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

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.

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 (CREATIVE THINKING)

https://youtu.be/uhy-Cu0xXkg

~Education, the Study of Everything

Video Solution

https://youtu.be/56N_nohFC0k

~savannahsolver

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.

@@ Line 1: / Line 1: @@
-== Problem 1 ==
+{{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?
@@ Line 5: / Line 7: @@
 <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 (CREATIVE THINKING)==
+https://youtu.be/uhy-Cu0xXkg
+~Education, the Study of Everything
+==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}}
-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.
+[[Category:Introductory Algebra Problems]]
+{{MAA Notice}}

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