FRACTIONS DECIMALS AND PERCENTAGES WORD PROBLEMS

About "Fractions decimals and percentages word problems"

Fractions decimals and percentages word problems :

Word problems on fractions decimals and percentages are much useful to the students who would like practice problems with fractions, decimals and percentages on real world situations. 

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. 

"The denominator of the fraction exceeds the numerator"

From the above information, fraction = x / (x+5) ------(1)

"If 3 be added to both, the fraction becomes 3/4"

From the above information, we have (x+3) / (x+5+3) = 3/4

(x+3)/(x+8) = 3/4 ---------> 4(x+3) = 3(x+8)

4x + 12 = 3x + 24 ---------> x = 12

(1)--------> x / (x+5) = 12 / (12+5) = 12/27

Hence, 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 :

No. of boys in the school = 450 x 2/3 = 300

Total no. of students = 450. 

Then no. of girls = 450 - 300 = 150

Hence the numbers girls in the school = 150 

Example 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 = (2/3)x

Perimeter = 80 cm -----> 2(l + w) = 80 -------> l + w = 40

l + w = 40 -------> x + (2/3)x = 40 -------> (3x+2x) / 3 = 40

(3x+2x) / 3 = 40 ------> 5x = 120 ------> x = 24

So, length = x = 24 cm

and width = (2/3)x = (2/3)24 = 16 cm

Area = l x w = 24x16 = 384 square cm.

Hence, area of the rectangle is 384 square cm

Problem 4 :

If good are purchased for $ 1500 and one fifth of them sold at a loss of 15%. Then at what profit percentage should the rest be sold to obtain a profit of 15%?

Solution :

As per the question, we need 15% profit on $1500.

Selling price for 15% on 1500

S.P  =115% x 1500 = 1.15x1500 = 1725

When all the good sold, we must have received $1725 for 15% profit.

When we look at the above picture, in order to reach 15% profit overall, the rest of the goods ($1200) has to be sold for $1470.

That is,

C.P = $1200,    S.P = $1470,    Profit = $270

Profit percentage  = (270/1200) x 100

Profit percentage  = 22.5 %

Hence, the rest of the goods to be sold at 22.5% profit in order to obtain 15% profit overall.

Problem 5 :

I purchased 120 books at the rate of $3 each and sold 1/3 of them at the rate of $4 each. 1/2 of them at the rate  of $ 5 each and rest at the cost price. Find my profit percentage.  

Solution :

Total money invested = 120x3 = $360 -------(1)

Let us see, how 120 books are sold in different prices.

From the above picture,

Total money received = 160 + 300 +60 = $ 520 --------(2)

Profit = (2) - (1) = 520 - 360 = $160

Profit percentage = (160/360)x100 % = 44.44%

Hence the profit percentage is 44.44

Problem 6 :

If A's salary is  20% less than B's salary. By what percent is B's salary more than A's salary ?

Solution :

Let us assume B's salary  =  $ 100 ----------(1)

Then, A's salary  =  $ 80 --------(2)

Now we have to find the percentage increase from (2) to (1). 

Difference between (1) and (2)  =  $ 20

Percentage increase from (2) to (1)  =  (20/80) x 100%  =  25%

Hence,  B's salary is 25% more than A's salary. 

Let us look at the next problem on "Fractions decimals and percentages word problems" 

Problem 7 :

In an election, a candidate who gets 84% of votes is elected by majority with 588 votes. What is the total number of votes polled ?

Solution :

Let "x" be the total number of votes polled. 

Given : A candidate who gets 84% of votes is elected by majority of 476 votes

From the above information, we have 

84% of x  =  588 ---------> 0.84x  =  588

x  =  588 / 0.84  =    700

Hence,  the total number of votes polled 700.

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Problem 8 :

When the price of a product was decreased by 10 % , the number sold increased by 30 %. What was the effect on the total revenue ?

Solution :

Before decrease in price and increase in sale, 

Let us assume that price per unit = $ 100.

Let us assume that the number of units sold = 100

Then the total revenue  =  100 x 100  =  10000 -----(1)

After decrease 10 % in price and increase 30 % in sale,

Price per unit = $ 90.

Number of units sold = 130

Then the total revenue  =  90 x 130  =  11700 -----(2)

From (1) and (2), it is clear that the revenue is increased. 

Difference between (1) and (2)  =  1700

Percent increase in revenue

 =  (Actual increase / Original revenue) x 100 %  

=  (1700/10000) x 100 %

=  17 %

Hence,  the net effect in the total revenue is 17 % increase. 

Let us look at the next problem on "Fractions decimals and percentages word problems" 

Problem 9 :

A student multiplied a number by 3/5 instead of 5/3. What is the percentage error in the calculation ?

Solution :

In the given two fractions, the denominators are 5 and 3. 

Let assume a number which is divisible by both 5 and 3. 

Least common multiple of (5, 3)  =  15.

So, let the number be 15. 

15 x 3/5  =  9  ----------(1) ---------incorrect

15 x 5/3  =  25  ---------(2) --------correct

Difference between (1) and (2) is 16

Percentage error = (Actual error / Correct answer ) x 100 % 

=  (16 / 25) x 100 % 

=  64 %

Hence,  the percentage error in the calculation is 64 %. 

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Problem 10 :

There are 15 boys and 12 girls in a section A of class 7. If 3 boys are transferred to section B of class 7,then find the percentage of boys in section A.      

Solution :

Before transfer :

No. of boys in section A  =  15

No. of boys in section B  =  12

Given : 3 boys are transferred from section A to B

After transfer :

No. of boys in section A  =  12

No. of boys in section B  =  12

Hence,  percentage of boys in section A is 50%

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Problem 11 :

A chemist mixed 6.35 grams of one compound with 2.45 grams of another compound. How many grams were there in the mixture ?

Solution : 

Quantity of one compound = 6.35 grams

Quantity of another compound = 2.45 grams 

Total quantity of mixture is

= Quantity of one comp + quantity of another comp 

=  6.35 + 2.45 

=  8.8 grams

Hence, the quantity of the mixture is 8.8 grams.

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Problem 12 : 

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 let us find the total bill. 

Total bill  =  3x1.5 + 4x0.75

Total bill  =  4.5 + 3

Total bill  =  $ 7.5

Given : discount 10%

The money that he has to pay is

=  (100 - 10)% of total bill 

=  90% x 7.5

=  0.9 x 7.5

=  6.75

Therefore, he has to pay $6.75

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Problem 13 : 

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  =  $24.60

Money earned in 1 hour  =  $24.60 / 6

Money earned in 1 hour  =  $4.10

Therefore money earned in 7.5 hours is 

=  7.5 x $4.10

=  $30.75

Let us look at the next problem on "Fractions decimals and percentages word problems" 

Problem 14 :

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 = 76.8 meters 

The length of each pipe piece = 8 meters

No of pieces where each piece has length 8 meters is

=  76.8 / 8

= 9.6

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

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Problem 15 :

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

The selling price  =  (100 + 20) % of cots price

=  120% x 22.5

=  120% x 22.5

=  1.2 x 22.5

=  27

Therefore the selling price of the toy is $27

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