# WORD PROBLEMS ON PERCENTAGE

Problem 1 :

John paid \$64 for an item after 20% discount. Find the price of the item without discount.

Solution :

Let x be the price of the item without discount.

Given : John paid \$64 for the item after 20% discount.

Then, we have

(100 - 20)% of x  =  64

0.8x  =  64

Divide each side by 0.8

x  =  80

So, the price of the item without discount is \$80.

Problem 2 :

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) is

=  (20/80)  100%

=  25%

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

Problem 3 :

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

x  =  700

Hence,  the total number of votes polled 700.

Problem 4 :

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  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  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 is

=  (Actual increase / Original revenue)  100 %

=  (1700/10000)  100 %

=  17 %

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

Problem 5 :

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 us 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  3/5  =  9  ----------(1) ---------incorrect

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

Difference between (1) and (2) is 16

Percentage error is

=  (Actual error/Correct answer)  100 %

=  (16 / 25)  100 %

=  64%

Hence,  the percentage error in the calculation is 64%. Apart from the problems on percentage given above, if you need more problems on percentage, please click the following links.

Percentage Word Problems - 1

Percentage Word Problems - 2

Percentage Word Problems - 3

Percentage Word Problems - 4

Percentage Word Problems - 5

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