Difference between revisions of "2012 AMC 10A Problems/Problem 1"
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Cagney can frost one in 20 seconds, and Lacey can frost one in 30 seconds. Working together, they can frost one in <math>\frac{20*30}{20+30} = \frac{600}{50} = 12</math> seconds. In 300 seconds (5 minutes), they can frost <math>\boxed{\textbf{(D)}\ 25 \text{ cupcakes }}</math>. | Cagney can frost one in 20 seconds, and Lacey can frost one in 30 seconds. Working together, they can frost one in <math>\frac{20*30}{20+30} = \frac{600}{50} = 12</math> seconds. In 300 seconds (5 minutes), they can frost <math>\boxed{\textbf{(D)}\ 25 \text{ cupcakes }}</math>. | ||
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+ | {{AMC10 box|year=2012|ab=A|before=First Problem|num-a=2}} |
Revision as of 22:40, 8 February 2012
Problem 1
Cagney can frost a cupcake every 20 seconds and Lacey can frost a cupcake every 30 seconds. Working together, how many cupcakes can they frost in 5 minutes?
Solution
Cagney can frost one in 20 seconds, and Lacey can frost one in 30 seconds. Working together, they can frost one in seconds. In 300 seconds (5 minutes), they can frost .
See Also
2012 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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 |