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>, so our answer is <math>\boxed{\textbf{(C)}}</math> and we are done.
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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 $x$ 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 $p = \frac{100(y-x)}{y}$, or \boxed{\textbf{(C)}}$.

See Also

1956 AHSME (ProblemsAnswer KeyResources)
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
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