**Word Problems on Simple Equations 1 :**

This is the continuity of our web page word problems on simple equations.

**Example 1 :**

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.

**Example 2 :**

Three persons A, B and C together have $51. B has $4 less than A. C has got $5 less than A. Find the money that A, B and C have.

**Solution :**

Let "x" be the money had by A,

Then, we have

A = x

B has $4 less than A ------> B = x - 4

C has got $5 less than A -------> C = x - 5

**Given : **A, B and C together have $51

So, we have

A + B + C = 51

x + (x - 4) + (x - 5) = 51

x + x - 4 + x - 5 = 51

Simplify.

3x - 9 = 51

Add 9 to both sides.

3x = 60

Divide both sides by 3.

x = 20

Then,

x - 4 = 20 - 4 = 16

x - 5 = 20 - 5 = 15

Hence, A, B and C have $20, $16 and $15 respectively.

**Example 3 :**

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.

**Example 4 :**

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.

After the above new admissions,

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.

3 ⋅ (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.

**Example 5 :**

The monthly incomes of two persons are in the ratio 4:5 and their monthly expenditures are in the ratio 7:9. If each saves $50 per month, find the monthly incomes of the two persons.

**Solution :**

From the income ratio 4 : 5,

Income of the 1st person = 4x

Income of the 2nd person = 5x

Using the fact

Expenditure = Income - Savings,

we can find the expenditure of the two persons as given below.

Then, expenditure of the 1st person is

= 4x - 50

Expenditure of the 2nd person is

= 5x - 50

**Given :**Expenditure ratio is 7 : 9

So, we have

(4x - 50) : (5x - 50) = 7 : 9

(4x - 50) / (5x - 50) = 7 / 9

Simplify.

9 ⋅ (4x - 50) = 7 ⋅ (5x - 50)

36x - 450 = 35x - 350

x = 100

Then, we have

4x = 4 ⋅ 100 = 400

5x = 5 ⋅ 100 = 500

Hence, the monthly incomes of the two persons are $400 and $500.

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