Difference between revisions of "2007 AMC 8 Problems/Problem 7"
(→Solution) |
Basketball8 (talk | contribs) (→Solution) |
||
| Line 6: | Line 6: | ||
<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. | ||
| Line 22: | Line 22: | ||
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
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 | ||
