# EXERCISE – 4.4

Q1. Set up equation 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 us assume the required number be x

Eight times a number = 8x

The given above statements can be written in the equation form as,

= 8x + 4 = 60

By transferring 4 from LHS TO RHS it becomes -4

= 8x = 60 – 4

= 8x = 56

Divide both side by 8.

Then we get,

= (8x/8) = 56/8

= x = 7

(b) One – fifth of

a number minus 4 gives 3.

SOLUTION:-

= Let us assume the required number be x.

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

The given above statements can be written in the equation form as,

= (x/5) – 4 = 3

By transferring – 4 from LHS to RHS it becomes 4

= x/5 = 3 + 4

= x/5 = 7

Multiply both sides by 5,

Then we get,

= (x/5) x 5 = 7 x 5

= x = 35

(c) If 1 take three – fourths of a number and add 3 to it, I get 21.

SOLUTION:-

Let us assume the required number be x

Three – fourths of a number = (3/4) x

The given above statements can be written by in the equation form as,

= (3/4)x + 3 = 21

By transferring 3 from LHS to RHS it becomes -3

= (3/4)x = 21 – 3

= (3/4) x =18

Multiply both sides by 4.

Then we get,

= (3x/4) x 4 = 18 x 4

= 3x = 72

Then,

Divide 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 us assume the required number be x

Twice a number = 2x

The given above statements can be written in the equation form as,

= 2x – 11 = 15

By transposing -11 from LHS to RHS it becomes 11

= 2x = 15 + 11

= 2x = 26

Then,

Divide 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 us assume the required number be x

Thrice the number = 3x

The given above statements can be written in the equation form as,

= 50 – 3x = 8

By transposing 50 from LHS to RHS it becomes -50

= -3x = 8 – 50

= -3x = -42

Then,

Divide 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 us assume the required number be x

The given above statements can be written in the equation form as,

= ( x+19)/5 = 8

Multiply both sides by 5,

= (( x+190)/5) x 5 = 8 x 5

= x + 19 = 40

Then,

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

SOLUTION:-

Let us assume the required number be x,

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

The given above statements can be written in the equation form as,

= (5/2)x – 7 = 23

By transposing -7 from LHS to RHS it becomes 7

= (5/2) x = 23 + 7

= (5/2) x = 30

Multiply both sides by 2,

= ((5/2 x) x 2 = 30 x 2

= 5x = 60

= x = 12

Q2. Solve the following:

(i) 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 us assume the lowest score be x

From the question it is given that,

The highest score is =  87

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

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

The given above statements can be written in the equation form as ,

Then,

= 2x + 7 = Highest score

= 2x + 7 = 87

By transposing 7 from LHS to RHS it becomes -7

= 2x = 87 – 7

= 2x = 80

= x = 80/2

= x = 40

Hence, the lowest score is 40

(B) In an isosceles triangle, the base angles are equal. The vertex angle is 400. What are the base angles of the triangle ?( Remember , the sum of three angles of a triangle is 180o).

SOLUTION:-

From the question it is given that,

We know that, the sum of angles of a triangle is 180o

Let base angle be b

Then,

= b + b + 400 = 1800

= 2b +400 = 1800

= 2b = 1800 – 400

= 2b = 1400

= b = 1400/2

= b = 700

Hence 700 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?

SOLUTION:-

Let us assume Rahul score be x

Then,

Sachin scored twice as many runs as Rahul is 2x

Together, their runs fell two short a double century.

= Rahul score + Sachin score = 200 – 2

= x + 2x = 198

= 3x = 198

Divide both the side by 3,

= 3x/3 = 198/3

= x= 66

So, Rahul score is 66

And Sachin score is 2x = 2 x 66 = 132

Q3. Solve the following:

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

SOLUTION:-

Let us assume number of Parmit marbles = m

From the question it is given that,

Then,

Ifran  has 7 marbles more than five times the marbles Parmit has

= 5 x Number of Parmit marbles + 7 = Total number of marbles Ifran having

= ( 5 x m) + 7 = 37

= 5m + 7  = 37

By transposing 7 from LHS to RHS it becomes -7

= 5m = 37 – 7

= 5m = 30

Divide both the side by 5

= 5m/5 = 30/5

= m = 6

So, Parmit has 6 marbles

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

SOLUTION:-

Let Laxmi age to be = y years old

From the question it is given that,

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

= 3 x Laxmi age + 4 = Age of Lakshmi father

= (3 x y) + 4  = 49

= 3y + 4 = 49

By transposing 4 from LHS to RHS it becomes -4

= 3y = 49 – 4

= 3y = 45

Divide both theside by 3

= 3y/3 = 45/3

= y = 15

So, Lakshmi age is 15 years.

(iii) People of sundargram planted trees in the village garden. Some of the trees were fruits trees. The number of non – fruits trees were two more than three times the number if fruit trees. What was the number of fruit trees planted if he number of non – fruits trees planted was 77 ?

SOLUTION:-

Let the number of fruit trees be f.

From the question it is given that,

3 x number of fruit trees + 2 = number of non – fruits trees

= 3f + 2 = 77

By transposing 2 from LHS to RHS it becomes -2

= 3f = 77 – 2

= 3f = 75

= f = 25

SO, NUMBER OF FRUIT TREE WAS 25.

Q4. Solve the following riddle:

I am a number,

Tell my identity!

Take me seven times over

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

= 7x/7 = 210 /7

= x = 30