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Difference between revisions of "2014 AMC 10B Problems/Problem 7"

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

Revision as of 17:18, 26 January 2016

Problem

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

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

Solution

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

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

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

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

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

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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