Difference between revisions of "2014 AMC 10B Problems/Problem 7"
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Suppose <math>A>B>0</math> and A is <math>x</math>% greater than <math>B</math>. What is <math>x</math>? | Suppose <math>A>B>0</math> and A is <math>x</math>% greater than <math>B</math>. What is <math>x</math>? | ||
| − | <math> \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})</math> | + | <math> \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)</math> |
==Solution== | ==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> | <math>\frac{A}{B}=\frac{100+x}{100}</math> | ||
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<math>100\frac{A}{B}=100+x</math> | <math>100\frac{A}{B}=100+x</math> | ||
| − | <math>100(\frac{A}{B}-1)=x</math> | + | <math>100\left(\frac{A}{B}-1\right)=x</math> |
| − | <math>\ | + | <math>100\left(\frac{A-B}{B}\right)=x</math>. <math>\boxed{\left(\textbf{A}\right)}</math> |
| − | ( | + | == Solution 2== |
| + | |||
| + | 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/decimal 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)== | ||
| + | |||
| + | Without loss of generality, let <math>A = 125</math> and <math>B = 100,</math> forcing <math>x</math> to be <math>25</math>. Plugging our values for <math>A</math> and <math>B</math> into these answer choices, we find that only <math>\boxed{\textbf{(A)}}</math> returns <math>25</math>. | ||
| + | |||
| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/NTqYTNTU130 | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | |||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/jj_rRTuWL14 | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2014|ab=B|num-b=6|num-a=8}} | {{AMC10 box|year=2014|ab=B|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 11:03, 2 July 2023
Contents
Problem
Suppose 
and A is 
% greater than 
. What is ?
Solution
We have that A is 
greater than B, so 
. We solve for . We get
. 
Solution 2
The question is basically asking the percentage increase from to 
. We know the formula for percentage increase is 
. We know the new is and the original is . We also must multiple by 
to get out of it's fractional/decimal form. Therefore, the answer is 
or 
.
Solution 3 (Answer Choices)
Without loss of generality, let 
and 
forcing to be 
. Plugging our values for and into these answer choices, we find that only 
returns .
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
