SOLVING WORD PROBLEMS WITH ONE VARIABLE

Solving Word Problems with One Variable :

In this section, we will learn, how to solve word problems using equations with one variable. 

Solving Word Problems with One Variable - Examples

Example 1 :

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 width of the pool ? 

Solution : 

Let w be the width of the pool.

The length is

l  =  2w + 2 -----(1)

Given : Perimeter of swimming pool is 154 m. 

Then, we have 

2(l + w)  =  154

Divide each side by 2. 

l + w  =  77

Substitute (2w + 2) for l. 

2w + 2 + w  =  77

3w + 2  =  77

Subtract 2 from each side. 

3w  =  75

Divide each side by 3. 

w  =  25

Substitute 25 for w in (1). 

(1)-----> l  =  2(25) + 2

l  =  50 + 2

l  =  52

So, the length and width of the pool are 52 m and 25 m  respectively.  

Example 2 :

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

Solution :

Let x be the smaller number. 

Then, the larger number is

=  x + 15 -----(1)

Given : Sum of the two numbers is 95. 

Then, we have 

x + (x + 15)  =  95

2x + 15  =  95

Subtract 15 from each side. 

2x  =  80

Divide each side by 2.

x  =  40

Substitute 40 for x in (1). 

(1)-----> x + 15  =  40 + 15

x + 15  =  55

So, the numbers are 40 and 55. 

Example 3 :

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

Solution :

Given : Two numbers are in the ratio 5 : 3.

Then, the two numbers are

5x  and  3x

Given : The two numbers differ by 18.

Then, we have

5x - 3x  =  18

2x  =  18

Divide each side by 2.

x  =  9

Find the two numbers :

5x  =  5(9)  =  45

3x  =  3(9)  =  27

So, the two numbers are 45 and 27.

Example 4 :

Three consecutive integers add up to 51. What are these integers ?

Solution :

Let the three consecutive integers be

x, (x + 1), and (x + 2)

Given : Three consecutive integers add up to 51.

Then, we have

x + (x + 1) + (x + 2)  =  51

3x + 3  =  51

Subtract 3 from each side. 

3x  =  48

Divide each side by 3. 

x  =  16

x + 1  =  16 + 1  =  17

x + 2  =  16 + 2  =  18

So, the consecutive integers are 16, 17, and 18.

Example 5 :

The sum of three consecutive multiples of 8 is 888. Find the multiples.

Solution :

Let the three consecutive multiples of 8 be

8x, 8(x + 1) and 8(x + 2)

Given : The sum of three consecutive multiples of 8 is 888.

Then, we have

8x + 8(x + 1) + 8(x + 2)  =  888

8x + 8x + 8 + 8x + 16  =  888

Simplify.

24x + 24  =  888

Subtract 24 from each side.

24x  =  864

Divide each side by 24. 

x  =  36

8x  =  8(36)  =  288

8(x + 1)  =  8(36 + 1)  =  8(37)  =  296

8(x + 2)  =  8(36 + 2)  =  8(38)  =  304

So, the three consecutive multiples of 8 are 288, 296, and 304.

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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