Difference between revisions of "1993 AIME Problems/Problem 3"
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== Solution == | == Solution == | ||
| − | {{ | + | Suppose that the number of fish is <math>x</math> and the number of contestants is <math>y</math>. Of those which caught 3 or more fish (<math>y - 21</math> did that), <math>x - \left(0(9) + 1(5) + 2(7)\right) = x - 19</math> fish were caught. Since they averaged 6 fish, <math>6 = \frac{x - 19}{y - 21} \Longrightarrow x - 19 = 6y - 126</math>. Similarily, of those which caught 12 or fewer fish averaged 5 fish per person, so <math>5 = \frac{x - (13(5) + 14(2) + 15(1))}{y - 8} = \frac{x - 108}{y - 8} \Longrightarrow x - 108 = 5y - 40</math>. Solving the two equation system, we find that <math>y = 175</math> and <math>x = 943</math>, the answer. |
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== See also == | == See also == | ||
{{AIME box|year=1993|num-b=2|num-a=4}} | {{AIME box|year=1993|num-b=2|num-a=4}} | ||
Revision as of 16:04, 26 March 2007
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 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 
and the number of contestants is 
. Of those which caught 3 or more fish (
did that), 
fish were caught. Since they averaged 6 fish, 
. Similarily, of those which caught 12 or fewer fish averaged 5 fish per person, so 
. Solving the two equation system, we find that 
and 
, the answer.
See also
| 1993 AIME (Problems • Answer Key • Resources) | ||
| 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 | ||
