Difference between revisions of "2011 AMC 12A Problems/Problem 4"
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== Solution == | == Solution == | ||
| − | Let us say that there are <math>f</math> fifth graders. According to the given information, there must be <math>2f</math> fourth graders and <math>4f</math> 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 <math>\frac{12\cdot 4f + 15\cdot 2f + 10\cdot f}{4f + 2f + f} = \frac{88f}{7f} = | + | Let us say that there are <math>f</math> fifth graders. According to the given information, there must be <math>2f</math> fourth graders and <math>4f</math> 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 <math>\frac{12\cdot 4f + 15\cdot 2f + 10\cdot f}{4f + 2f + f} = \frac{88f}{7f} = \frac{88}{7} \Rightarrow \boxed{C}</math> |
== 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}} | ||
Revision as of 17:07, 10 February 2011
Problem
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 us say that there are 
fifth graders. According to the given information, there must be 
fourth graders and 
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 
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 | |
