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>, | + | 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)}}$. |
== See Also == | == See Also == |
Revision as of 21:04, 13 January 2023
Given two positive integers and with . The percent that is less than is:
Solution
Suppose that is percent less than . Then , so that . Solving for , we get , or \boxed{\textbf{(C)}}$.
See Also
1956 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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All AHSME Problems and Solutions |
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