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

## 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 $\frac{A}{B}=\frac{100+x}{100}$ $100\frac{A}{B}=100+x$ $100\left(\frac{A}{B}-1\right)=x$ $100\left(\frac{A-B}{B}\right)=x$. $\boxed{\left(\textbf{A}\right)}$

## Solution 2

The question is basically asking the percentage increase from $B$ to $A$. We know the formula for percentage increase is $\frac{\text{New-Original}}{\text{Original}}$. We know the new is $A$ and the original is $B$. We also must multiple by $100$ to get $x$ out of it's fractional/percentage form. Therefore, the answer is $100\left(\frac{A-B}{B}\right)$ or $\boxed{\text{A}}$.

~savannahsolver

## See Also

 2014 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 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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