Well-designed AP Board Solutions Class 7 Maths Chapter 4 Simple Equations Exercise 4.3 offers step-by-step explanations to help students understand problem-solving strategies.

## Simple Equations Class 7 Exercise 4.3 Solutions – 7th Class Maths 4.3 Exercise Solutions

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.

Solution:

Let the number required be x

Eight times of the number = 8x

When 4 is added to 8x, the sum = 8x + 4

According to the problem

8x + 4 = 60 ⇒ 8x = 60 – 4

8x = 56 ⇒ x = \(\frac { 56 }{ 8 }\) ⇒ x = 7

∴ The number required = 7

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

Solution:

Let the number required be x

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

According to the problem

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

\(\frac { x }{ 5 }\) = 7 ⇒ x = 7 × 5 ⇒ x = 35

∴ The number required = 35

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

Solution:

Let the number required be x

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

According to the problem

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

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

x = \(\frac{18 \times 4}{3}\) ⇒ x = 6 × 4 ⇒ x = 24

∴ The required number = 24

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

Solution:

Let the number required be x

Twice of the number = 2x

According to the problem

2x – 11 = 15 ⇒ 2x = 15 + 11

⇒ 2x = 26 ⇒ x = \(\frac { 26 }{ 2 }\) ⇒ x = 13

∴ The required number = 13

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

Solution:

Let the number required be x

Thrice of the number = 3x

According to the problem

50 – 3x = 8 ⇒ 50 – 8 = 3x

⇒ 3x = 42 ⇒ x = \(\frac { 42 }{ 3 }\) ⇒ x = 14

∴ The required number = 14

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

Solution:

Let the number required be x

When 19 is added to x then the sum = x + 19

According to the problem

\(\frac { x+19 }{ 5 }\) = 8 ⇒ x + 19 = 40

x = 40 – 19 ⇒ x = 21

∴ The required number = 21.

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

Solution:

Let the number required be x\(\frac { 5 }{ 2 }\) of a number = \(\frac { 5x }{ 2 }\)

According to the problem

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

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

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

Solution:

Let the lowest score be x

Twice the lowest score = 2x

Highest marks = 87

According to the problem

2x + 7 = 87

2x = 87 – 7 (Transposing 7 to RHS)

2x = 80

x = \(\frac { 80 }{ 2 }\) (divide both sides by 2)

x = 40

∴ Lowest score = 40

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°).

Solution:

Let the equal angle is Isosceles triangle = a

Vertex angle = 40°

We know that the sum of the angles of a triangle = 180°

40° + a° + a° = 180°

2a = 180° – 40°

(Transposing 40° to RHS)

2a = 140°

a = \(\frac{140^{\circ}}{2}\) (divide both sides by 2 )

a = 70°

∴ Base angles of the triangle = 70°

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?

Solution:

Let the score of Rahul = x

Score of Sachin = 2x

Two short of double century = 200 – 2 = 198

According to the problem

2x + x = 198

3x = 198

x = \(\frac{198}{3}\) (divide by 3 on both sides)

x = 66

∴ Rahul’s score = 66

Sachin’s score = 2 × 66 = 132

Question 3.

Solve the following :

i) 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?

Solution:

Let the number of marbles of Parmit has = x

5 times of marbles of Parmit = 5x

According to the problem

5x + 7 = 37

5x = 37 – 7 (Transposing 7 to RHS)

5x = 30

x = \(\frac{30}{5}\) (Divide both sides by 5)

x = 6

∴ There are 6 marbles with Parmit.

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

Solution:

Let the age of Laxmi be x years

3 times the age of Laxmi = 3x years

According to the problem

3x + 4 = 49

3x = 49 – 4 (Transposing 4 to RHS)

3x = 45

x = \(\frac{45}{3}\) (Divide both sides by 3 )

x = 15

∴ Age of Laxmi = 15.

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 ?

Solution:

Let the number of fruit trees be x

3 times the number = 3x

According to the problem

3x + 2 = 77

3x = 77 – 2 (Transposing 2 to RHS)

3x = 75 (Divide both sides by 3)

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

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 !

Solution:

Let the number = x

7 times the number = 7x

7 x is added to 50 , then sum = 7x + 50.

According to the problem

7x + 50 + 40 = 300

7x + 90 = 300

7x = 300 – 90 (Transposing 90 to RHS)

7x = 210

x = \(\frac { 210 }{ 7 }\) (Divide both sides by 7 )

x = 30

∴ The number required = 30