Question #2d1614a1 - SAT Suite Question Bank

ID: 2d1614a1

Modded SAT Question Bank by Abdullah Mallik

$upper P left parenthesis t right parenthesis equals 290 left parenthesis 1.04 right parenthesis Superscript left parenthesis four sixths right parenthesis t$

The function $P$ models the population, in thousands, of a certain city $t$ years after $2005$ . According to the model, the population is predicted to increase by $n %$ every $18$ months. What is the value of $n$ ?

$0.38$
$1.04$
$4$
$6$

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Correct Answer: C

Rationale

Choice C is correct. It's given that the function $P$ models the population of the city $t$ years after $2005$ . Since there are $12$ months in a year, $18$ months is equivalent to $StartFraction 18 Over 12 EndFraction$ years. Therefore, the expression $StartFraction 18 Over 12 EndFraction x$ can represent the number of years in $x$ $18$ -month periods. Substituting $StartFraction 18 Over 12 EndFraction x$ for $t$ in the given equation yields $P (\frac{18}{12} x) = 290 {(1.04)}^{(\frac{4}{6}) (\frac{18}{12} x)}$ , which is equivalent to $P (\frac{18}{12} x) = 290 {(1.04)}^{x}$ . Therefore, for each $18$ -month period, the predicted population of the city is $1.04$ times, or $104 percent sign$ of, the previous population. This means that the population is predicted to increase by $4 percent sign$ every $18$ months.

Choice A is incorrect and may result from conceptual or calculation errors.

Choice B is incorrect. Each year, the predicted population of the city is $1.04$ times the previous year's predicted population, which is not the same as an increase of $1.04 percent sign$ .

Choice D is incorrect and may result from conceptual or calculation errors.

Question Difficulty: Hard