2014 AMC 10B Problems/Problem 7: Difference between revisions
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==Solution== | ==Solution== | ||
We have that A is x% greater than B, so <math>A=\frac{100+x}{100}(B)</math>. We solve for B. We get | |||
<math>\frac{A}{B}=\frac{100+x}{100}</math> | |||
<math>100\frac{A}{B}=100+x</math> | |||
<math>100(\frac{A}{B}-1)=x</math> | |||
<math>\boxed{100(\frac{A-B}{B}) (\textbf{A})}=x</math>. | |||
==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}} | ||
Revision as of 15:30, 20 February 2014
Problem
Suppose 
and A is 
% greater than 
. What is ?
Solution
We have that A is x% greater than B, so 
. We solve for B. We get
.
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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