Difference between revisions of "1997 AHSME Problems/Problem 14"
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<math> \textbf{(A)}\ 81\qquad\textbf{(B)}\ 84\qquad\textbf{(C)}\ 87\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 102 </math> | <math> \textbf{(A)}\ 81\qquad\textbf{(B)}\ 84\qquad\textbf{(C)}\ 87\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 102 </math> | ||
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| + | ==Solution== | ||
| + | |||
| + | Let <math>x</math> be the population in <math>1996</math>, and let <math>k</math> be the constant of proportionality. | ||
| + | |||
| + | If <math>n=1994</math>, then the difference in population between <math>1996</math> and <math>1994</math> is directly proportional to the population in <math>1995</math>. | ||
| + | |||
| + | Translating this sentence, <math>(x - 39) = k(60)</math> | ||
| + | |||
| + | Similarly, letting <math>n=1995</math> gives the sentence <math>(123 - 60) = kx</math> | ||
| + | |||
| + | Since <math>kx = 63</math>, we have <math>k = \frac{63}{x}</math> | ||
| + | |||
| + | Plugging this into the first equation, we have: | ||
| + | |||
| + | <math>(x - 39) = \frac{60\cdot 63}{x}</math> | ||
| + | |||
| + | <math>x - 39 = \frac{3780}{x}</math> | ||
| + | |||
| + | <math>x^2 - 39x - 3780 = 0</math> | ||
| + | |||
| + | <math>(x - 84)(x + 45) = 0</math> | ||
| + | |||
| + | Since <math>x>0</math>, we must have <math>x=84</math>, and the answer is <math>\boxed{B}</math>. | ||
| + | |||
| + | == See also == | ||
| + | {{AHSME box|year=1997|num-b=13|num-a=15}} | ||
| + | {{MAA Notice}} | ||
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 
and year 
is directly proportional to the population in year 
. If the populations in the years 
, 
, and 
were 
, 
, and 
, respectively, then the population in 
was
Solution
Let 
be the population in , and let 
be the constant of proportionality.
If 
, then the difference in population between and is directly proportional to the population in .
Translating this sentence, 
Similarly, letting 
gives the sentence 
Since 
, we have 
Plugging this into the first equation, we have:
Since 
, we must have 
, and the answer is 
.
See also
| 1997 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 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. 
