Difference between revisions of "1995 AHSME Problems/Problem 4"
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We are given: <math>M=\frac{3Q}{10}</math>, <math>Q=\frac{P}{5}</math>, <math>N=\frac{P}{2}</math>. We want M in terms of N, so we substitute N into everything: | We are given: <math>M=\frac{3Q}{10}</math>, <math>Q=\frac{P}{5}</math>, <math>N=\frac{P}{2}</math>. We want M in terms of N, so we substitute N into everything: | ||
− | <math>\frac{2}{5}N=\frac{ | + | <math>\frac{2}{5}N=\frac{P}{5}=Q</math> |
<math>M=\frac{3N}{25}</math> | <math>M=\frac{3N}{25}</math> |
Revision as of 10:07, 29 April 2008
Problem
If is of , is of , and is of , then
Solution
We are given: , , . We want M in terms of N, so we substitute N into everything:
See also
1995 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |