NCERT Solutions Class 7 Maths Chapter 4 Simple Equations Ex 4.4

NCERT Solutions Class 7 Mathematics 
Chapter – 4 (Simple Equations)

The NCERT Solutions in English Language for Class 7 Mathematics Chapter – 4 Simple Equations Exercise 4.4 has been provided here to help the students in solving the questions from this exercise. 

Chapter : 4 Simple Equations

• NCERT Solution Class 7 Maths Exercise – 4.1
• NCERT Solution Class 7 Maths Exercise – 4.2
• NCERT Solution Class 7 Maths Exercise – 4.3

Exercise – 4.4 

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  of the numbers, the result is 23.

Solution –
(a) Add 4 to eight times a number; you get 60.
Let us assume the required number be x
Eight times a number = 8x
According to the Question
⇒ 8x + 4 = 60
By transposing 4 from LHS to RHS it becomes – 4
⇒ 8x = 60 – 4
⇒ 8x = 56
Divide both side by 8,
⇒ =
⇒ x = 7

(b) One-fifth of a number minus 4 gives 3.
Let us assume the required number be x
One-fifth of a number = x =
According to the Question
⇒ – 4 = 3
By transposing – 4 from LHS to RHS it becomes 4
⇒ = 3 + 4

⇒ = 7
Multiply both side by 5,
⇒ × 5 = 7 × 5
⇒ x = 35

(c) If I take three-fourths of a number and add 3 to it, I get 21.
Let us assume the required number be x
Three-fourths of a number = x
According to the Question
⇒ x + 3 = 21
By transposing 3 from LHS to RHS it becomes – 3

⇒ x = 21 – 3

⇒ x = 18
Multiply both side by 4,
⇒  × 4 = 18 × 4
⇒ 3x = 72
Divide both side by 3,
⇒ =
⇒ x = 24

(d) When I subtracted 11 from twice a number, the result was 15.
Let us assume the required number be x
Twice a number = 2x
According to the Question
⇒ 2x –11 = 15
By transposing -11 from LHS to RHS it becomes 11
⇒ 2x = 15 + 11
⇒ 2x = 26
Divide both side by 2,
⇒ =

⇒ x = 13

(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
Let us assume the required number be x
Thrice the number = 3x
According to the Question
⇒ 50 – 3x = 8
By transposing 50 from LHS to RHS it becomes – 50
⇒ – 3x = 8 – 50
⇒ -3x = – 42
Divide both side by -3,
⇒ =
⇒ x = 14

(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
Let us assume the required number be x
According to the Question
⇒ = 8
Multiply both side by 5,
⇒  × 5 = 8 × 5
⇒ x + 19 = 40
By transposing 19 from LHS to RHS it becomes – 19
⇒ x = 40 – 19
⇒ x = 21

(g) Anwar thinks of a number. If he takes away 7 from of the number, the result is 23.
Let us assume the required number be x
of the number = x
According to the Question
⇒ x – 7 = 23
By transposing -7 from LHS to RHS it becomes 7
⇒  x = 23 + 7

⇒ x = 30
Multiply both side by 2,
⇒  × 2 = 30 × 2
⇒ 5x = 60
Divide both the side by 5
⇒ =
⇒ x = 12

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?

Solution –

(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?
Let us assume the lowest score be x
According to the Question
The highest score is = 87
⇒ 2x + 7 = Highest score
⇒ 2x + 7 = 87
By transposing 7 from LHS to RHS it becomes -7
⇒ 2x = 87 – 7
⇒ 2x = 80
Divide both the side by 2
⇒ =
⇒ x = 40
Hence, the lowest score is 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°).
Let each base angle be x degrees.
According to the Question
We know that, the sum of angles of a triangle is 180^o
⇒ x + x + 40^o = 180^o
⇒ 2x + 40 = 180^o
By transposing 40 from LHS to RHS it becomes -40
⇒ 2x = 180 – 40
⇒ 2x = 140
Divide both the side by 2
⇒ =
⇒ b = 70^o
Hence, 70^o is the base angle of an isosceles triangle.

(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?
Let us assume Rahul’s score be x
According to the Question
Sachin scored = 2x
⇒ Rahul’s score + Sachin’s score = 200 – 2
⇒ x + 2x = 198
⇒ 3x = 198
Divide both the side by 3,
⇒ =
⇒ x = 66
So, Rahul’s score (x) = 66
And Sachin’s score is (2x) = 2 × 66 = 132

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) 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?
Let us assume number of Parmit’s marbles = x
According to the Question
Irfan has 7 marbles more than five times the marbles Parmit has
⇒ 5 × Number of Parmit’s marbles + 7 = Total number of marbles Irfan having
⇒ (5 × x) + 7 = 37
⇒ 5x + 7 = 37
By transposing 7 from LHS to RHS it becomes -7
⇒ 5x = 37 – 7
⇒ 5x = 30
Divide both the side by 5
⇒  =
⇒ x = 6
So, Permit has 6 marbles

(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age?
Let Laxmi’s age to be = x years old
According to the Question
Lakshmi’s father is 4 years older than three times of her age
⇒ 3 × Laxmi’s age + 4 = Age of Lakshmi’s father
⇒ (3 × x) + 4 = 49
⇒ 3x + 4 = 49
By transposing 4 from LHS to RHS it becomes -4
⇒ 3x = 49 – 4
⇒ 3x = 45
Divide both the side by 3
⇒  =
⇒ x = 15
So, Lakshmi’s age is 15 years.

(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?
Let the number of fruit tress be x.
According to the Question
3 × number of fruit trees + 2 = number of non-fruit trees
⇒ 3x + 2 = 77
By transposing 2 from LHS to RHS it becomes -2
⇒ 3x = 77 – 2
⇒ 3x = 75
Divide both the side by 3
⇒ =
⇒ x = 25
So, number of fruit tree was 25.

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 us assume the number be x.
Take me seven times over and add a fifty = 7x + 50
To reach a triple century you still need forty = (7x + 50) + 40 = 300
⇒ 7x + 50 + 40 = 300
⇒ 7x + 90 = 300
By transposing 90 from LHS to RHS it becomes -90
⇒ 7x = 300 – 90
⇒ 7x = 210
Divide both side by 7
⇒ =
⇒ x = 30
Hence the number is 30.