Difference between revisions of "2014 AMC 10B Problems/Problem 7"
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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- | + | 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== | ||
+ | https://youtu.be/jj_rRTuWL14 | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2014|ab=B|num-b=6|num-a=8}} | {{AMC10 box|year=2014|ab=B|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 23:18, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.