# PERCENT WORD PROBLEMS

Percent Word Problems :

In this section, we will learn how to solve word problems on percentage step by step.

## Percent Word Problems - Examples

Example 1 :

Jacob got 50% of the questions correct on a 30-question test and 90% on a 50-question test. What percent of all questions did Jacob get correct ?

Solution :

Total numbers of questions in both the tests :

=  30 + 50

=  80

No. of questions he got correct on a 30-question test :

=  50% ⋅ 30

=  0.5 ⋅ 30

=  15

No. of questions he got correct on a 50-question test :

=  90% ⋅ 50

=  0.9 ⋅ 50

=  45

Total number of questions he got correct in both the tests :

=  15 + 45

=  60

Percent of questions Jacob got correct :

=  (60/80) ⋅ 100 %

=  75 %

So, Jacob got 75% of all questions correct.

Example 2 :

This year, the chickens on a farm laid 30% less eggs than they did last year. If they had laid 3500 eggs this year, how many eggs did they lay last year ?

Solution :

Let x be the number of eggs laid last year.

Given : This year, the chickens laid 30% less eggs than they did last year and they laid 3500 eggs this year.

Then, we have

(100 - 30)% of x  =  3500

70% of x  =  3500

0.7 ⋅ x  =  3500

Divide each side by 0.7

x  =  5000

So, the chickens laid 5000 eggs last year.

Example 3 :

Miguel is following a recipe for marinara sauce that requires half a tablespoon of vinegar. If one cup is equivalent to 16 tablespoons, approximately what percent of a cup of vinegar is the amount by the recipe ?

Solution :

Percent of a cup of vinegar is the amount by the recipe :

=  (Amount of vinegar required / 1 cup of vinegar) ⋅ 100 %

=  (Half a tablespoon / 16 tablespoons) ⋅ 100 %

=  (0.5 / 16) ⋅ 100 %

3.1 %

So, approximately 3.1% of a cup of vinegar is the amount by the recipe.

Example 4 :

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

Solution :

Let \$100 be the retail price of the book.

The price of the book after 20% discount :

=  (100 - 20)% ⋅ 100

=  80% ⋅ 100

=  0.8 ⋅ 100

=  \$80

The price of the book at 30% off the discount price :

=  (100 - 30)% ⋅ 80

=  70% ⋅ 80

=  0.7 ⋅ 80

=  \$56

The retail price is \$100 and Alex bought the book for \$56.

So, the percent of the retail price Alex paid is 56%.

Example 5 :

Each day, Jacob eats 40% of the pistachios left in his jar at that 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 ?

Solution :

Let x be the number of pistachios at the start of the first day.

Number of pistachios at the end of the first day :

=  (100 - 40)% ⋅ x

=  60% ⋅ x

=  0.6x

=  \$80

Number of pistachios at the end of the second day :

=  (100 - 40)% ⋅ 0.6x

=  60% ⋅ 0.6x

=  0.6 ⋅ 0.6x

=  0.36x

Given : Number of pistachios remain at the end of second day is 27.

Then, we have

0.36x  =  27

Divide each side by 0.36

x  =  75

So, the number of pistachios were in the jar at the start of the first day is 75.

Example 6 :

Lily buys a doll at a 10 percent discount off the original price of \$105.82. However she has to pay a sales tax of x% on the discounted price. If the total amount she pays for the doll is \$100, then find the value of x.

Solution :

The price of the doll after 10% discount :

=  (100 - 10)% ⋅ 105.82

=  90⋅ 105.82

=  0.9 ⋅ 105.82

=  \$95.238

The price of the doll after x% sales tax on the discounted price :

=  (100 + x)% ⋅ 95.238

=  (100 + x) / 100 ⋅ 95.238

=  (1 + 0.01x) ⋅ 95.238

=  95.238 + 0.95238x

Given : The total amount Lily pays for the doll is \$100

Then, we have

95.238 + 0.95238x  =  100

Subtract 95.238 from each side.

