# SIMPLE EQUATIONS WORD PROBLEMS WORKSHEET

Simple Equations Word Problems Worksheet :

Worksheet given in this section will be much useful for the students who would like to practice solving word problems on simple equations.

## Simple Equations Word Problems Worksheet - Problems

Problem 1 :

The denominator of a fraction exceeds the numerator by 5. If 3 be added to both, the fraction becomes 3/4. Find the fraction.

Problem 2 :

If thrice of A's age 6 years ago be subtracted from twice his present age, the result would be equal to his present age. Find A's present age.

Problem 3 :

The fourth part of a number exceeds the sixth part by 4. Find the number.

Problem 4 :

The width of the rectangle is 2/3 of its length. If the perimeter of the rectangle is 80 cm. Find its area.

Problem 5 :

In a triangle, the second angle is 5° more than the first angle. And the third angle is three times of the first angle. Find the three angles of the triangle.

Problem 6 :

If a number of which the half is greater than 1/5 th of the number by 15, find the number.

Problem 7 :

The average age of three boys is 25 years and their ages are in the proportion 3 : 5 : 7. Find the age of the youngest boy.

Problem 8 :

The ratio of the no. of boys to the no. of girls in a school of 720 students is 3 : 5. If 18 new girls are admitted in the school, find how many new boys may be admitted so that the ratio of the no. of boys to the no. of girls may change to 2:3.

## Simple Equations Word Problems Worksheet - Solutions

Problem 1 :

The denominator of a fraction exceeds the numerator by 5. If 3 be added to both, the fraction becomes 3/4. Find the fraction.

Solution :

Let "x" be the numerator.

"The denominator of the fraction exceeds the numerator"

From the above information,

Fraction  =  x / (x + 5) ----------(1)

"If 3 be added to both, the fraction becomes 3 / 4"

From the above information, we have

(x+3) / (x + 5 + 3)  =  3 / 4

Simplify.

(x + 3) / (x + 8)  =  3/4

4(x + 3)  =  3(x + 8)

4x + 12  =  3x + 24

x  =  12

Plug x  = 12 in (1)

Fraction  =  12 / (12 + 5)

Fraction  =  12 / 17

Hence, the required fraction is 12 / 17.

Problem 2 :

If thrice of A's age 6 years ago be subtracted from twice his present age, the result would be equal to his present age. Find A's present age.

Solution :

Let "x" be A's present age.

A's age 6 years ago  =  x - 6

Thrice of A's age 6 years ago  =  3(x-6)

Twice his present age  =  2x

Given : Thrice of A's age 6 years ago be subtracted from twice his present age, the result would be equal to his present age.

So, we have

2x - 3(x - 6)  =  x

Simplify.

2x - 3x + 18  =  x

- x + 18  =  x

18  =  2x

Divide both sides by 2.

9  =  x

Hence, A's present age is 9 years.

Problem 3 :

The fourth part of a number exceeds the sixth part by 4. Find the number.

Solution :

Let "x" be the required number.

Fourth part of the number  =  x/4

Sixth part of the number  =  x/6

Given : The fourth part of a number exceeds the sixth part by 4.

x/4 - x/6  =  4

L.C.M of (4, 6) is 12.

(3x/12) - (2x/12)  =  4

Simplify.

(3x - 2x) / 12  =  4

x / 12  =  4

Multiply both sides by 12.

x  =  48

Hence, the required number is 48.

Problem 4 :

The width of the rectangle is 2/3 of its length. If the perimeter of the rectangle is 80 cm. Find its area.

Solution :

Let "x" be the length of the rectangle.

Then, width of the rectangle  is 2x / 3

Given : Perimeter is 80cm.

Perimeter  =  80 cm

⋅ (l + w)  =  80

Divide both sides by 2.

l + w  =  40

Plug l  =  x and w  =  2x / 3.

x + 2x / 3  =  40

Simplify.

(3x + 2x) / 3  =  40

5x / 3  =  40

Multiply both sides by 3/5.

x  =  24

The length is 24 cm.

Then, the width is

=  2x / 3

=  (2 ⋅ 24) / 3

=  16 cm

Formula to find the area of a rectangle is

=  l ⋅ w

Plug l  =  24 and w  =  16.

=  24 ⋅ 16

=  384

Hence, area of the rectangle is 384 square cm.

Problem 5 :

In a triangle, the second angle is 5° more than the first angle. And the third angle is three times of the first angle. Find the three angles of the triangle.

Solution :

Let x° be the first angle.

Then, we have

the second angle  =  x° + 5°

third angle  =  3 ⋅ x°

We know that the sum of three angle in any triangle is 180°.

x° + (x° + 5°) + (3 ⋅ x°)  =  180°

x + x + 5 + 3x  =  180

Simplify.

5x + 5  =  180

Subtract 5 from both sides.

5x  =  175

Divide both sides by 5.

x  =  35

The first angle is 35°.

The second angle is

=  35° + 5°

=  40°

The third angle is

=  3 ⋅ 45°

=  135°

Hence, the three angles of the triangle are 35°, 40° and 135°.

Problem 6 :

If a number of which the half is greater than 1/5 th of the number by 15, find the number.

Solution :

Let "x" be the required number.

Half of the number  =  x/2

1/5 the of the number  =  1/5 ⋅ x  =  x/5

Given : Half of a number is greater than 1/5 th of the number by 15.

So, we have

x/2 - x/5  =  15

L.C.M of (2 , 5) is 10.

Make each denominator as 10 sing multiplication.

5x/10 - 2x/10  =  15

(5x - 2x) / 10  =  15

3x / 10  =  15

Multiply both sides by 10/3.

x  =  50

Hence, the required number is 50.

Problem 7 :

The average age of three boys is 25 years and their ages are in the proportion 3 : 5 : 7. Find the age of the youngest boy.

Solution :

From the ratio 3 : 5 : 7, the ages of three boys are

3x, 5x and 7x.

Given : Average age of the three boys is 25 years.

So, we have

(3x + 5x + 7x) / 3  =  25

Simplify.

15x / 3  =  25

5x  =  25

Divide both sides by 5.

x  =  5

Then, ages of the three boys are

3x  =  3 ⋅ 5  =  15

5x  =  5 ⋅ 5  =  25

7x  =  7 ⋅ 5  =  35

Hence, the age of the youngest boy is 15 years.

Problem 8 :

The ratio of the no. of boys to the no. of girls in a school of 720 students is 3 : 5. If 18 new girls are admitted in the school, find how many new boys may be admitted so that the ratio of the no. of boys to the no. of girls may change to 2:3.

Solution :

Sum of the terms in the given ratio 3 : 5 is

=  3 + 5

=  8

So, no. of boys in the school is

=  3/8 of total students

=  3/8 ⋅ 720

=  270

No. of girls in the school is

=  5/8 of total students

=  5/8 ⋅ 720

=  450

Let "x" be the no. of new boys admitted in the school.

Given : No. of new girls admitted is 18.

no. of boys in the school  =  270 + x

no. of girls in the school  =  450 + 18  =  468

Given : The ratio after the new admission is 2 : 3.

So, we have

(270 + x) : 468  =  2 : 3

(270 + x) / 468  =  2 / 3

Simplify.

⋅ (270 + x)  =  2 ⋅ 468

810 + 3x  =  936

Subtract 810 from both sides.

3x  =  126

Divide both sides by 3.

x  =  42

Hence the no. of new boys admitted in the school is 42.

After having gone through the stuff given above, we hope that the students would have understood, how to solve word problems using simple equations.

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