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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{...)
 
(→‎Problem: fix answer choice)
 
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The percent that <math>M</math> is greater than <math>N</math> is:
 
The percent that <math>M</math> is greater than <math>N</math> is:
  
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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>
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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}{M} \qquad (\mathrm{E})\ \frac{100(M+N)}{N}</math>
  
 
== Solution ==
 
== Solution ==
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<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>.
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<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>.
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== See Also ==
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{{AHSME 50p box|year=1951|before=First Question|num-a=2}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 14:40, 13 November 2024

Problem

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

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

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 $\boxed{\mathrm{(B)}\ \dfrac{100(M-N)}{N}}$.


See Also

1951 AHSC (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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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