SOLVING WORD PROBLEMS WITH ONE VARIABLE

Problem 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.  

Problem 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. 

Problem 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.

Problem 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.

Problem 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.

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