Difference between revisions of "2013 AMC 12B Problems/Problem 5"
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| + | {{duplicate|[[2013 AMC 12B Problems|2013 AMC 12B #5]] and [[2013 AMC 10B Problems|2013 AMC 10B #6]]}} | ||
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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? | ||
<math>\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23.25 \qquad \textbf{(C)}\ 24.75 \qquad \textbf{(D)}\ 26.25 \qquad \textbf{(E)}\ 28</math> | <math>\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23.25 \qquad \textbf{(C)}\ 24.75 \qquad \textbf{(D)}\ 26.25 \qquad \textbf{(E)}\ 28</math> | ||
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| + | ==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>. | ||
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| + | == See also == | ||
| + | {{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 17:05, 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. 
