Difference between revisions of "1956 AHSME Problems/Problem 13"
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== Solution == | == Solution == | ||
| − | Suppose that <math>x</math> is <math>p</math> percent less than <math>y</math>. Then <math>x = \frac{100 - p}{100}y</math>, so that <math>y - x = \frac{p}{100}y</math>. Solving for <math>p</math>, we get <math>p = \frac{100(y-x)}{y}</math>, or \boxed{\textbf{(C)}} | + | Suppose that <math>x</math> is <math>p</math> percent less than <math>y</math>. Then <math>x = \frac{100 - p}{100}y</math>, so that <math>y - x = \frac{p}{100}y</math>. Solving for <math>p</math>, we get <math>p = \frac{100(y-x)}{y}</math>, or <math>\boxed{\textbf{(C)}}</math>. |
== See Also == | == See Also == | ||
Latest revision as of 22:05, 13 January 2023
Given two positive integers 
and 
with 
. The percent that is less than is:
Solution
Suppose that is 
percent less than . Then 
, so that 
. Solving for , we get 
, or 
.
See Also
| 1956 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 | ||
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