1993 AIME Problems/Problem 3
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[hide]Problem
The table below displays some of the results of last summer's Frostbite Falls Fishing Festival, showing how many contestants caught fish for various values of
.

In the newspaper story covering the event, it was reported that
- (a) the winner caught
fish;
- (b) those who caught
or more fish averaged
fish each;
- (c) those who caught
or fewer fish averaged
fish each.
What was the total number of fish caught during the festival?
Solution 1
Suppose that the number of fish is and the number of contestants is
. The
fishers that caught
or more fish caught a total of
fish. Since they averaged
fish,

Similarily, those who caught or fewer fish averaged
fish per person, so

Solving the two equation system, we find that and
, the answer.
Solution 2
Let be the total number of fish caught by the contestants who didn't catch
, or
fish and let
be the number of contestants who didn't catch
, or
fish. From
, we know that
. From
we have
. Using these two equations gets us
. Plug this back into the equation to get
. Thus, the total number of fish caught is
- Heavytoothpaste
See also
1993 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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