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Difference between revisions of "2011 AMC 10A Problems/Problem 5"
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=== Solution ===
 
=== Solution ===
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Without loss of generality, let there be 1 fifth grader. It follows that there are 2 fourth graders and 4 third graders.  
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Let there be <math>x</math> fifth graders. It follows that there are <math>2x</math> fourth graders and <math>4x</math> third graders.  
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We have <math>\frac{(1)(10)+(2)(15)+(4)(12)}{1+2+4} = \boxed{\textbf{(C)}\frac{88}{7}}</math>.
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We have <math>\frac{(1x)(10)+(2x)(15)+(4x)(12)}{1x+2x+4x} = \boxed{\textbf{(C)}\frac{88}{7}}</math>.

Revision as of 21:31, 18 February 2011

Problem 5

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 there be $x$ fifth graders. It follows that there are $2x$ fourth graders and $4x$ third graders. We have $\frac{(1x)(10)+(2x)(15)+(4x)(12)}{1x+2x+4x} = \boxed{\textbf{(C)}\frac{88}{7}}$.

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