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

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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>.
 
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>.
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==See Also==
 
==See Also==

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 (Problems • Answer Key • Resources)
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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