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 $\boxed{\text{A}}$ .

Solution 3 (Answer Choices)

Without loss of generality, let $A = 125$ and $B = 100,$ so $x = 25$ . Plugging $A$ and $B$ into these answer choices, we find that only $\boxed{\text{A}}$ returns $25$ .

Video Solution

https://youtu.be/jj_rRTuWL14

~savannahsolver

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

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

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