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

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Suppose <math>A>B>0</math> and A is <math>x</math>% greater than <math>B</math>. What is <math>x</math>?
 
Suppose <math>A>B>0</math> and A is <math>x</math>% greater than <math>B</math>. What is <math>x</math>?
  
<math> \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})</math>
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<math> \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)</math>
  
 
==Solution==
 
==Solution==

Revision as of 21:10, 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

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

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

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

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

See Also

2014 AMC 10B (ProblemsAnswer KeyResources)
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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