2014 AMC 10B Problems/Problem 7: Difference between revisions
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== Solution 2== | == Solution 2== | ||
The question is basically asking the percentage increase from <math>B</math> to <math>A</math>. We know the formula for percentage increase is <math>\frac{\text{New-Original}}{\text{Original}}</math>. We know the new is <math>A</math> and the original is <math>B</math>. We also must multiple by <math>100</math> to get <math>x</math> out of it's fractional/percentage form. Therefore, the answer is <math>100(\frac{A-B}{B})</math> or <math>\boxed{\text{A}}</math>. | The question is basically asking the percentage increase from <math>B</math> to <math>A</math>. We know the formula for percentage increase is <math>\frac{\text{New-Original}}{\text{Original}}</math>. We know the new is <math>A</math> and the original is <math>B</math>. We also must multiple by <math>100</math> to get <math>x</math> out of it's fractional/percentage form. Therefore, the answer is <math>100\left(\frac{A-B}{B}\right)</math> or <math>\boxed{\text{A}}</math>. | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 22:18, 24 October 2020
Problem
Suppose
and A is
% greater than
. What is
?
Solution
We have that A is
greater than B, so
. We solve for
. We get
.
Solution 2
The question is basically asking the percentage increase from
to
. We know the formula for percentage increase is
. We know the new is
and the original is
. We also must multiple by
to get
out of it's fractional/percentage form. Therefore, the answer is
or
.
Video Solution
~savannahsolver
See Also
| 2014 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
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