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

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

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

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

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

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

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: Unable to save thumbnail to destination

@@ Line 7: / Line 7: @@
 ==Solution==
-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
+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>

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

Revision as of 17:18, 26 January 2016

Problem

Solution

See Also