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. Note that the answer key says C, but the answer is actually $\boxed{\text{(E)}}$ , please notice this.

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

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1994 AJHSME Problems/Problem 14

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