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

 
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== Problem 1 ==
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{{duplicate|[[2012 AMC 12A Problems|2012 AMC 12A #2]] and [[2012 AMC 10A Problems|2012 AMC 10A #1]]}}
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== 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?
 
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?
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<math> \textbf{(A)}\ 10\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 30 </math>
 
<math> \textbf{(A)}\ 10\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 30 </math>
  
== Solution ==
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== Solution 1 ==
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Cagney can frost one in <math>20</math> seconds, and Lacey can frost one in <math>30</math> seconds. Working together, they can frost one in <math>\frac{20\cdot30}{20+30} = \frac{600}{50} = 12</math> seconds. In <math>300</math> seconds (<math>5</math> minutes), they can frost <math>\boxed{\textbf{(D)}\ 25}</math> cupcakes.
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== Solution 2 ==
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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.
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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>
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==Video Solution==
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https://youtu.be/56N_nohFC0k
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~savannahsolver
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== See Also ==
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{{AMC10 box|year=2012|ab=A|before=First Problem|num-a=2}}
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{{AMC12 box|year=2012|ab=A|num-b=1|num-a=3}}
  
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>\qquad\textbf{(D)}\ 25</math> cupcakes.
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 21:19, 14 September 2020

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 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 $5\times5 = 25 \Rightarrow \boxed{\textbf{(D)}}$

Video Solution

https://youtu.be/56N_nohFC0k

~savannahsolver

See Also

2012 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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