Mathematics – Class 7 – Chapter 4 – Simple Equation – Exercise 4.4 – NCERT Exercise Solution

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 required number be x

Eight times a number = 8x

According to above statement we can  write in the equation form as,

8x + 4 = 60

Transposing (4) from LHS to RHS it becomes (– 4)

8x = 60 – 4

8x = 56

Dividing both side by 8,

We get,

(8x/8) = 56/8

x = 7

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

Solution:

Let the required number be x

One-fifth of a number = (1/5) x

                                        = x/5

According to the above statement, we can write in the equation form as,

(x/5) – 4 = 3

Transposing (– 4) from LHS to RHS it becomes (4)

x/5 = 3 + 4

x/5 = 7

Multiplying both sides by 5,

we get,

(x/5) × 5 = 7 × 5

x = 35

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

Solution:

Let the required number to be x

Three-fourths of a number = (3/4) x

According above statement can be written in the equation form as,

(3/4) x + 3 = 21

By transposing 3 from LHS to RHS it becomes – 3

(3/4) x = 21 – 3

(3/4) x = 18

Multiplying both side by 4,

We get,

(3x/4) × 4 = 18 × 4

3x = 72

Dividing both side by 3,

(3x/3) = 72/3

x = 24

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

Solution:

Let the required number be x

Twice a number = 2x

According to above statement we can write in the equation form as,

2x –11 = 15

Transposing (-11) from LHS to RHS it becomes (11)

2x = 15 + 11

2x = 26

Dividing both side by 2,

(2x/2) = 26/2

x = 13

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

Solution:

Let the required number be x

Thrice the number = 3x

According to the above statement, we can write in the equation form as,

50 – 3x = 8

Transposing (50) from LHS to RHS it becomes (– 50)

– 3x = 8 – 50

-3x = – 42

Dividing both side by (-3)

(-3x/-3) = – 42/-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.

Solution:

Let the required number to be x

According to above statement we can write in the equation form as,

(x + 19)/5 = 8

Multiplying both side by 5,

((x + 19)/5) × 5 = 8 × 5

x + 19 = 40

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 5/2 of the number, the result is 23.

Solution:

Let the required number be x

5/2 of the number = (5/2) x

According to above statement we can write in the equation form as,

(5/2) x – 7 = 23

Transposing -7 from LHS to RHS it becomes 7

(5/2) x = 23 + 7

(5/2) x = 30

Multiplying both side by 2,

((5/2) x) × 2 = 30 × 2

5x = 60

Dividing both the side by 5

5x/5 = 60/5

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?

Solution:

Let the lowest score be x

According to the question, it is given,

The highest score is = 87

The highest marks obtained by a student in her class is twice the lowest marks plus 7= 2x + 7

5/2 of the number = (5/2) x

According to above statement we can write in the equation form as,

2x + 7 = Highest score

2x + 7 = 87

Transposing (7) from LHS to RHS it becomes (-7)

2x = 87 – 7

2x = 80

Dividing both the side by 2

2x/2 = 80/2

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°).

Solution:

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

According to the question it is given,

Vertex angle = 40°

The base angle is equal

Let the base angle be b

b + b + 40° = 180°

2b + 40°= 180°

Transposing (40) from LHS to RHS it becomes (-40)

2b = 180 – 40

2b = 140

Dividing both the side by 2

2b/2 = 140/2

b = 70o

Hence, the base angle of an isosceles triangle is 70o

(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  Rahul’s score be x

Sachin scored twice as many runs as Rahul is 2x

Together, their runs fell two short of a double century,

Rahul’s score + Sachin’s score = 200 – 2

x + 2x = 198

3x = 198

Dividing both the side by 3,

3x/3 = 198/3

x = 66

So, Rahul’s score is 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?

Solution:

Let the number of Parmit’s marbles = m

According to the question, it is given,

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 × m) + 7 = 37

5m + 7 = 37

Transposing (7) from LHS to RHS it becomes (-7)

5m = 37 – 7

5m = 30

Dividing both the side by 5

5m/5 = 30/5

m = 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?

Solution:

Let Laxmi’s age be = ‘y’ years old

According to the question, it is given,

Lakshmi’s father is 4 years older than three times of her age

3 × Laxmi’s age + 4 = Age of Lakshmi’s father

(3 × y) + 4 = 49

3y + 4 = 49

Transposing (4) from LHS to RHS it becomes (-4)

3y = 49 – 4

3y = 45

Dividing both the side by 3

3y/3 = 45/3

y = 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?

Solution:

Let the number of fruit trees be f.

According to the question it is given that,

3 × number of fruit trees + 2 = number of non-fruit trees

3f + 2 = 77

Transposing 2 from LHS to RHS it becomes -2

3f = 77 – 2

3f = 75

Dividing both the side by 3

3f/3 = 75/3

f = 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 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

Transposing (90) from LHS to RHS it becomes (-90)

7x = 300 – 90

7x = 210

Dividing both side by 7

7x/7 = 210/7

x = 30

Hence the required number is 30.

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