Difference between revisions of "2013 AMC 12B Problems/Problem 5"
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==Problem== | ==Problem== | ||
The average age of <math>33</math> fifth-graders is <math>11</math>. The average age of <math>55</math> of their parents is <math>33</math>. What is the average age of all of these parents and fifth-graders? | The average age of <math>33</math> fifth-graders is <math>11</math>. The average age of <math>55</math> of their parents is <math>33</math>. What is the average age of all of these parents and fifth-graders? | ||
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== See also == | == See also == | ||
{{AMC12 box|year=2013|ab=B|num-b=4|num-a=6}} | {{AMC12 box|year=2013|ab=B|num-b=4|num-a=6}} | ||
+ | {{AMC10 box|year=2013|ab=B|num-b=5|num-a=7}} | ||
+ | {{MAA Notice}} |
Latest revision as of 16:05, 27 January 2015
- The following problem is from both the 2013 AMC 12B #5 and 2013 AMC 10B #6, so both problems redirect to this page.
Problem
The average age of fifth-graders is . The average age of of their parents is . 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 12B (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 12 Problems and Solutions |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.