Difference between revisions of "2012 AMC 10A Problems/Problem 1"
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− | == Problem | + | {{duplicate|[[2012 AMC 12A Problems|2012 AMC 12A #2]] and [[2012 AMC 10A Problems|2012 AMC 10A #1]]}} |
+ | |||
+ | == 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 == | + | == Solution 1 == |
+ | |||
+ | 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. | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | 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. | ||
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | 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> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/56N_nohFC0k | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | == See Also == | ||
+ | |||
+ | {{AMC10 box|year=2012|ab=A|before=First Problem|num-a=2}} | ||
+ | {{AMC12 box|year=2012|ab=A|num-b=1|num-a=3}} | ||
− | + | [[Category:Introductory Algebra Problems]] | |
+ | {{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?
Solution 1
Cagney can frost one in seconds, and Lacey can frost one in seconds. Working together, they can frost one in seconds. In seconds ( minutes), they can frost cupcakes.
Solution 2
In seconds ( minutes), Cagney will frost cupcakes, and Lacey will frost cupcakes. Therefore, working together they will frost cupcakes.
Solution 3
Since Cagney frosts cupcakes a minute, and Lacey frosts cupcakes a minute, they together frost cupcakes a minute. Therefore, in minutes, they frost
Video Solution
~savannahsolver
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.