Difference between revisions of "2013 AMC 10B Problems/Problem 6"
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==Solution== | ==Solution== | ||
The sum of the ages of the fifth graders is <math>33 * 11</math>, while the sum of the ages of the parents is <math>55 * 33</math>. Therefore, the total sum of all their ages must be <math>2178</math>, and given <math>33 + 55 = 88</math> people in total, their average age is <math>\frac{2178}{88} = \frac{99}{4} = \boxed{\textbf{(C)}\ 24.75}</math>. | The sum of the ages of the fifth graders is <math>33 * 11</math>, while the sum of the ages of the parents is <math>55 * 33</math>. Therefore, the total sum of all their ages must be <math>2178</math>, and given <math>33 + 55 = 88</math> people in total, their average age is <math>\frac{2178}{88} = \frac{99}{4} = \boxed{\textbf{(C)}\ 24.75}</math>. | ||
| + | == See also == | ||
| + | {{AMC10 box|year=2013|ab=B|num-b=5|num-a=7}} | ||
Revision as of 17:00, 27 March 2013
Problem
The average age of 33 fifth-graders is 11. The average age of 55 of their parents is 33. What is the average age of all of these parents and fifth-graders?
Solution
The sum of the ages of the fifth graders is 
, while the sum of the ages of the parents is 
. Therefore, the total sum of all their ages must be 
, and given 
people in total, their average age is 
.
See also
| 2013 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 | ||
