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.

2(l + w) = 80

Divide both sides 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)

= 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

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.

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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