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} = \frac{88}{7} \Rightarrow \boxed{C}</math> | ||
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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. | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/3LWBLXzcSKo | ||
| + | |||
| + | ~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}} | ||
| + | {{AMC10 box|year=2011|ab=A|num-b=4|num-a=6}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 17:53, 22 December 2020
Contents
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 
If you want to simplify the problem even more, just imagine/assume that only 
fifth grader existed. Then you can simply get rid of the variables.
Video Solution
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
