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KPMG Past Questions And Answers


56)

Charlie bought a $60 radio on sale at 5% off. How much did he pay, including 5% sale tax?

Solution
Since 5% of 60 is 3, Charlie saved $3, and thus paid $57 for the radio. He then had to pay 5% sale tax on the $57: .05 × 57 = 2.85, so the total cost was $57 + $2.85 = $59.85. The correct answer is D.
57)

In 1970 the populations of town A and town B were the same. From 1970 to 1980, however, the population of town A increased by 60% while the population of town B decrease by 60%. In 1980, the population of town B was what percent of the population of town A?

In your math class, you would let x be the population of town A in 1970 and then proceed to set an algebra problem. Don’t do that on SAT. Assume that the populations of both towns were 100 in 1970. Then, since 60% of 100 is 60, in 1980 the populations were 100 + 60 = 160 (town A) and 100 – 60 = 40 (town B). Then, in 1980, town B’s population was 40/160= 1/4 = 25% of town A’s. Choice A is correct.
58)

Brian gave 20% of his baseball cards to Scott and 15% to Adam. If he still had 520 cards, how many did he have originally?

Originally, Brian had 100% of the cards (all of them). After he gave away 35% of
them, he had 100% - 35% = 65% of them left. Then 520 is 65% of what number?
520 = 65x ⇒x = 520 ÷ 65 = 800.
59)

One day at Central High School, 1/12 of the students were absent, and of those present went on a field trip. If the number of students staying in School was 704, how many students are enrolled in Central High?

If s is the number of student enrolled, 1s/12 is the numbers who were absent, and 11s/12 is the numbers who were present. Since 1/5 of those present went on a field trip, 4/5 of them stayed in Schoo.l Therefore, 704 =(4/5 × 11/12)s= 11s/15 ⇒ s = 704 ÷ 11/15 = 704 × 15/11 = 960. (960).
60)

Billy won some goldfish at the state fair. During the first week,1/5 of them died; and during the second week, 3/8 of those still alive at the end of the first week died. What fraction of the original goldfish were still alive after 2 weeks?

Algebraic solution: Let x = number of goldfish Billy won. During the first week, 1x/5 died, so 4x/5 were still alive. During the second week, 3/8 of those died and 5/8 survived: 5/8 x 4/5 x =1x/2. Assume that the original number of goldfish was 40, the LCM of the denominators. Then, 8 died the first week (1/5 of 40) and 12 of the 32 survivors ( 3/8 of 32) died the second week. In all, 8 + 12 =
20 died, the other 20 , 1/2 of the original number were still alive. (C).

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