Difference between revisions of "2008 AMC 12A Problems/Problem 3"

(New page: ==Problem== Suppose that <math>\frac {2}{3}</math> of <math>10</math> bananas are worth as much as <math>8</math> oranges. How many oranges are worth as much is <math>\frac {1}{2}</math> ...)
 
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==Problem==  
 
==Problem==  
Suppose that <math>\frac {2}{3}</math> of <math>10</math> bananas are worth as much as <math>8</math> oranges. How many oranges are worth as much is <math>\frac {1}{2}</math> of <math>5</math> bananas?
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Suppose that <math>\tfrac{2}{3}</math> of <math>10</math> bananas are worth as much as <math>8</math> oranges. How many oranges are worth as much as <math>\tfrac{1}{2}</math> of <math>5</math> bananas?
  
<math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ \frac {5}{2} \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ \frac {7}{2} \qquad \textbf{(E)}\ 4</math>
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<math>\mathrm{(A)}\ 2\qquad\mathrm{(B)}\ \frac{5}{2}\qquad\mathrm{(C)}\ 3\qquad\mathrm{(D)}\ \frac{7}{2}\qquad\mathrm{(E)}\ 4</math>
  
 
==Solution==  
 
==Solution==  
If <math>\frac {2}{3} \cdot 10 \ \text{bananas} = 8 \ \text{oranges}</math>, then <math>\frac{1}{2} \cdot 5 \ \text{bananas} = \left(\frac{1}{2} \cdot 5 \ \text{bananas}\right) \cdot \left(\frac{8 \ \text{oranges}}{\frac {2}{3} \cdot 10 \ \text{bananas}}\right) = 3 \ \text{oranges} \Rightarrow C</math>.  
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If <math>\frac{2}{3}\cdot10\ \text{bananas}=8\ \text{oranges}</math>, then <math>\frac{1}{2}\cdot5\ \text{bananas}=\left(\frac{1}{2}\cdot 5\ \text{bananas}\right)\cdot\left(\frac{8\ \text{oranges}}{\frac{2}{3}\cdot10\ \text{bananas}}\right)=3\ \text{oranges}\Longrightarrow\mathrm{(C)}</math>.  
  
 
==See Also==  
 
==See Also==  
 
{{AMC12 box|year=2008|ab=A|num-b=2|num-a=4}}
 
{{AMC12 box|year=2008|ab=A|num-b=2|num-a=4}}

Revision as of 22:50, 25 April 2008

Problem

Suppose that $\tfrac{2}{3}$ of $10$ bananas are worth as much as $8$ oranges. How many oranges are worth as much as $\tfrac{1}{2}$ of $5$ bananas?

$\mathrm{(A)}\ 2\qquad\mathrm{(B)}\ \frac{5}{2}\qquad\mathrm{(C)}\ 3\qquad\mathrm{(D)}\ \frac{7}{2}\qquad\mathrm{(E)}\ 4$

Solution

If $\frac{2}{3}\cdot10\ \text{bananas}=8\ \text{oranges}$, then $\frac{1}{2}\cdot5\ \text{bananas}=\left(\frac{1}{2}\cdot 5\ \text{bananas}\right)\cdot\left(\frac{8\ \text{oranges}}{\frac{2}{3}\cdot10\ \text{bananas}}\right)=3\ \text{oranges}\Longrightarrow\mathrm{(C)}$.

See Also

2008 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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