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Difference between revisions of "2004 AMC 8 Problems/Problem 9"

(Created page with "== Problem == The average of the five numbers in a list is <math>54</math>. The average of the first two numbers is <math>48</math>. What is the average of the last three numbers...")
 
 
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numbers is <math>48</math>. What is the average of the last three numbers?
 
numbers is <math>48</math>. What is the average of the last three numbers?
  
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<math> \mathrm{(A)\ 55 }\qquad\mathrm{(B)\ 56 }\qquad\mathrm{(C)\ 57 }\qquad\mathrm{(D)\ 58 }\qquad\mathrm{(E)\ 59 } </math>
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<math>\textbf{(A)}\ 55 \qquad \textbf{(B)}\ 56 \qquad \textbf{(C)}\ 57 \qquad \textbf{(D)}\ 58 \qquad \textbf{(E)}\ 59</math>
  
 
== Solution ==
 
== Solution ==
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Let the <math>5</math> numbers be <math>a, b, c, d</math>, and <math>e</math>. Thus <math>\frac{a+b+c+d+e}{5}=54</math> and <math>a+b+c+d+e=270</math>. Since <math>\frac{a+b}{2}=48</math>, <math>a+b=96</math>. Substituting back into our original equation, we have <math>96+c+d+e=270</math> and <math>c+d+e=174</math>. Dividing by <math>3</math> gives the average of <math>58</math> <math>\boxed{\textbf{(D)}\ 58}</math>
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Let the <math>5</math> numbers be <math>a, b, c, d</math>, and <math>e</math>. Thus <math>\frac{a+b+c+d+e}{5}=54</math> and <math>a+b+c+d+e=270</math>. Since <math>\frac{a+b}{2}=48</math>, <math>a+b=96</math>. Substituting back into our original equation, we have <math>96+c+d+e=270</math> and <math>c+d+e=174</math>. Dividing by <math>3</math> gives the average of <math>\boxed{\textbf{(D)}\ 58}</math>.
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==See Also==
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{{AMC8 box|year=2004|num-b=8|num-a=10}}
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{{MAA Notice}}

Latest revision as of 23:55, 4 July 2013

Problem

The average of the five numbers in a list is $54$. The average of the first two numbers is $48$. What is the average of the last three numbers?

$\textbf{(A)}\ 55 \qquad \textbf{(B)}\ 56 \qquad \textbf{(C)}\ 57 \qquad \textbf{(D)}\ 58 \qquad \textbf{(E)}\ 59$

Solution

Let the $5$ numbers be $a, b, c, d$, and $e$. Thus $\frac{a+b+c+d+e}{5}=54$ and $a+b+c+d+e=270$. Since $\frac{a+b}{2}=48$, $a+b=96$. Substituting back into our original equation, we have $96+c+d+e=270$ and $c+d+e=174$. Dividing by $3$ gives the average of $\boxed{\textbf{(D)}\ 58}$.

See Also

2004 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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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