1998 AHSME Problems/Problem 9

Problem

A speaker talked for sixty minutes to a full auditorium. Twenty percent of the audience heard the entire talk and ten percent slept through the entire talk. Half of the remainder heard one third of the talk and the other half heard two thirds of the talk. What was the average number of minutes of the talk heard by members of the audience?

$\mathrm{(A) \ } 24 \qquad \mathrm{(B) \ } 27\qquad \mathrm{(C) \ }30 \qquad \mathrm{(D) \ }33 \qquad \mathrm{(E) \ }36$

Solution

Assume that there are $100$ people in the audience.

$20$ people heard $60$ minutes of the talk, for a total of $20\cdot 60 = 1200$ minutes heard.

$10$ people heard $0$ minutes.

$\frac{70}{2} = 35$ people heard $20$ minutes of the talk, for a total of $35\cdot 20 = 700$ minutes.

$35$ people heard $40$ minutes of the talk, for a total of $1400$ minutes.

Altogether, there were $1200 + 0 + 700 + 1400 = 3300$ minutes heard among $100$ people.

Thus, the average is $\frac{3300}{100} = 33$ minutes, and the answer is $\boxed{D}$ .

1998 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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1998 AHSME Problems/Problem 9

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