Difference between revisions of "1951 AHSME Problems/Problem 1"

Revision as of 07:45, 29 April 2012

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 AHSC (Problems • Answer Key • Resources)
Preceded by First Question	Followed by Problem 2
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All AHSME Problems and Solutions

@@ Line 8: / Line 8: @@
-== See also ==
+== See Also ==
-{{AHSME box|year=1951|before=First Question|num-a=2}}
+{{AHSME 50p box|year=1951|before=First Question|num-a=2}}
+[[Category:Introductory Algebra Problems]]

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Difference between revisions of "1951 AHSME Problems/Problem 1"

Revision as of 07:45, 29 April 2012

Problem

Solution

See Also