Difference between revisions of "2007 AMC 8 Problems/Problem 7"
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− | The total ages would be | + | The total ages would be 30*5=150. Then, if one 18 year old leaves, we subtract 18 from 150 and get 132 Then we divide 132 by 4 to get the new average, D:33 |
==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}} | {{MAA Notice}} |
Revision as of 17:31, 1 August 2021
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
Solution 3
The total ages would be 30*5=150. Then, if one 18 year old leaves, we subtract 18 from 150 and get 132 Then we divide 132 by 4 to get the new average, D:33
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 |
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