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Difference between revisions of "2012 AMC 10A Problems/Problem 1"

(Undo revision 44567 by Minimario (talk))
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== Solution ==
 
== Solution ==
  
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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{\qquad\textbf{(D)}\ 25}</math> cupcakes.
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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>.

Revision as of 20:37, 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?

$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 30$

Solution

Cagney can frost one in 20 seconds, and Lacey can frost one in 30 seconds. Working together, they can frost one in $\frac{20*30}{20+30} = \frac{600}{50} = 12$ seconds. In 300 seconds (5 minutes), they can frost $\boxed{\textbf{(D)}\ 25 \text{ cupcakes }}$.

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