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: | ||
− | <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> | + | <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> |
== 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 <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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.