Difference between revisions of "2013 AMC 12B Problems/Problem 5"
Turkeybob777 (talk | contribs) |
m (→See also: added amc 10 template) |
||
(5 intermediate revisions by 4 users not shown) | |||
Line 1: | Line 1: | ||
+ | {{duplicate|[[2013 AMC 12B Problems|2013 AMC 12B #5]] and [[2013 AMC 10B Problems|2013 AMC 10B #6]]}} | ||
+ | |||
==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? | ||
Line 6: | Line 8: | ||
==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 == | ||
+ | {{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.