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

Revision as of 17:20, 18 July 2011

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
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

@@ Line 5: / Line 5: @@
 == 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> is increased in the form of a decimal, and then multiply by <math>100</math> to make it a percentage. Therefore, the answer is <math>\boxed{\mathrm{(B)}\ \dfrac{100(M-N)}{N}}</math>.
+== See also ==
+{{AHSME box|year=1951|before=First Question|num-a=2}}

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

Revision as of 17:20, 18 July 2011

Problem

Solution

See also