PERCENTAGE CHANGE WORD PROBLEMS

Problem 1 :

In a particular store the number of TV's sold the week of Black Friday was 685. The number of TVs sold the following week was 500. TV sales the week following Black Friday were what percent less than TV sales the week of Black Friday?

(A)  17%   (B)  27%   (C)  37%   (D)  47%

Solution :

Number of TV's sold on Black Friday = 685.

Number of TV's sold on the following week = 500.

Percentage change

  =  (New value - old value)/Old value ⋅ 100%

=  [ (685-500)/685] ⋅ 100%

=  (185/685) ⋅ 100%

=  27%

The sales dropped 27% over the period.

Problem 2 :

In March, a city zoo attracted 32000 visitors to its polar bear exhibit. In April, the number of visitors to the exhibit increased by 15%. How many visitors did the zoo attract to its polar bear exhibit in April?

(A)  32150   (B)  32480   (C)  35200  (D)  36800

Solution :

Number of visitors attracted in the month March

=  32000

In April, 15% of visitors increased.

Number of visitors attracted by polar bear exhibit

=  115% of 32000

=  36800

Problem 3 :

A charity organization collected 2140 donations last month. With the help of 50 additional volunteers, the organization collected 2690 donations this month. To the nearest tenth of a percent, what was the percent increase in the number of donations the charity organization collected?

(A) 20.4%   (B)  20.7%    (C)  25.4%   (D)  25.7%

Solution :

Donation collected by the organization last month = 2140

Donation collected by the organization this month = 2690

Percentage change  =  [(2690-2140)/2690] ⋅ 100%

=  (550/2690) ⋅ 100%

=  20.4%

Problem 4 :

The discount price of a book is 20% less than the retail price. James manages to purchase the book at 30% off the discount price at the special book sale. What percent of the retail price did James pay?

(A)  42%   (B)  48%    (C)  50%   (D)  56%

Solution :

Let x be the retail price.

After discount  =  80% of x

=  70% of 80% of x

=  0.70(0.80 of x)

=  0.56 of x

=  56% of x

James will pay 56% of retail price.

Problem 5 :

Each day, Robert eats 40% of the pistachios left in his jar at the time. At the end of the second day, 27 pistachios remain. How many pistachios were in the jar at the start of the first day ?

(A)  75   (B)  80   (C)  85  (D)  95

Solution :

Let x be the number of pistachios in the jar on the first day. 

At the end of first day number of pistachios  =  60% of x

At the end of second day number of pistachios

=  60% of (60% of x)

=  0.60(0.60 x)

=  0.36x

Number of pistachios at the second day  =  27

0.36 x  =  27

x  =  27/0.36

x  =  75

Problem 6 :

Joanne bought a doll at a 10 percent discount off the original price of $105.82. However, she had to pay a sales tax of x% on the discounted price. If the total amount she paid for the doll was $100, what is the value of x ?

(A)  2   (B)  3   (C)  4  (D)  5

Solution :

Original price of doll  =  $105.82

90% of original price  =   0.90(105.82)

=  95.23

x% of discounted price + 90% of the total price  =  $100

x% of 95.23 + 95.23  =  100

95.23(x% + 1)  =  100

95.23 (x+100)/100  =  100

x+100  =  105

x  =  5

So, she has to pay 5% sales tax.

Problem 7 :

In 2010, the number of houses built in Town A was 25 percent greater than the number of houses built in Town B. If 70 houses were built in Town A during 2010, how many were built in Town B ?

(A)  56   (B)  50    (C)  48   (D)  20

Solution :

Number of houses were in Town A on 2010  =  70

Number of houses in Town A  =  25% greater than number of houses in Town B

Let x be the number of houses in Town B.

70  =  125% of x

70  =  1.25x

x  =  70/1.25

x  =  56

So, the number of houses in Town B is 56.

Problem 8 :

Over two week span, John ate 20 pounds of chicken wings and 15 pounds of hot dogs. Kyle ate 20 percent more chicken wings and 40 percent more hot dogs. Considering only chicken wings and hot dogs, Kyle ate approximately x percent more food, by weight, than John, what is x (rounded to the nearest percent) ?

(A)  25   (B)  27    (C)  29   (D)  30

Solution :

Quantity of chicken wings ate by John  =  20 pounds

Quantity of hot dogs ate by John  =  15 pounds

Quantity of food at by John  =  20+15

=  35 pounds

Quantity of chicken wings ate by Kyle  =  120% of 20

=  1.20(20)

=  24 pounds

Quantity of hot dogs ate by Kyle  =  140% of 15 pounds

=  1.40(15)

=  21

Quantity of food at by Kyle  =  24+21

=  45 pounds

Percentage change  =  [(45-35)/35] ⋅ 100%

=  28.57% 

Approximately 29%.

Problem 9 :

Due to deforestation, researchers, expect the deer population to decline by 6 percent every year. If the current deer population is 12000, what is the approximate expected population size in 10 years from now ?

(A)  25000   (B)  48000    (C)  56000   (D)  30000

Solution :

In 10 years the population will be declined 60% of the initial population.

Expected population after 10 years  =  40% of 12000

=  48000

So, the expected population size after 10 years from now is 48000.

Problem 10 :

In 2000 the price of a house was $72600. By 2010 the price of the house has increased to 125598.

(A)  70%    (B)  62%    (C)  73%    (D)  90%

Solution :

Percentage increase  =  [(125598 - 72600)/72600] ⋅ 100%

=  (52998/72600) ⋅ 100%

=  0.73 ⋅ 100%

=  73%

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