Difference between revisions of "2012 AMC 10A Problems/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 == | == Problem == | ||
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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> | ||
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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 (CREATIVE THINKING)== | ||
| + | https://youtu.be/uhy-Cu0xXkg | ||
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| + | ~Education, the Study of Everything | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/56N_nohFC0k | ||
| + | |||
| + | ~savannahsolver | ||
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| + | == See Also == | ||
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| + | {{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 12:57, 1 July 2023
- The following problem is from both the 2012 AMC 12A #2 and 2012 AMC 10A #1, so both problems redirect to this page.
Contents
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 (CREATIVE THINKING)
~Education, the Study of Everything
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. 
