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

(Solution)
m (Edit)
Line 7: Line 7:
  
 
==Solution==
 
==Solution==
We have that A is x% greater than B, so <math>A=\frac{100+x}{100}(B)</math>. We solve for B. We get  
+
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  
  
 
<math>\frac{A}{B}=\frac{100+x}{100}</math>
 
<math>\frac{A}{B}=\frac{100+x}{100}</math>
Line 16: Line 16:
  
 
<math>\boxed{100(\frac{A-B}{B}) (\textbf{A})}=x</math>.
 
<math>\boxed{100(\frac{A-B}{B}) (\textbf{A})}=x</math>.
 +
 +
(Edited by TrueshotBarrage)
  
 
==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 18:32, 20 February 2014

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

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

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

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

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

(Edited by TrueshotBarrage)

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