FRACTIONS DECIMALS AND PERCENTAGES WORD PROBLEMS

Problem 1 : 

The denominator of a fraction exceeds the numerator by 5. If 3 be added to both, the fraction becomes 3/4. Find the fraction. 

Solution :

Let x be the numerator. 

Given :The denominator of the fraction exceeds the numerator.

Then, the required fraction is 

x / (x + 5) -----(1)

Given : If 3 be added to both, the fraction becomes 3/4.

(x + 3) / (x + 5 + 3)  =  3 / 4

Simplify and solve for x. 

(x + 3 ) / (x + 8)  =  3 / 4

4(x + 3)  =  3(x + 8)

4x + 12  =  3x + 24

x  =  12

To get the required fraction, substitute 12 for x in (1). 

(1)-----> x / (x + 5)  =  12 / (12 + 5)

x / (x + 5)  =  12 / 27

So, the required fraction is 12/27.

Problem 2 :

In a school, there are 450 students in total. If 2/3 of the total strength are boys, find the number of girls in the school. 

Solution :

Given : Total no. of students in the school is 450 and 2/3 of the total strength is boys. 

Then, no. of boys in the school is

=  450 ⋅ 2/3  =  300

Out of the total students 450, there are 300 boys.

Then no. of girls is

=  450 - 300  =  150

So, the numbers girls in the school is 150. 

Problem 3 :

The width of the rectangle is 2/3 of its length. If the perimeter of the rectangle is 80 cm. Find its area. 

Solution :

Let x be the length of the rectangle. 

Then, width of the rectangle is

=  2x / 3

Given : Perimeter  =  80 cm

That is, 

2(l + w)  =  80

Divide each side by 2. 

l + w  =  40

Substitute x for l and 2x/3 for w.

x + 2x/3  =  40

(3x + 2x) / 3  =  40

5x  =  120

Divide each side by 5.

x  =  24

2x / 3  =  2(24) / 3  =  16

Therefore, length and width of the rectangle are 24 cm and 16 cm. 

Area of the rectangle is

=  l ⋅ w

=  24 ⋅ 16

=  384 square cm.

So, the area of the rectangle is 384 square cm.

Problem 4 :

A chemist mixed 20 percent of 6.36 grams of one compound with 60 percent of 2.48 grams of another compound. How many grams were there in the mixture ?

Solution : 

Quantity taken from the first compound is

=  20% of 6.36

=  0.2 ⋅ 6.36

=  1.272

Quantity taken from the first compound is

=  60% of 2.48

=  0.6 ⋅ 2.48

=  1.488

Total quantity of mixture is

=  Quantity from 1st comp + Quantity from 2nd comp 

=  1.272 + 1.488

=  2.76 

So, the quantity of the mixture is 2.76 grams.

Problem 5 :

David buys 3 pens where the price of each pen is $1.5 and the 4 pencils where the price of each pen is $0.75. If he gets a discount of 10% in the total bill, how much does he have to pay ?

Solution : 

First, find the total bill. 

Total bill  =  (3 ⋅ 1.5) + (4 ⋅ 0.75)

Total bill  =  4.5 + 3

Total bill  =  $ 7.5

Given : Discount is 10%

The money that he has to pay is

=  90% of the total bill

=  0.9  7.5

=  6.75

So, David has to pay $6.75.

Problem 6 :

Joseph earned $24.60 for working 6 hours. How much will earn, if he works for 7.5 hours ? 

Solution : 

Given : Money earned in 6 hours is $24.60

Then, money earned in 1 hour is  

=  24.60 / 6

=  $4.10

Therefore money earned in 7.5 hours is 

=  7.5  4.10

=  $30.75

So, money earned in 7.5 hours is $30.75.

Problem 7 :

A pipe is 76.8 meters long. What will the greatest number of pieces of pipe each 8 meters long that can be cut from this pipe ?

Solution :

The original length of the pipe is 76.8 meters. 

The length of each pipe piece is 8 meters.

No of pieces where each piece has length 8 meters is

=  76.8 / 8

= 9.6

So, the greatest number of pieces of pipe each 8 meters long is 9.

Problem 8 :

The cost of a toy is  $22.5. If the profit is 20%, what is the selling price of the toy ? 

Solution :

The cost of a toy is  $22.5

If the profit is 20%, then the selling price is 120% of cost price. That is,  

Selling price  =  120 % of cots price

Selling price  =  1.2 ⋅ 22.5

Selling price  =  27

So, the selling price of the toy is $27

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