PERCENT WORD PROBLEMS FOR SAT

Problem 1 :

Alma bought a laptop computer at a store that gave a 20 percent discount off its original price. The total amount she paid to the cashier was p dollars, including an 8 percent sales tax on the discounted price.

Which of the following represents the original price of the computer in terms of p ?

a)  0.88p   b)  p/0.88    c)  (0.8) (1.88p)   d)  p/(0.8) (1.88) 

Solution :

80% of original price + 8% of sales tax = p

80% of original price + 8% of (80% of original price) = p

80% of original price (1 + 8%) = p

80% of original price (88%) = p

0.8 (original price) 0.88 = p

Original price = p/(0.8)(0.88)

Problem 2 :

Katarina is a botanist studying the production of pears by two types of pear trees. She noticed that Type A trees produced 20 percent more pears than Type B trees did. Based on Katarina’s observation, if the Type A trees produced 144 pears, how many pears did the Type B trees produce?

A) 115     B) 120     C) 124       D) 173

Solution :

Number of trees produced in Type A = 144

Let x be the number of trees produced in Type B.

120% of x = 144

1.20x = 144

x = 144/1.20

x = 120

So, number of trees produced in Type B is 120.

Problem 3 :

A radioactive substance decays at an annual rate of 13 percent. If the initial amount of the substance is 325 grams, which of the following functions f models the remaining amount of the substance, in grams, t years later?

a) f(t) = 325(0.87)^t      b) f(t) = 325(0.13)^t

c) f(t) = 0.87(325)^t      d) f(t) = 0.13(325)^t

Solution :

Initial amount = 325

Percentage increase = 13%

f(t) = P(1 - r%)^t

= 325(1 - 13%)^t

= 325(87%)^t

= 325(0.87)^t

Problem 4 :

The amount of money a performer earns is directly proportional to the number of people attending the performance. The performer earns $120 at a performance where 8 people attend.

The performer uses 43% of the money earned to pay the costs involved in putting on each performance. The rest of the money earned is the performer’s profit. What is the profit the performer makes at a performance where 8 people attend?

a) $51.60     b) $57.00     c) $68.40     d) $77.00

Solution :

Amount earning = 120

Profit percentage = (100 - 43)%

= 57%

= 57% of 120

= 0.57 (120)

= 68.4

So, the answer is option c.

Problem 5 :

The atomic weight of an unknown element, in atomic mass units (amu), is approximately 20% less than that of calcium. The atomic weight of calcium is 40 amu. Which of the following best approximates the atomic weight, in amu, of the unknown element?

A) 8     B) 20     C) 32     D) 48

Solution :

Atomic weight of calcium = 40 amu

atomic weight = 80% of 40 amu

= 0.80 x 40

= 32

So, option c is correct.

Problem 6 :

The sum of three numbers is 855. One of the numbers, x, is 50% more than the sum of the other two numbers. What is the value of x ?

a) 570      b) 513     c) 214     d) 155

Solution :

Sum of the three numbers = 855

Let the three numbers be x, y and z

x + y + z = 855

x = 150% of (y + z)

x = (150/100) (y + z)

x = (3/2)(y + z)

y + z = 2x/3

x + (2x/3) = 855

(3x + 2x)/3 = 855

x = 855 (3/5)

x = 513

So, the answer is option b.

Problem 7 :

A rectangle was altered by increasing its length by 10 percent and decreasing its width by p percent. If these alterations decreased the area of the rectangle by 12 percent, what is the value of p ?

a) 12       b) 15        c) 20        d) 22

Solution :

Let l and w be the length and width of rectangle.

After alteration :

New length = 110% of l

New width = w - p% of w

= w(1 - p%)

Area of the old rectangle = l w

Area of the new rectangle = 88% of lw

Area after alteration = 110% of l x w(1 - p%)

88% of lw = 110% of l x w(1 - p%)

0.88 = 1.10 (1 - p%)

0.88/1.10 = 1 - p%

1 - p% = 0.8

p% = 1 - 0.8

p% = 0.2

p = 20%

So, the answer is option c

Problem 8 :

In planning maintenance for a city’s infrastructure, a civil engineer estimates that, starting from the present, the population of the city will decrease by 10 percent every 20 years. If the present population of the city is 50,000, which of the following expressions represents the engineer’s estimate of the population of the city t years from now?

A) 50,000(0.1)^20t      B) 50,000(0.1)^t/20

C) 50,000(0.9)^20t      D) 50,000(0.9)^t/20

Solution :

Initial population = 50000

Decreasing rate = 10% 

Decay function = 50000(1 - 10%)^t/20

= 50000(90%)^t/20 

= 50000(0.9)^t/20