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1993 AIME Problems/Problem 3

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Problem

The table below displays some of the results of last summer's Frostbite Falls Fishing Festival, showing how many contestants caught n\, fish for various values of n\,.

\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline n & 0 & 1 & 2 & 3 & \dots & 13 & 14 & 15 \\\hline \...

In the newspaper story covering the event, it was reported that

(a) the winner caught 15 fish;
(b) those who caught 3 or more fish averaged 6 fish each;
(c) those who caught 12 or fewer fish averaged 5 fish each.

What was the total number of fish caught during the festival?

Solution

Suppose that the number of fish is x and the number of contestants is y. The y-21 fishers that caught 3 or more fish caught a total of x - \left(0(9) + 1(5) + 2(7)\right) = x - 19 fish. Since they averaged 6 fish,
6 = \frac{x - 19}{y - 21} \Longrightarrow x - 19 = 6y - 126.
Similarily, those whom caught 12 or fewer fish averaged 5 fish per person, so
5 = \frac{x - (13(5) + 14(2) + 15(1))}{y - 8} = \frac{x - 108}{y - 8} \Longrightarrow x - 108 = 5y - 40.
Solving the two equation system, we find that y = 175 and x = \boxed{943}, the answer.

See also

1993 AIME (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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