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 $n\,$ .

$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline n & 0 & 1 & 2 & 3 & \dots & 13 & 14 & 15 \\ \hline \text{number of contestants who caught} \ n \ \text{fish} & 9 & 5 & 7 & 23 & \dots & 5 & 2 & 1 \\ \hline \end{array}$

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 1

Suppose that the number of fish is $x$ and the number of contestants is $y$ . The $y-(9+5+7)=y-21$ fishers that caught $3$ or more fish caught a total of $x - \left(0\cdot(9) + 1\cdot(5) + 2\cdot(7)\right) = x - 19$ fish. Since they averaged $6$ fish,

$6 = \frac{x - 19}{y - 21} \Longrightarrow x - 19 = 6y - 126.$

Similarily, those who 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.

Solution 2

Let $f$ be the total number of fish caught by the contestants who didn't catch $0, 1, 2, 3, 13, 14$ , or $15$ fish and let $a$ be the number of contestants who didn't catch $0, 1, 2, 3, 13, 14$ , or $15$ fish. From $\text{(b)}$ , we know that $\frac{69+f+65+28+15}{a+31}=6\implies f+177=6(a+31)$ . From $\text{(c)}$ we have $\frac{f+69+14+5}{a+44}=5\implies f+88=5(a+44)$ .

1993 AIME (Problems • Answer Key • Resources)
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1993 AIME Problems/Problem 3

Contents

Problem

Solution 1

Solution 2

See also