Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2013 AMC 12B Problems/Problem 5"
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 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 $33$ 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?

$\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23.25 \qquad \textbf{(C)}\ 24.75 \qquad \textbf{(D)}\ 26.25 \qquad \textbf{(E)}\ 28$

Solution

The sum of the ages of the fifth graders is $33 * 11$, while the sum of the ages of the parents is $55 * 33$. Therefore, the total sum of all their ages must be $2178$, and given $33 + 55 = 88$ people in total, their average age is $\frac{2178}{88} = \frac{99}{4} = \boxed{\textbf{(C)}\ 24.75}$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_5&oldid=67061"