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

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In <math>300</math> seconds (<math>5</math> minutes), Cagney will frost <math>\dfrac{300}{20} = 15</math> cupcakes, and Lacey will frost <math>\dfrac{300}{30} = 10</math> cupcakes. Therefore, working together they will frost <math>15 + 10 = \boxed{\textbf{(D)}\ 25}</math> cupcakes.
 
In <math>300</math> seconds (<math>5</math> minutes), Cagney will frost <math>\dfrac{300}{20} = 15</math> cupcakes, and Lacey will frost <math>\dfrac{300}{30} = 10</math> cupcakes. Therefore, working together they will frost <math>15 + 10 = \boxed{\textbf{(D)}\ 25}</math> cupcakes.
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'''Solution 3:'''
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Since Cagney frosts <math>3</math> cupcakes a minute, and Lacey frosts <math>2</math> cupcakes a minute, they together frost <math>3+2=5</math> cupcakes a minute. Therefore, in <math>5</math> minutes, they frost <math>5\times5 = 25 \Rightarrow \boxed{\textbf{(D)}}</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 22:39, 7 February 2015

The following problem is from both the 2012 AMC 12A #2 and 2012 AMC 10A #1, so both problems redirect to this page.

Problem

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

Solution 1:

Cagney can frost one in $20$ seconds, and Lacey can frost one in $30$ seconds. Working together, they can frost one in $\frac{20\cdot30}{20+30} = \frac{600}{50} = 12$ seconds. In $300$ seconds ($5$ minutes), they can frost $\boxed{\textbf{(D)}\ 25}$ cupcakes.

Solution 2:

In $300$ seconds ($5$ minutes), Cagney will frost $\dfrac{300}{20} = 15$ cupcakes, and Lacey will frost $\dfrac{300}{30} = 10$ cupcakes. Therefore, working together they will frost $15 + 10 = \boxed{\textbf{(D)}\ 25}$ cupcakes.

Solution 3:

Since Cagney frosts $3$ cupcakes a minute, and Lacey frosts $2$ cupcakes a minute, they together frost $3+2=5$ cupcakes a minute. Therefore, in $5$ minutes, they frost $5\times5 = 25 \Rightarrow \boxed{\textbf{(D)}}$

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
2012 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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

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