# 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/decimal form. Therefore, the answer is $100\left(\frac{A-B}{B}\right)$ or $\boxed{\text{A}}$.

Without loss of generality, let $A = 125$ and $B = 100,$ forcing $x$ to be $25$. Plugging our values for $A$ and $B$ into these answer choices, we find that only $\boxed{\textbf{(A)}}$ returns $25$.

## Video Solution (CREATIVE THINKING)

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