Difference between revisions of "1951 AHSME Problems/Problem 1"
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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> | ||
| + | == 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> 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 50p box|year=1951|before=First Question|num-a=2}} | |
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 12:19, 5 July 2013
Problem
The percent that 
is greater than 
is:
Solution
is the amount by which is greater than . We divide this by to get the percent by which is increased in the form of a decimal, and then multiply by 
to make it a percentage. Therefore, the answer is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
