Difference between revisions of "2011 AMC 10A Problems/Problem 5"
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<math> \textbf{(A)}\ 12 \qquad\textbf{(B)}\ \frac{37}{3} \qquad\textbf{(C)}\ \frac{88}{7} \qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 14 </math> | <math> \textbf{(A)}\ 12 \qquad\textbf{(B)}\ \frac{37}{3} \qquad\textbf{(C)}\ \frac{88}{7} \qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 14 </math> | ||
| − | + | == Solution == | |
Let there be <math>x</math> fifth graders. It follows that there are <math>2x</math> fourth graders and <math>4x</math> third graders. | Let there be <math>x</math> fifth graders. It follows that there are <math>2x</math> fourth graders and <math>4x</math> third graders. | ||
We have <math>\frac{(1x)(10)+(2x)(15)+(4x)(12)}{1x+2x+4x} = \boxed{\textbf{(C)}\frac{88}{7}}</math>. | We have <math>\frac{(1x)(10)+(2x)(15)+(4x)(12)}{1x+2x+4x} = \boxed{\textbf{(C)}\frac{88}{7}}</math>. | ||
| + | == See Also == | ||
| + | {{AMC10 box|year=2011|ab=A|num-b=4|num-a=6}} | ||
Revision as of 10:55, 8 May 2011
Problem 5
At an elementary school, the students in third grade, fourth grade, and fifth grade run an average of 
, 
, and 
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?
Solution
Let there be 
fifth graders. It follows that there are 
fourth graders and 
third graders.
We have 
.
See Also
| 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 | ||
