Difference between revisions of "2014 AMC 10B Problems/Problem 7"
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<math>100\frac{A}{B}=100+x</math> | <math>100\frac{A}{B}=100+x</math> | ||
− | <math>100(\frac{A}{B}-1)=x</math> | + | <math>100\left(\frac{A}{B}-1\right)=x</math> |
− | <math>100(\frac{A-B}{B})=x</math>. <math>\boxed{(\textbf{A})}</math> | + | <math>100\left(\frac{A-B}{B}\right)=x</math>. <math>\boxed{\left(\textbf{A}\right)}</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 20:13, 13 October 2018
Problem
Suppose and A is % greater than . What is ?
Solution
We have that A is greater than B, so . We solve for . 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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