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

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The question is basically asking the percentage increase from <math>B</math> to <math>A</math>. We know the formula for percentage increase is <math>\frac{\text{New-Original}}{\text{Original}}</math>. We know the new is <math>A</math> and the original is <math>B</math>. We also must multiple by <math>100</math> to get <math>x</math> out of it's fractional/percentage form. Therefore, the answer is <math>100\left(\frac{A-B}{B}\right)</math> or <math>\boxed{\text{A}}</math>.
 
The question is basically asking the percentage increase from <math>B</math> to <math>A</math>. We know the formula for percentage increase is <math>\frac{\text{New-Original}}{\text{Original}}</math>. We know the new is <math>A</math> and the original is <math>B</math>. We also must multiple by <math>100</math> to get <math>x</math> out of it's fractional/percentage form. Therefore, the answer is <math>100\left(\frac{A-B}{B}\right)</math> or <math>\boxed{\text{A}}</math>.
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==Solution 3 (Answer Choices)==
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Without loss of generality, let <math>A = 125</math> and <math>B = 100,</math> so <math>x = 25</math>. Plugging <math>A</math> and <math>B</math> into these answer choices, we find that only <math>\boxed{\text{A}}</math> returns <math>25</math>.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 15:21, 24 April 2022

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

See Also

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

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