Track / Stage 3 / 114 of 260 #114 of 2000

Problem 114

AMC 10/12, early questions

Combinatorics Difficulty 3.4 Find the answer

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?

Your answer

Official solution

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.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.