Difference between revisions of "2007 AMC 12B Problems/Problem 4"
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== Problem == | == Problem == | ||
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| + | At Frank's Fruit Market, <math>3</math> bananas cost as much as <math>2</math> apples, and <math>6</math> apples cost as much as <math>4</math> oranges. How many oranges cost as much as <math>18</math> bananas? | ||
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| + | <math>\textbf{(A) } 6 \qquad\textbf{(B) } 8 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } 18</math> | ||
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| + | == Solution == | ||
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| + | <math>3</math> bananas cost as much as <math>2</math> apples, so <math>18</math> bananas cost as much as <math>12</math> apples. Since <math>6</math> apples cost as much as <math>4</math> oranges, <math>12</math> apples cost as much as <math>8</math> oranges, giving <math>(B)</math>. | ||
==See Also== | ==See Also== | ||
Latest revision as of 21:56, 6 June 2025
Problem
At Frank's Fruit Market, 
bananas cost as much as 
apples, and 
apples cost as much as 
oranges. How many oranges cost as much as 
bananas?
Solution
bananas cost as much as apples, so bananas cost as much as 
apples. Since apples cost as much as oranges, apples cost as much as 
oranges, giving 
.
See Also
| 2007 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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
