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

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==Solution 3==
 
==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>
 
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}}
 
{{MAA Notice}}

Latest revision as of 16:36, 28 October 2024

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

Video Solution by WhyMath

https://youtu.be/hL0TX6diuvY

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. AMC logo.png

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