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

Revision as of 17:33, 1 August 2021

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 12 over 4 = 3 years over 30.

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, D:33

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.

@@ Line 24: / Line 24: @@
 ==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>\dfrac{12}{4} = 3</math> years over <math>30</math>. Giving us <math> \boxed{\textbf{(D)}\ 33} </math>
+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 12 over 4 = 3 years over 30.
 ==Solution 3==

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