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

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<math>\mathrm{(A)}\ 25 \qquad\mathrm{(B)}\ 26 \qquad\mathrm{(C)}\ 29 \qquad\mathrm{(D)}\ 33 \qquad\mathrm{(E)}\ 36</math>
 
<math>\mathrm{(A)}\ 25 \qquad\mathrm{(B)}\ 26 \qquad\mathrm{(C)}\ 29 \qquad\mathrm{(D)}\ 33 \qquad\mathrm{(E)}\ 36</math>
  
== Solution ==
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== Solution 1==
  
 
Let <math>x</math> be the average of the remaining <math>4</math> people.
 
Let <math>x</math> be the average of the remaining <math>4</math> people.
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<math>x = 33</math>
 
<math>x = 33</math>
  
The answer is <math>\boxed{D}</math>
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Therefore, the answer is <math> \boxed{\textbf{(D)}\ 33} </math>
  
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==Solution 2==
  
OR
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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>
  
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==Solution 3==
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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>
  
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>\frac{12}{4} = 3</math> years over <math>30</math>. Giving us <math>33</math>. <math>\boxed{D}</math>
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==Video Solution by SpreadTheMathLove==
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https://www.youtube.com/watch?v=omFpSGMWhFc
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==Video Solution by WhyMath==
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https://youtu.be/hL0TX6diuvY
 +
 
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==See Also==
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{{AMC8 box|year=2007|num-b=6|num-a=8}}
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{{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 (ProblemsAnswer KeyResources)
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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