2021 CSAT — Q8

Quant Logical & quantitative reasoning 2.5 marks Easy

A biology class at high school predicted that a local population of animals will double in size every $12$ years. The population at the beginning of the year $2021$ was estimated to be $50$ animals. If $P$ represents the population after $n$ years, then which one of the following equations represents the model of the class for the population?

A $P = 12 + 50n$
B $P = 50 + 12n$
C $P = 50\,(2)^{12n}$
D $P = 50\,(2)^{n/12}$ Answer

Worked rationale

“Doubles every $12$ years” is exponential, not linear, so the base is $2$ and the exponent counts how many doubling periods have elapsed. In $n$ years the number of $12$ -year periods is $n/12$ , so

P = 50\,(2)^{n/12}.

Sanity check the exponent: at $n=12$ , $P = 50\cdot 2^{1} = 100$ (doubled once ✓); at $n=24$ , $P=50\cdot 2^2 = 200$ (doubled twice ✓).

Answer: (d) $P = 50\,(2)^{n/12}$ .

Why the other options miss

A

wrong model shape: builds a linear model and even swaps the roles of $12$ and $50$ .
B

linear, not doubling: grows $+12$ per year instead of doubling; ignores “double.”
C

exponent inverted: doubles every year and then $12$ times more — at $n=1$ it gives $50\cdot 2^{12}$ , absurdly fast.

Specialist insight

The decisive check is a single test value: plug $n=12$ and demand $P=100$ . Only $50\,(2)^{n/12}$ passes — $(2)^{12n}$ explodes and the two linear forms creep. Whenever a model has “every $T$ units it $\times k$ ,” the exponent is (elapsed time) $/T$ , i.e. $k^{\,n/T}$ ; the $/T$ in the exponent is exactly what separates the right answer from the trap option $2^{12n}$ .

The trap, in one line

Doubling every $12$ yr $\Rightarrow$ exponent is $n/12$ , giving $P=50(2)^{n/12} \Rightarrow$ (d), not $2^{12n}$ .

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