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Difference between revisions of "2007 AMC 8 Problems/Problem 7"

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<math>x = 33</math>
 
<math>x = 33</math>
  
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The answer is <math>\boxed{D}</math>
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Therefore, the answer is <math> \boxed{\textbf{(D)}\ 33} </math>
  
 
==Solution 2==
 
==Solution 2==

Revision as of 12:12, 9 December 2012

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 $\frac{12}{4} = 3$ years over $30$. Giving us $33$. $\boxed{D}$

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
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