2007 AMC 8 Problems/Problem 7

Problem

The average age of $5$ people in a room is $30$ years. An $18$ -year-old person leaves the room. What is the average age of the four remaining people?

$\mathrm{(A)}\ 25 \qquad\mathrm{(B)}\ 26 \qquad\mathrm{(C)}\ 29 \qquad\mathrm{(D)}\ 33 \qquad\mathrm{(E)}\ 36$

Solution 1

Let $x$ be the average of the remaining $4$ people.

The equation we get is $\frac{4x + 18}{5} = 30$

Simplify,

$4x + 18 = 150$

$4x = 132$

$x = 33$

Therefore, the answer is $\boxed{\textbf{(D)}\ 33}$

Solution 2

Since an $18$ year old left from a group of people averaging $30$ , The remaining people must total $30 - 18 = 12$ years older than $30$ . Therefore, the average is $\dfrac{12}{4} = 3$ years over $30$ . Giving us $\boxed{\textbf{(D)}\ 33}$

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, $\boxed{\textbf{(D)}\ 33}$

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=omFpSGMWhFc

2007 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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