1998 AHSME Problem 9

Algebra Level 2

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?

24 33 36 27

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1 solution

By the question we have,

20% heard 60 minutes.

10% heard 0 minutes.

Half the remainder , = 70 % 2 = 35 % = \frac{70\%}{2} = 35\% heard 20 minutes

And the other 35% heard 40 minutes.

Therefore average number of minutes heard = 20 % × 60 + 10 % × 0 + 35 % × 20 + 35 % × 40 = 20\% \times 60 + 10\% \times 0 + 35\% \times 20 + 35\% \times 40

= 0.2 × 60 + 0 + . 35 × 20 + . 35 × 40 = 0.2 \times 60 + 0 + .35 \times 20 + .35 \times 40

= 12 + 7 + 14 = 12 + 7 + 14

= 33 = \boxed{33}

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