Difference between revisions of "1999 AMC 8 Problems/Problem 13"

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==problem==
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==Problem==
  
 
The average age of the 40 members of a computer science camp is 17 years. There are 20 girls, 15 boys, and 5 adults. If the average age of the girls is 15 and the average age of the boys is 16, what is the average age of the adults?
 
The average age of the 40 members of a computer science camp is 17 years. There are 20 girls, 15 boys, and 5 adults. If the average age of the girls is 15 and the average age of the boys is 16, what is the average age of the adults?
<math> \text{(A)}\ 26\qquad\text{(B)}\ 27\qquad\text{(C)}\ 28\qquad\text{(D)}\ 29\qquad\text{(E)}\ 30 </math>
 
  
==solution==
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<math>\text{(A)}\ 26 \qquad \text{(B)}\ 27 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 29 \qquad \text{(E)}\ 30</math>
  
The answer is (C). First, find the total amount of the girl's ages and add it to the total amount of the boy's ages. It equals 540. You have to see what can give you 17 when you divide by 40. So, you do <math>17\cdot40=680</math> Then you do <math>680-540=140</math> The 5 adult's ages should add up to 140. You then do <math>140\div5=24</math>. So, the answer is (C).
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==Solution==
  
==see also==
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First, find the total amount of the girl's ages and add it to the total amount of the boy's ages. It equals <math>(20)(15)+(15)(16)=540</math>. The total amount of everyone's ages can be found from the average age, <math>17\cdot40=680</math>. Then you do <math>680-540=140</math> to find the sum of the adult's ages. The average age of an adult is divided among the five of them, <math>140\div5=\boxed{\text{(C)}\ 28}</math>.
  
{{AJHSME box|year=1999|num-b=12|num-a=14}}
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==See Also==
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{{AMC8 box|year=1999|num-b=12|num-a=14}}
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{{MAA Notice}}

Latest revision as of 23:32, 20 November 2019

Problem

The average age of the 40 members of a computer science camp is 17 years. There are 20 girls, 15 boys, and 5 adults. If the average age of the girls is 15 and the average age of the boys is 16, what is the average age of the adults?

$\text{(A)}\ 26 \qquad \text{(B)}\ 27 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 29 \qquad \text{(E)}\ 30$

Solution

First, find the total amount of the girl's ages and add it to the total amount of the boy's ages. It equals $(20)(15)+(15)(16)=540$. The total amount of everyone's ages can be found from the average age, $17\cdot40=680$. Then you do $680-540=140$ to find the sum of the adult's ages. The average age of an adult is divided among the five of them, $140\div5=\boxed{\text{(C)}\ 28}$.

See Also

1999 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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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