0.95238x  =  4.762

Divide each side by 0.95238 (Use calculator).

x    5

So, the value of x is about 5.

Example 7 :

Over a two week span, James ate 20 pounds of chicken wings and 15 pounds of hot dogs. Anderson ate 20 percent more chicken wings 40 percent more hot dogs. Considering only chicken wings and hot dogs, what percent of food did Anderson eat more than James, by weight (rounded to the nearest percent).

Solution :

Total pounds of chicken wings and hot dogs eaten by James :

=  20 + 15

=  35

No. of pounds of chicken wings eaten by Anderson :

=  (100 + 20)% ⋅ 20

=  120⋅ 20

=  1.2 ⋅ 20

=  24

No. of pounds of hot dogs eaten by Anderson :

=  (100 + 40)% ⋅ 15

=  140⋅ 15

=  1.4 ⋅ 15

=  21

Total pounds of chicken wings and hot dogs eaten by Anderson :

=  24 + 21

=  45

No. of pounds of food eaten by Anderson more than James :

=  45 - 35

=  10

Percent increase from James to Anderson :

=  (10 / 35) ⋅ 100 %

29 %

So, Anderson ate about 29 percent of more food than James, by weight.

Example 8 :

William is playing a board game in which he has to collect as many cards as possible. On his first turn, he loses 18 percent of his cards. On the second turn, he increases his card count by 36 percent. If his final card count after these two turns is n, which of the following represents his starting card count in terms of n ?

A)  n / [(1.18)(0.64)]

B)  (1.18)(0.64)n

C)  n / [(1.36)(0.82)]

D)  (0.82)(1.36)n

Solution :

Let x be the starting card count.

Number of cards after losing 18% of cards on the first turn :

=  (100 - 18)% ⋅ x

=  82% ⋅ x

=  0.82x

Number of cards after increasing 36% of cards on the second turn :

=  (100 + 36)% ⋅ 0.82x

=  136% ⋅ 0.82x

=  1.36 ⋅ 0.82x

=  (1.36)(0.82)x

Given : Final card count after these two turns is n.

Then, we have

(1.36)(0.82)x  =  n

Divide each side by (1.36)(0.82).

x  =  n / [(1.36)(0.82)]

So, the starting card count in terms of n is

n / [(1.36)(0.82)]

Therefore, option C is correct.

Example 9 :

Due to deforestation, researchers expect the deer population to decline by 6 percent every year. if the current deer population is 12,000, then, find the approximate expected population size 10 years from now (rounded to the nearest tens).

Solution :

Population of deer after 1 year from now  :

=  (100 - 6)% ⋅ 12000

=  94% ⋅ 12000

=  0.94 ⋅ 12000

Population of deer after 2 years from now  :

=  0.94 ⋅ 0.94 ⋅ 12000

=  (0.94)⋅ 12000

In this way, population of deer after 10 years from now :

=  (0.94)10 ⋅ 12000

≈  6460

So, the expected population size of deer 10 years from now is about 6460.

Example 10 :

A small clothing store sells 3 different types of accessories. 20% are scarves, 60% are ties and the other 40 accessories are belts. If half of the ties are replaced with scarves, how many scarves will the store have ?

Solution :

Let x be the total number of accessories in the store.

Scarves and ties make up 80% of all accessories.

So, the remaining 20% of the accessories must be belts.

Then, we have

20% of x  =  40

20% ⋅ x  =  40

0.2x  =  40

Divide each side by 0.2

x  =  200

In all, there are 200 accessories in the store.

Number of scarves in the store :

=  20% ⋅ 200

=  0.2 ⋅ 200

=  40

Number of ties in the store :

=  60% ⋅ 200

=  0.6 ⋅ 200

=  120

Half of 120 ties (60 ties) are replaced with scarves

Now, the number of scarves in the store :

=  40 + 60

=  100

So, after replacing half of the ties with scarves, the store will have 100 scarves. After having gone through the stuff given above, we hope that the students would have understood, how to solve percent word problems.

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