1993 AIME Problems/Problem 3

Revision as of 17:04, 26 March 2007 by Azjps (talk | contribs) (Solution: -of)

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, 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 $x = 943$, the answer.

See also

1993 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions