# 1997 AHSME Problems/Problem 14

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem

The number of geese in a flock increases so that the difference between the populations in year $n+2$ and year $n$ is directly proportional to the population in year $n+1$. If the populations in the years $1994$, $1995$, and $1997$ were $39$, $60$, and $123$, respectively, then the population in $1996$ was $\textbf{(A)}\ 81\qquad\textbf{(B)}\ 84\qquad\textbf{(C)}\ 87\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 102$

## Solution

Let $x$ be the population in $1996$, and let $k$ be the constant of proportionality.

If $n=1994$, then the difference in population between $1996$ and $1994$ is directly proportional to the population in $1995$.

Translating this sentence, $(x - 39) = k(60)$

Similarly, letting $n=1995$ gives the sentence $(123 - 60) = kx$

Since $kx = 63$, we have $k = \frac{63}{x}$

Plugging this into the first equation, we have: $(x - 39) = \frac{60\cdot 63}{x}$ $x - 39 = \frac{3780}{x}$ $x^2 - 39x - 3780 = 0$ $(x - 84)(x + 45) = 0$

Since $x>0$, we must have $x=84$, and the answer is $\boxed{B}$.

## See also

 1997 AHSME (Problems • Answer Key • Resources) Preceded byProblem 13 Followed byProblem 15 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. Invalid username
Login to AoPS