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Difference between revisions of "1956 AHSME Problems/Problem 13"

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Given two positive integers <math>x</math> and <math>y</math> with <math>x < y</math>. The percent that <math>x</math> is less than <math>y</math> is:
 
Given two positive integers <math>x</math> and <math>y</math> with <math>x < y</math>. The percent that <math>x</math> is less than <math>y</math> is:
  
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<math>\textbf{(A)}\ \frac{100(y-x)}{x}\qquad \textbf{(B)}\ \frac{100(x-y)}{x}\qquad \textbf{(C)}\ \frac{100(y-x)}{y}\qquad \\ \textbf{(D)}\ 100(y-x) \textbf{(E)}\ 100(x - y)</math>
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<math>\textbf{(A)}\ \frac{100(y-x)}{x}\qquad \textbf{(B)}\ \frac{100(x-y)}{x}\qquad \textbf{(C)}\ \frac{100(y-x)}{y}\qquad \\ \textbf{(D)}\ 100(y-x)\qquad \textbf{(E)}\ 100(x - y)</math>
  
  

Revision as of 21:04, 13 January 2023

Given two positive integers $x$ and $y$ with $x < y$. The percent that $x$ is less than $y$ is:

$\textbf{(A)}\ \frac{100(y-x)}{x}\qquad \textbf{(B)}\ \frac{100(x-y)}{x}\qquad \textbf{(C)}\ \frac{100(y-x)}{y}\qquad \\ \textbf{(D)}\ 100(y-x)\qquad \textbf{(E)}\ 100(x - y)$


Solution

Suppose that $x$ is $p$ percent less than $y$. Then $x = \frac{100 - p}{100}y$, so that $y - x = \frac{p}{100}y$. Solving for $p$, we get $p = \frac{100(y-x)}{y}$, so our answer is $\boxed{\textbf{(C)}}$ and we are done.

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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