Difference between revisions of "1997 AHSME Problems/Problem 14"

Latest revision as of 13:13, 5 July 2013

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 $\boxed{B}$ .

1997 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 == See also ==
 {{AHSME box|year=1997|num-b=13|num-a=15}}
+{{MAA Notice}}

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Difference between revisions of "1997 AHSME Problems/Problem 14"

Latest revision as of 13:13, 5 July 2013

Problem

Solution

See also