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

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