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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| + | {{duplicate|[[2008 AMC 12A Problems|2008 AMC 12A #3]] and [[2008 AMC 10A Problems/Problem 4|2008 AMC 10A #4]]}} | ||
==Problem== | ==Problem== | ||
| − | Suppose that <math>\ | + | 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>\ | + | <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} \ | + | 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}} | ||
| + | {{AMC10 box|year=2008|ab=A|num-b=3|num-a=5}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 21:35, 3 July 2013
- The following problem is from both the 2008 AMC 12A #3 and 2008 AMC 10A #4, so both problems redirect to this page.
Problem
Suppose that 
of 
bananas are worth as much as 
oranges. How many oranges are worth as much as 
of 
bananas?
Solution
If 
, then 
.
See Also
| 2008 AMC 12A (Problems • Answer Key • Resources) | |
| 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 | |
| 2008 AMC 10A (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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
