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Difference between revisions of "2012 AMC 12B Problems/Problem 4"

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==Solution==
 
==Solution==
  
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So convert everything to dollars; 400(euros) x 1.3 = 520 dollars. now, 520 divided by 500 = 1.04, which means a value that is 4% greater; B.
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The ratio <math>\frac{400 \text{ euros}}{500 \text{ dollars}}</math> can be simplified using conversion factors:<cmath>\frac{400 \text{ euros}}{500 \text{ dollars}} \cdot \frac{1.3 \text{ dollars}}{1 \text{ euro}} = \frac{520}{500} = 1.04</cmath> which means the money is greater by <math>\boxed{ \textbf{(B)} \ 4 }</math> percent.
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==Solution 2==
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If we divide each of Etienne's and Diana's values by <math>100</math>, the problem stays the same. Then, Etienne has <math>1.3</math> times the amount of money Diana has, so Etienne has <math>5.2</math> dollars. Since <math>\dfrac{5.2}{5} = 1.04</math>, Etienne has <math>\boxed{ \textbf{(B)} \ 4 }</math> percent more money than Diana. ~Extremelysupercooldude
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2012|ab=B|num-b=3|num-a=5}}
 
{{AMC12 box|year=2012|ab=B|num-b=3|num-a=5}}
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{{MAA Notice}}

Latest revision as of 06:18, 29 June 2023

Problem

Suppose that the euro is worth 1.3 dollars. If Diana has 500 dollars and Etienne has 400 euros, by what percent is the value of Etienne's money greater that the value of Diana's money?

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6.5\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 13$

Solution

The ratio $\frac{400 \text{ euros}}{500 \text{ dollars}}$ can be simplified using conversion factors:\[\frac{400 \text{ euros}}{500 \text{ dollars}} \cdot \frac{1.3 \text{ dollars}}{1 \text{ euro}} = \frac{520}{500} = 1.04\] which means the money is greater by $\boxed{ \textbf{(B)} \ 4 }$ percent.

Solution 2

If we divide each of Etienne's and Diana's values by $100$, the problem stays the same. Then, Etienne has $1.3$ times the amount of money Diana has, so Etienne has $5.2$ dollars. Since $\dfrac{5.2}{5} = 1.04$, Etienne has $\boxed{ \textbf{(B)} \ 4 }$ percent more money than Diana. ~Extremelysupercooldude

See Also

2012 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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 12 Problems and Solutions

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