**Simple Equations**

## 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.

Let the given number be *x*

By the given condition of the problem

8*x* + 4 = 60

Transposing 4 to RHS

8*x* = 60 – 4

8*x* = 56

Transposing 8 to RHS

*x* = ^{56}/_{8}

*x* = 7

Ans: the given number is 7.

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

Let the given number be *x*

By the given condition of the problem

^{1}/_{5}* x* – 4 = 3

Transposing 4 to RHS

^{1}/_{5}* x* = 3 + 4

^{1}/_{5}* x* = 7

Transposing 5 to RHS

*x* = 7 X 5

*x* = 35

Ans: the given number is 35.

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

Let the number be *x*

By the given condition of the problem

^{3}/_{4}* x* + 3 = 21

Transposing 3 to RHS

^{3}/_{4}* x* = 21 – 3

^{3}/_{4}* x* = 18

Transposing ¾ to RHS

*x* = 18 X ^{4}/_{3}

*x* = ^{72}/_{3}

*x* = 24

Ans: the given number is 24.

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

Let the number be *x*

By the given condition of the problem

2*x* – 11 = 15

Transposing 11 to RHS

2*x* = 15 + 11

2*x* = 26

Transposing 2 to RHS

*x* = ^{26}/_{2}

*x* = 13

Ans: the given number is 13.

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

Let the number of notebooks be *x*

By the given condition of the problem

50 – 3*x* = 8

Transposing 50 to RHS

-3*x* = 8 – 50

-3*x* = -42

Transposing -3 to RHS

*x* = ^{-42}/_{-3}

*x* = 14

Ans: the number of notebooks with Munna are 14.

(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.

Let the number be *x*

By the given condition of the problem

(*x* + 19)/5 = 8

Transposing 5 to RHS

(*x* + 19) = 8 X 5

*x* +19 = 40

Transposing 19 to RHS

*x* = 40 – 19

*x* = 21

Ans: the given number is 21.

(g) Anwar thinks of a number. If he takes away 7 from 5/2 of the number, the result is 23.

Let the number be *x*

By the given condition of the problem

^{5}/_{2 }*x* – 7 = 23

Transposing 7 to RHS

^{5}/_{2 }*x* = 23 + 7

^{5}/_{2 }*x* = 30

Transposing ^{5}/_{2 } to RHS

*x* = 30 X ^{2}/_{5}

*x* = 60/5

*x* = 12^{ }

Ans: the given number is 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?

Let the lowest score be *x*

Highest score = 87

By the given condition of the problem

2 *x* + 7 = 87

Transposing 7 to RHS

2 *x* = 87 – 7

2 *x* = 80

Transposing 2 to RHS

*x* = ^{80}/_{2}

*x* = 40

Ans: 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 the base angle of an isosceles triangle be *x*^{0}

The vertex angle = 40^{0}

*The Sum of the three angles of a triangle is 180 ^{0} and base angles of isosceles triangles are equal.*

By the given condition of the problem

*x* + *x* + 40 = 180

2*x* + 40 = 180

Transposing 40 to RHS

2*x* = 180 – 40

2*x* = 140

Transposing 2 to RHS

*x* = ^{140}/_{2}

*x* = 70

Ans: The base angles are 70^{0} each.

(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 the runs scored by Rahul be *x*

Runs scored by Sachin = 2*x*

By the given condition of the problem

*x* + 2*x* = 200 – 2

*x* + 2*x* = 180

3*x* = 180

Transposing 3 to RHS

*x* = ^{180}/_{3}

*x* = 60

Transposing 2 to RHS

*x* = ^{140}/_{2}

*x* = 70

Ans: The runs scored by Rahul = 60.

The runs scored by Sachin = 2*x* = 2 X 60 = 120.

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?

Let the marbles with Parmit be *x*

The marbles with Irfan = 37

By the given condition of the problem

5*x* + 7 = 37

Transposing 7 to RHS

5*x* = 37 – 7

5*x* = 30

Transposing 5 to RHS

*x* = ^{30}/_{5}

*x* = 6

Ans: The marbles with Parmit are = 6.

(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 the age of Laxmi be *x* years

Age of Laxmi’s father = 49 years

By the given condition of the problem

3*x* + 4 = 49

Transposing 4 to RHS

3*x* = 49 – 4

3*x* = 45

Transposing 3 to RHS

*x* = ^{45}/_{3}

*x* = 15

Ans: The age of Laxmi 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 trees be *x*

the number of non-fruit trees = 77

By the given condition of the problem

3*x* + 2 = 77

Transposing 2 to RHS

3*x* = 77 – 2

3*x* = 75

Transposing 3 to RHS

*x* = ^{75}/_{3}

*x* = 25

Ans: The number of fruit trees = 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!

Ans: 30

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