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

m (Problem)
m (Solution)
Line 13: Line 13:
 
<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 21:13, 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 (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS