Difference between revisions of "2007 AMC 8 Problems/Problem 7"
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<math>\mathrm{(A)}\ 25 \qquad\mathrm{(B)}\ 26 \qquad\mathrm{(C)}\ 29 \qquad\mathrm{(D)}\ 33 \qquad\mathrm{(E)}\ 36</math> | <math>\mathrm{(A)}\ 25 \qquad\mathrm{(B)}\ 26 \qquad\mathrm{(C)}\ 29 \qquad\mathrm{(D)}\ 33 \qquad\mathrm{(E)}\ 36</math> | ||
− | == Solution == | + | == Solution 1== |
Let <math>x</math> be the average of the remaining <math>4</math> people. | Let <math>x</math> be the average of the remaining <math>4</math> people. | ||
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The answer is <math>\boxed{D}</math> | The answer is <math>\boxed{D}</math> | ||
+ | ==Solution 2== | ||
− | + | Since an <math>18</math> year old left from a group of people averaging <math>30</math>, The remaining people must total <math>30 - 18 = 12</math> years older than <math>30</math>. Therefore, the average is <math>\frac{12}{4} = 3</math> years over <math>30</math>. Giving us <math>33</math>. <math>\boxed{D}</math> | |
− | + | ==See Also== | |
− | + | {{AMC8 box|year=2007|num-b=6|num-a=8}} |
Revision as of 22:35, 12 November 2012
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,
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 |