Difference between revisions of "1986 AHSME Problems/Problem 20"
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| − | We see that <math>x</math> is multiplied by <math>\frac{100+p}{100}</math> when it is increased by <math>p | + | We see that <math>x</math> is multiplied by <math>\frac{100+p}{100}</math> when it is increased by <math>p</math>%. Therefore, <math>y</math> is multiplied by <math>\frac{100}{100+p}</math>, and so it is decreased by <math>\frac{p}{100+p}</math> times itself. Therefore, it is decreased by <math>\frac{100p}{100+p}</math>%, and so the answer is <math>\boxed{E}</math>. |
== See also == | == See also == | ||
Latest revision as of 18:50, 9 October 2017
Problem
Suppose 
and 
are inversely proportional and positive. If increases by 
, then decreases by
Solution
We see that is multiplied by 
when it is increased by 
%. Therefore, is multiplied by 
, and so it is decreased by 
times itself. Therefore, it is decreased by 
%, and so the answer is 
.
See also
| 1986 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
