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Difference between revisions of "2011 AMC 12A Problems/Problem 4"

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If you want to simplify the problem even more, just imagine/assume that only <math>1</math> fifth grader existed. Then you can simply get rid of the variables.
 
If you want to simplify the problem even more, just imagine/assume that only <math>1</math> fifth grader existed. Then you can simply get rid of the variables.
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==Video Solution==
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https://youtu.be/3LWBLXzcSKo
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~savannahsolver
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|num-b=3|num-a=5|ab=A}}
 
{{AMC12 box|year=2011|num-b=3|num-a=5|ab=A}}
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{{AMC10 box|year=2011|ab=A|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:53, 22 December 2020

Problem

At an elementary school, the students in third grade, fourth grade, and fifth grade run an average of $12$, $15$, and $10$ minutes per day, respectively. There are twice as many third graders as fourth graders, and twice as many fourth graders as fifth graders. What is the average number of minutes run per day by these students?

$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ \frac{37}{3} \qquad \textbf{(C)}\ \frac{88}{7} \qquad \textbf{(D)}\ 13 \qquad \textbf{(E)}\ 14$

Solution

Let us say that there are $f$ fifth graders. According to the given information, there must be $2f$ fourth graders and $4f$ third graders. The average time run by each student is equal to the total amount of time run divided by the number of students. This gives us $\frac{12\cdot 4f + 15\cdot 2f + 10\cdot f}{4f + 2f + f} = \frac{88f}{7f} = \frac{88}{7} \Rightarrow \boxed{C}$


If you want to simplify the problem even more, just imagine/assume that only $1$ fifth grader existed. Then you can simply get rid of the variables.

Video Solution

https://youtu.be/3LWBLXzcSKo

~savannahsolver

See also

2011 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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
2011 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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

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