1993 AIME Problems/Problem 3

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 $n\,$ .

$n\,$	$0\,$	$1\,$	$2\,$	$3\,$	$\dots\,$	$13\,$	$14\,$	$15\,$
$\mbox{number of contestants who caught }n\mbox{ fish}\,$	$9\,$	$5\,$	$7\,$	$23\,$	$\dots\,$	$5\,$	$2\,$	$1\,$

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 $y$ . Of those which caught 3 or more fish ( $y - 21$ did that), $x - \left(0(9) + 1(5) + 2(7)\right) = x - 19$ fish were caught. Since they averaged 6 fish, $6 = \frac{x - 19}{y - 21} \Longrightarrow x - 19 = 6y - 126$ . Similarily, of those which 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 $x = 943$ , the answer.

1993 AIME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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1993 AIME Problems/Problem 3

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