Difference between revisions of "1951 AHSME Problems/Problem 1"
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== Solution == | == 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 | + | <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 == | ||
+ | {{AHSME box|year=1951|before=First Question|num-a=2}} |
Revision as of 18:20, 18 July 2011
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 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 |