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

**Ex 4.4 Class 7 Maths 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 5/2 of the numbers, the result is 23.**

**Solution:**

(a) Let the required number be x.

Step I: 8x + 4

Step II: 8x + 4 = 60 is the required equation

Solving the equation, we have

8x + 4 = 60

⇒ 8x = 60 – 4 (Transposing 4 to RHS)

⇒ 8x = 56

⇒ 8 x/8 = 56/8 (Dividing both sides by 8)

⇒ x = 7

Thus, x – 7 is the required unknown number.

(b) Let the required number be x.

Step I: 1/5 x – 4

Step II: 1/5 x – 4 = 3 is the required equation. 5

Solving the equation, we get

1/5 x – 4 = 3

⇒ 1/5 x = 4 + 3 (Transposing 4 to RHS)

⇒ 1/5 x = 7

⇒ 1/ 5 x × 5 = 7 × 5 (Multiplying both sides by 5)

⇒ x = 35 is the required unknown number,

(c) Let the required number be x.

Step I: 3 /4 x + 3

Step II: 3 /4 x + 3 = 21 is the required equation.

Solving the equation, we have

⇒ x = 24 is the required unknown number.

(d) Let the required unknown number be x.

Step I: 2x – 11

Step II: 2x -11 = 15 is the required equations.

Solving the equation, we have

2x – 11= 15

⇒ 2x = 15 + 13 (Transposing 11 to RHS)

⇒ 2x = 28

⇒ 2 x/2 = 28/2 (Dividing both sides by 2)

⇒ x = 14 is the required unknown number,

(e) Let the required number be x.

Step I: 50 – 3x

Step II: 50 – 3x = 8 is the required equations.

Solving the equation, we have

50 – 3x = 8

⇒ -3x = 8 – 50 (Transposing 50 to RHS)

⇒ -3x = -42

⇒ -3 x/-3 = -42/-3 (Dividing both sides by -3)

⇒ x = 14 is the required unknown number.

(f) Let the required number be x.

Step I: x + 19

Step II: x + 19/5

Step III: x + 19/5 = 8 is the required equation.

Solving the equation, we have

x + 19/5 = 8

⇒ x + 19/5 × 5 = 8 × 5(Multiplying both sides by 5)

⇒ x + 19 = 40

⇒ x = 40 – 19 (Transposing 19 to RHS)

∴ x = 21 is the required unknown number.

(g) Let the required number be x.

Step I: 5/2x – 7

Step II: 5/5 – 7 = 23 is the required equation.

Solving the equation, we have

⇒ x = 12 is the required unknown number.

**Ex 4.4 Class 7 Maths 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 angle 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?**

**Solution:**

(a) Let the lowest score be x.

Step I: Highest marks obtained = 2x + 7

Step II: 2x + 7 = 87 is the required equation. Solving the equation, we have

2x + 7 = 87

⇒ 2x = 87 – 7 (Transposing 7 to RHS)

⇒ 2x = 80

⇒ 2 x/2 = 80/2 (Dividing both sides by 2)

⇒ x = 40 is the required lowest marks.

(b) Let each base angle be x degrees.

Step I: Sum of all angles of the triangle (x + x + 40) degrees.

Step II: x + x + 40 = 180°

⇒ 2x + 40° = 180°

Solving the equation, we have

2x + 40° = 180°

2x = 180° – 40° (Transposing 40° to RHS)

2x = 140°

⇒ 2 x/2 =140°/2 (Dividing both sides by 2)

⇒ x = 70°

Thus the required each base angle = 70°

(c) Let the runs scored by Rahul = x

Runs scored by Sachin = 2x

Step I: x + 2x = 3x

Step II: 3x + 2 = 200

Solving the equation, we have

3x + 2 = 200

⇒ 3x = 200 – 2 (Transposing 2 to RHS)

⇒ 3x = 198

⇒ 3 x/ 3 = 198/3 (Dividing both sides by 3)

⇒ x = 66

Thus, the runs scored by Rahul is 66 and the runs scored by Sachin = 2 × 66 = 132

**Ex 4.4 Class 7 Maths 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?**

**(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 a 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:**

(i) Let the number of marbles with Parmit be

Step I: Number of marbles that Irfan has = 5x + 7

Step II: 5x + 7 = 37 Solving the equation, we have 5x + 7 = 37

⇒ 5x = 37 – 7 (Transposing 7 to RHS)

⇒ 5x = 30

⇒5 x/5 = 30/5 (Dividing both sides by 5)

⇒ x = 6

Thus, the required number of marbles = 6.

(ii) Let Laxmi’s age be x years.

Step I: Father’s age = 3x + 4

Step II: 3x + 4 = 49

Solving the equation, we get

3x + 4-= 49

⇒ 3x = 49 – 4 (Transposing to RHS)

⇒ 3x = 45

⇒ 3 x/3 = 45/3 (Dividing both sides by 3)

⇒ x = 15

Thus, the age of Laxmi = 15 years

(iii) Let the number of planted fruit tree be x.

Step I: Number of non-fruit trees = 3x + 2

Step II: 3x + 2 = 77

Solving the equation, we have

3x + 2 = 77

⇒ 3x = 77 – 2 (Transposing 2 to RHS)

⇒ 3x = 75

⇒ 3 x/ 3 = 75/3 (Dividing both sides by 3)

⇒ x = 25

Thus, the required number of fruit tree planted = 25

**Ex 4.4 Class 7 Maths 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:**

Suppose my identity number is x.

Step I: 7 + 50

Step II: 7x + 50 + 40 = 300

or 7x + 90 = 300

Solving the equation, we have

7x + 90 = 300

⇒ 7x = 300 – 90 (Transforming 90 to RHS)

⇒ 7x = 210

⇒ 7x/7 = 210/7 (Dividing both sides by 7)

⇒ x = 30

Thus, my identity is 30.