2}\) x

According to the given question, we get

\(\frac{5}{2}\) x – 7 = 23

Transposing -7 to R.H.S

\(\frac{5}{2}\)x = 23 + 7

\(\frac{5}{2}\)x = 30

Multiplying both sides by 2, we get

\(\frac{5}{2}\) x × 2 = 30 × 2

5x = 60

Dividing both sides by 5, we get

\(\frac{5 x}{5}=\frac{60}{5}\)

x = 12

The required number = 12.

Question 2.

Solve the following:

(a) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. What is the lowest score?

(b) In an isosceles triangle, the base angles are equal. The vertex angle is 40°. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180°).

(c) 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 scored be x

Twice the lowest marks = 2x

According to the given question, we get

2x + 7 = 87

Transposing 7 to R.H.S

2x = 87 – 7

2x = 80

Dividing both sides by 2, we get

\(\frac{2 x}{2}=\frac{80}{2}\)

x = 40

∴ The lowest marks = 40

(b) Let the base angle be x°

∴ The other base angle = x°

The vertex angle = 40°

∴ Sum of the angles = x + x + 40

= 2x + 40

According to the given question, we get

2x + 40 = 180°

(Sum of the three angles of a triangle)

2x = 180° – 40°

2x = 140°

Dividing both sides by 2, we get

\(\frac{2 x}{2}=\frac{140}{2}\)

x = 70°

∴ The base angle of the triangle

= 70°

(c) Let the runs scored by Rahul be x

∴ Sachins runs = 2x

According to the given question, we get

x + 2x =2(100) -2

3x = 200 – 2

3x = 198

Dividing both sides by 3, we get

\(\frac{3 x}{x}=\frac{198}{3}\)

x = 66

∴ Rahul score = 66 runs and Sachins score = 2 × 66 = 132 runs

Question 3.

Solve the following:

(i) Ifran says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?

(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age?

(iii) People of Sundargram planted 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 if the number of non-fruit trees planted was 77?

Answer:

(i) Let the number of marbles with Parmit be x

Number of marbles with Irfan = 37

According to the given question, we get

5x + 7 = 37

Transposing 7 to R.H.S

5x = 37 – 7

5x = 30

Dividing both sides by 5, we get

\(\frac{5 x}{5}=\frac{30}{5}\)

\(\frac{5 x}{5}=\frac{30}{5}\) ⇒ x = 6

∴ Parmit has 6 marbles.

(ii) Let Laxmi’s age be x years

Three times Laxmi s age = 3x

According to the given question, we get

3x + 4 = 49

Transposing 4 to R.H.S

3x = 49 – 4

3x = 45

Dividing both sides by 3, we get

\(\frac{3 x}{3}=\frac{45}{3}\)

x = 15

∴ Laxmi’s age= 15 years

(iii) Let the number of fruit trees be x

Three times fruit trees = 3x

Number of non-fruit trees = 2 + 3x

According to the given question, we get

2 + 3 x = 77

Transposing 2 to R.H.S

3x = 77 – 2

3x = 75

Dividing both sides by 3, we get

\(\frac{3 x}{3}=\frac{75}{3}\)

⇒ x = 25

∴ The number of fruit trees = 25

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 required number be x

7 times the number = 7x

each triple century = 3 × 100 = 300

According to the given riddle, we get

7x + 50 = 300 – 40

7x + 50 = 260

Transposing 50 to R.H.S

7x = 260 – 50

7x = 210

Dividing both sides by 7, we get

\(\frac{7 \mathrm{x}}{7}=\frac{210}{7}\)

x = 30

∴ The required number is 30.