**Word Problems on Linear Equations in One Variable : **

In this section, we are going to learn, how to solve word problems on linear equations in one variable.

**Example 1 : **

Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

**Solution : **

Let x be one of the two numbers.

Then, the other number is (x + 15).

**Given :** Sum of two numbers is 95.

So, we have

x + (x + 15) = 95

Simplify.

x + x + 15 = 95

2x + 15 = 95

Subtract 15 from each side.

2x + 15 - 15 = 95 - 15

2x = 80

Divider each side by 2.

2x/2 = 80/2

x = 40

x + 15 = 40 + 15

x + 15 = 55

Hence, the two numbers are 40 and 55.

**Example 2 :**

Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers ?

**Solution : **

From the ratio 5 : 3, the two numbers can be assumed as

5x and 3x

**Given :** The two numbers differ by 18.

So, we have

5x - 3x = 18

2x = 18

Divide each side by 2.

2x/2 = 18/2

x = 9

5x = 5(9) = 45

3x = 3(9) = 27

Hence, the two numbers are 45 and 27.

**Example 3 :**

If you subtract 1/2 from a number and multiply the result by 1/2, you get 1/8. What is the number.

**Solution : **

Let x be the required number.

From, the given information, we have

(x - 1/2) ⋅ 1/2 = 1/8

Multiply each side by 2.

(x - 1/2) ⋅ 1/2 ⋅ 2 = 1/8 ⋅ 2

x - 1/2 = 1/4

Add 1/2 to each side.

x - 1/2 + 1/2 = 1/4 + 1/2

x = 1/4 + 2/4

x = (1 + 2)/4

x = 3/4

Hence, the required number is 3/4.

**Example 4 : **

The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its width. What are the length and the width of the pool ?

**Solution : **

Let l be the length and w be the width of the swimming pool.

**Given :** Length is 2 m more than twice its width.

So, the length is

l = 2w + 2

**Given :** The perimeter of the swimming pool is 154 m.

2l + 2w = 154

Plug l = 2w + 2

2(2w + 2) + 2w = 154

Simplify.

4w + 4 + 2w = 154

6w + 4 = 154

Subtract 4 from each side.

6w + 4 - 4 = 154 - 4

6w = 150

Divide each side by 6.

6w / 6 = 150 / 6

w = 25

Then, the length is

l = 2(25) + 2

l = 50 + 2

l = 52

Hence, the length and width of the rectangular swimming pool are 52 m and 25 m respectively.

**Example 5 : **

The base of an isosceles triangle is 4/3 cm. The perimeter of the triangle is 4 2/15 cm. What is the length of either of the remaining equal sides ?

**Solution : **

Let x be the length of each of the remaining two equal sides.

So, the sides of the triangle are

x, x and 4/3

**Given :** The perimeter of the triangle is 4 2/15 cm.

x + x + 4/3 = 4 2/15

2x + 4/3 = 62/15

Subtract 4/3 from each side.

2x + 4/3 - 4/3 = 62/15 - 4/3

2x = 62/15 - 20/15

2x = (62 - 20)/15

2x = 42/15

2x = 14/5

Divide each side by 2.

2x ÷ 2 = (14/5) ÷ 2

x = 7/5

x = 1⅖

Hence, the length of either of the remaining equal sides is 1⅖ cm.

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

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