# SAT PROBLEMS ON PERCENTAGE AND DECREASE

Problem 1 :

A painter charges \$1700 to paint the exterior of a house. The painter must use 57% of this money to purchase supplies for the job. What is the profit that the painter makes?

A) \$731     B) \$969    C) \$1643      D) \$2669

Solution :

Amount that painter is charging = \$1700

Amount uses to purchase supplies for the job

= 57% of 1700

So, the rest of the money can be considered as profit.

Rest of the money = (100 - 57)% of 1700

= 43% of 1700

= 0.43 (1700)

= 731

Then the painter will receive \$731. So, option A is correct.

Problem 2 :

At South Side Elementary school, approximately 25% of the second graders and 13% of the third graders participated in the science fair. If there are 168 second graders and 201 third graders at South Side, approximately how many second and third grade students participated in the science fair?

A) 68     B) 72     C) 140     D) 369

Solution :

Number of second graders = 168

Number of third graders = 201

Number of second graders who are participating in science fair

= 25% of 168

= 0.25(168)

= 42

Number of second graders who are participating in science fair

= 13% of 201

= 0.13(201)

= 26.13

Approximately 26 students.

Number of students who are participating in the science fair

= 42 + 26

= 68 students

So, option A is correct.

Problem 3 :

Dalilah surveyed a random sample of students in the freshman class to determine which theme they preferred for the homecoming dance. Of the 80 students surveyed, 70% preferred “Dancing with the Stars” over the other options. Based on this information, about how many students in the entire 374 person class would be expected to prefer “Dancing with the Stars” as the theme for the homecoming dance?

A) 56      B) 262      C) 300      D) 374

Solution :

In the total of 80 students surveyed, 70% of them are wishing in Dancing with stars.

70% of 80 = 0.70(80)

= 56

Using the concept of proportion, we can find the number of students who are preferring in this.

80 : 56 = 374 : x

80/56 = 374/x

80x = 374(56)

x = 20944/80

x = 261.2

Approximately 262 students. So, option B is correct.

Problem 4 :

Use the bar graph to answer the following question.

Of the following, which best approximates the percent increase in solar power between 2005 and 2015?

A) 50%      B) 75%       C) 100%        D) 125%

Solution :

Use of solar in 2005 = 1.25

Use of solar in 2015 = 2.5

Percentage increase

= {(New value - old value) / old value} x 100%

= {(2.5 - 1.25) / 1.25} x 100%

= (1.25 / 1.25) x 100%

= 100%

So, the required percentage increase is 100%. So, option C is correct.

Problem 5 :

In the town of Southbury, Mrs. Olson’s third grade class has 28 students. Mrs. Olson conducted a survey and found that approximately 71.4 percent of her students had at least one pet at home. If the students in Mrs. Olson’s class are representative of the students in Southbury, and there are 5,832 third grade students in Southbury, which of the following best estimates the number of third graders in Southbury who have no pets at home?

A) 1166        B) 1668         C) 4164        D) 8168

Solution :

Number of students who are not having at least one pet at home = 71.4

Number of students who are not having no pets at home

= 100 - 71.4

= 28.6 %

Ratio between the number of students and number of students who are not having pets at home

= 28.6% of 28

= 0.286(28)

= 8.008

Approximately 8 students.

In 28 students, 8 students they are not having at least one pet at home

28 : 8 = 5832 : x

28/8 = 5832/x

28x = 5832 (8)

x = 5832 (8) / 28

= 1666.28

Approximately 1666 students who are not having pets. So, option B is correct.

Problem 6 :

Kathie bought a teapot at a kitchen supply store that gave a 25% discount off the original price. The total amount she paid was t dollars, including a 6% sales tax on the discounted price. Which of the following represents the original price of the teapot in terms of t?

A) 0.81𝑡    B) 𝑡 /0.81    C) (0.75)(1.06)𝑡   D) 𝑡 / (0.75)(1.06)

Solution :

Let x be the original price of the tea pot. Since she is receiving 25% discount.

After discount the price of the tea pot = 75% of x

Tax = 6% of the price after discount

106% of 75% of x = t

1.06 (0.75 x) = t

x = t/(1.06) (0.75)

So, option D is correct.

Problem 7 :

Ryan bought a stock at the beginning of 2017 valued at s dollars. At the beginning of 2018 the stock had increased in value by 5%. Between the beginning of 2018 and the beginning of 2019, the stock increased by an additional 7%. What was the value of Ryan’s stock in the beginning of 2019 in terms of s?

A) 𝑠(1.05)(1.07)      B) 𝑠/(1.05)(1.07)

C) 𝑠(0.95)(0.97)       D) 𝑠/(0.05)(0.07)

Solution :

Value of stock in 2017 = s

At 2018, the value of the stock = 105% of s

At 2019, the value of stock = 107% of (105 of s)

= 1.07 (1.05 s)

So, option A is correct.

Problem 8 :

Aditi is a marine biologist studying the mating patterns of whales. According to her research, schools of beluga whales produce 25% more offspring than what schools of orca whales produce. Based on this research, if a school of beluga whales had 30 offspring, how many offspring would Aditi predict a school of orca whales of the same size would produce?

A) 22      B) 24     C) 33     D) 38

Solution :

beluga whales produce 25% more offspring than what schools of orca whales produce

If beluga whales produce 30.

Let x be the number of orca whales produce

125% of x = 30

1.25x = 30

x = 30/1.25

x = 24

So, option B is correct.

Problem 9 :

At the Hillside School, the two 8th grade classes are raising young trout in their classrooms. The tank in Mr. Lucha’s room has 20% more trout than the tank in Mrs. Echol’s room. If there are 84 trout in Mr. Lucha’s tank, how many are in Mrs. Echol’s?

A) 105    B) 101    C) 70     D) 67

Solution :

The tank in Mr. Lucha’s room has 20% more trout than the tank in Mrs. Echol’s room

Number of trout in Mr. Lucha’s tank = 84

120 % of number of trout in Mrs. Echol’s room = 84

1.20 of  number of trout in Mrs. Echol’s room = 84

number of trout in Mrs. Echol’s room = 84 / 1.20

= 70

So, option C is correct.

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