SAT Mathematics Practice Test 3

Question : 48 of 58

Marks: +1, -0

Choice A is incorrect and may be the result of confusing the 12 percent decrease in area with the percent decrease in width. Choice B is incorrect because decreasing the width by 15 percent results in a 6.5 percent decrease in area, not a 12 percent decrease. Choice D is incorrect and may be the result of adding the percents given in the question (10 + 12).

$50,000(0.1)^{20t}$
$50,000(0.1)^{t/20}$
$50,000(0.9)^{20t}$
$50,000(0.9)^{t/20}$

Validate

Solution:

For the present population to decrease by 10 percent, it must be multiplied by the factor 0.9.
Since the engineer estimates that the population will decrease by 10 percent every 20 years,
the present population, 50,000, must be multiplied by (0.9)n ,
where n is the number of 20-year periods that will have elapsed t years from now. After t years, the number of 20-year periods that have elapsed is

t/20

.
Therefore

50,000(0.9)^{t/20}

represents the engineer’s estimate of the population of the city t years from now.
Choices A, B, and C are incorrect because each of these choices either confuses the percent decrease with the multiplicative factor that represents the percent decrease or mistakenly multiplies t by 20 to find the number of 20-year periods that will have elapsed in t years.

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