1951 AHSME Problems/Problem 1

Problem

The percent that $M$ is greater than $N$ is:

$(\mathrm{A})\ \frac{100(M-N)}{M} \qquad (\mathrm{B})\ \frac{100(M-N)}{N} \qquad (\mathrm{C})\ \frac{M-N}{N} \qquad (\mathrm{D})\ \frac{M-N}{N} \qquad (\mathrm{E})\ \frac{100(M+N)}{N}$

Solution

$M-N$ is the amount by which $M$ is greater than $N$ . We divide this by $N$ to get the percent by which $N$ is increased in the form of a decimal, and then multiply by $100$ to make it a percentage. Therefore, the answer is $\boxed{\mathrm{(B)}\ \dfrac{100(M-N)}{N}}$ .

1951 AHSME (Problems • Answer Key • Resources)
Preceded by First Question	Followed by Problem 2
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1951 AHSME Problems/Problem 1

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