1995 AHSME Problems/Problem 4
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Contents
[hide]Problem
If is of , is of , and is of , then
Solution 1
We are given: , , . We want M in terms of N, so we substitute N into everything:
Solution 2
Alternatively, picking an arbitrary value for of , we find that .
We find that , meaning , giving .
Finally, since is of , we have .
Thus, and , so their ratio
This method does not prove that the answer must be constant, but it proves that if the answer is a constant, it must be .
See also
1995 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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