Difference between revisions of "2007 AMC 8 Problems/Problem 7"
(→Solution 2) |
|||
Line 28: | Line 28: | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2007|num-b=6|num-a=8}} | {{AMC8 box|year=2007|num-b=6|num-a=8}} | ||
+ | {{MAA Notice}} |
Revision as of 00:24, 5 July 2013
Contents
[hide]Problem
The average age of people in a room is years. An -year-old person leaves the room. What is the average age of the four remaining people?
Solution 1
Let be the average of the remaining people.
The equation we get is
Simplify,
Therefore, the answer is
Solution 2
Since an year old left from a group of people averaging , The remaining people must total years older than . Therefore, the average is years over . Giving us
See Also
2007 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.