Difference between revisions of "2007 AMC 8 Problems/Problem 7"
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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>\dfrac{12}{4} = 3</math> years over <math>30</math>. Giving us <math> \boxed{\textbf{(D)}\ 33} </math> | 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>\dfrac{12}{4} = 3</math> years over <math>30</math>. Giving us <math> \boxed{\textbf{(D)}\ 33} </math> | ||
| + | |||
| + | ==Solution 3== | ||
| + | The total ages would be <math>30*5=150</math>. Then, if one <math>18</math> year old leaves, we subtract <math>18</math> from <math>150</math> and get <math>132</math>. Then, we divide <math>132</math> by <math>4</math> to get the new average, <math> \boxed{\textbf{(D)}\ 33} </math> | ||
| + | |||
| + | ==Video Solution by SpreadTheMathLove== | ||
| + | https://www.youtube.com/watch?v=omFpSGMWhFc | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/hL0TX6diuvY | ||
==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}} | ||
Latest revision as of 16:36, 28 October 2024
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,
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 
. Then, if one year old leaves, we subtract from 
and get 
. Then, we divide by to get the new average,
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=omFpSGMWhFc
Video Solution by WhyMath
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. 
