1994 AJHSME Problems/Problem 14

Problem

Two children at a time can play pairball. For $90$ minutes, with only two children playing at time, five children take turns so that each one plays the same amount of time. The number of minutes each child plays is

$\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ 36$

Solution

There are $2 \times 90 = 180$ minutes of total playing time. Divided equally among the five children, each child gets $180/5 = \boxed{\text{(E)}\ 36}$ minutes.

1994 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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1994 AJHSME Problems/Problem 14

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