# NCERT Solution class 7 Mathematics EXERCISE- 4.4

# CHAPTER – 4

# 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 40 ^{0}. What are the base angles of the triangle ?( Remember , the sum of three angles of a triangle is 180^{o}).**

**SOLUTION:-**

From the question it is given that,

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

Let base angle be b

Then,

= b + b + 40^{0 }= 180^{0}

= 2b +40^{0} = 180^{0}

= 2b = 180^{0} – 40^{0}

= 2b = 140^{0}

= b = 140^{0}/2

= b = 70^{0}

Hence 70^{0} 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**

**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

= 7x/7 = 210 /7

= x = 30