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2014 AMC 10B Problems/Problem 7: Difference between revisions

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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 $A>B>0$ and A is $x$% greater than $B$. What is $x$?

$\textbf {(A) } 100\left(\frac{A-B}{B}\right) \qquad \textbf {(B) } 100\left(\frac{A+B}{B}\right) \qquad \textbf {(C) } 100\left(\frac{A+B}{A}\right)\qquad \textbf {(D) } 100\left(\frac{A-B}{A}\right) \qquad \textbf {(E) } 100\left(\frac{A}{B}\right)$

Solution

We have that A is $x\%$ greater than B, so $A=\frac{100+x}{100}(B)$. We solve for $x$. We get

$\frac{A}{B}=\frac{100+x}{100}$

$100\frac{A}{B}=100+x$

$100\left(\frac{A}{B}-1\right)=x$

$100\left(\frac{A-B}{B}\right)=x$. $\boxed{\left(\textbf{A}\right)}$

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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