# FRACTIONS DECIMALS AND PERCENTAGES WORD PROBLEMS

Problem 1 :

The numerator and denominator of a fraction add up to 10. Adding 3 to both numerator and denominator of the fraction results . Find the fraction.

Solution :

Let x be the numerator and y be the denominator of the fraction.

Given : The numerator and denominator add up to 8.

x + y = 10 ----(1)

Given : When 3 is added to both numerator and denominator, the fraction becomes .

⁽ˣ ⁺ ³⁾⁄₍y ₊ ₃₎ =

5(x + 3) = 3(y + 3)

5x + 15 = 3y + 9

5x - 3y = -6 ----(2)

3(1) + (2) :

8x = 24

Divide both sides by 8.

x = 3

Substitute x = 3 in (1).

3 + y = 10

Subtract 3 from both sides.

y = 7

ˣ⁄y = ³⁄₇

The fraction is ³⁄₇.

Problem 2 :

Three-fifth of the students in a school are girls. If there is a total of 750 students in the school, find the number of boys.

Solution :

If three-fifth of the students are girls, then two-fifth of ths students are boys.

Because, in total 5 parts, 3 parts are girls and the remaning two parts are boys.

Number of boys in the school :

= 750 ⋅

= 450

Problem 3 :

Sum of the length and width of a rectangle is 55 cm. If the length is 1.75 times of the width, find the area of the rectangle.

Solution :

Let l be the length and w be the width of the rectangle.

Given : The length and width add up to 55 cm.

l + w = 55 ----(1)

Given : The length is 1.75 times of the width.

l = 1.75w ----(2)

Substitute l = 1.75w in (1).

1.75w + w = 55

2.75w = 55

Divide both sides by 2.75.

w = 20 cm

Substitute w = 20 in (2).

l = 1.75(20)

l = 35 cm

Area of the rectangle :

= l ⋅ w

= 35 ⋅ 20

= 700 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 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)

= 4.5 + 3

= \$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

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

Problem 7 :

The length of a rope is 92.4 meters. If the rope is cut into pieces of length 7 meters each, find the maximum number of peices can be received.

Solution :

The original length of the rope is 92.4 meters.

The length of each piece is 7 meters.

Number of pieces with length 7 meters each can be cut from the rope :

= 92.4/7

= 13.2

Number of pieces can not be decimal.

Therefore, maximum number of pieces of length 7 meters each can be cut from the rope is 13.

Problem 8 :

The original price of a suit is \$1000. A salesman decides to discount it by 30%. Later on, the manager decides to give a 15% discount off the salesman's price. What is the final price of the suit?

Solution :

Price of the suit after 30% discount given by salesman :

= (100 - 30)% of 1000

= 70% of 1000

= 0.7 ⋅ 1000

= 700

Final price of the suit after 15% discount given by the manager :

= (100 - 15)% of 700

= 85% of 700

= 0.85 ⋅ 700

= \$595

Problem 9 :

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.

Selling price = 120 % of cots price

= 1.2 ⋅ 22.5

= 27

Selling price of the toy is \$27.

Problem 10 :

A sports equipment manufacturer produced 3,600 footballs and 2,200 basketballs during the fall. In the winter, it produced 3,060 footballs and a certain number of basketballs. If the manufacturer decreased the production of basketballs by the same percentage as it did foir footballs, how many basketballs did it produce in the winter?

Solution :

Percentage decrease in the production of footbals from the fall to the winter :

= [(3600 - 3060)/3600] ⋅ 100%

= 540/3600 ⋅ 100%

= 15%

Given : From the fall to the winter, the manufacturer decreased the production of basketballs by the same percentage as it did foir footballs.

Number of basketballs produced in the winter :

= (100 - 15)% of 2200

= 85% of 2200

= 0.85 ⋅ 2200

= 1870

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