Difference between revisions of "1999 AMC 8 Problems/Problem 13"
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| − | The | + | ==Problem== |
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| + | 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? | ||
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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> | ||
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| + | ==Solution== | ||
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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>. | ||
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| + | ==See Also== | ||
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| + | {{AMC8 box|year=1999|num-b=12|num-a=14}} | ||
| + | {{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?
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 
. The total amount of everyone's ages can be found from the average age, 
. Then you do 
to find the sum of the adult's ages. The average age of an adult is divided among the five of them, 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
