Difference between revisions of "1951 AHSME Problems/Problem 1"
(New page: == Problem == The percent that <math>M</math> is greater than <math>N</math> is: <math>(\mathrm{A})\ \frac{100(M-N)}{M} \qquad (\mathrm{B})\ \frac{100(M-N)}{N} \qquad (\mathrm{C})\ \frac{...) |
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<math>(\mathrm{A})\ \frac{100(M-N)}{M} \qquad (\mathrm{B})\ \frac{100(M-N)}{N} \qquad (\mathrm{C})\ \frac{M-N}{N} \qquad (\mathrm{D})\ \frac{M-N}{N} \qquad (\mathrm{E})\ \frac{100(M+N)}{N}</math> | <math>(\mathrm{A})\ \frac{100(M-N)}{M} \qquad (\mathrm{B})\ \frac{100(M-N)}{N} \qquad (\mathrm{C})\ \frac{M-N}{N} \qquad (\mathrm{D})\ \frac{M-N}{N} \qquad (\mathrm{E})\ \frac{100(M+N)}{N}</math> | ||
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== Solution == | == Solution == | ||
<math>M-N</math> is the amount by which <math>M</math> is greater than <math>N</math>. We divide this by <math>N</math> to get the percent by which <math>N</math> increased expressed as a decimal, and then multiply by <math>100</math> to make it a percentage. Therefore, the answer is <math>\mathrm{B}</math>. | <math>M-N</math> is the amount by which <math>M</math> is greater than <math>N</math>. We divide this by <math>N</math> to get the percent by which <math>N</math> increased expressed as a decimal, and then multiply by <math>100</math> to make it a percentage. Therefore, the answer is <math>\mathrm{B}</math>. | ||
Revision as of 03:05, 9 February 2011
Problem
The percent that 
is greater than 
is:
DICKKSSS
Solution
is the amount by which is greater than . We divide this by to get the percent by which increased expressed as a decimal, and then multiply by 
to make it a percentage. Therefore, the answer is 
.
