WORD PROBLEMS ON LINEAR EQUATIONS IN ONE VARIABLE

About "Word Problems on Linear Equations in One Variable"

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. 

Word Problems on Linear Equations in One Variable - Examples

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 2/15 cm.

x + x + 4/3  =  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  =  1

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, "Word Problems on Linear Equations in One Variable". 

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