1998 AMC 12 Problem 9

Attempt Problem 9 of the 1998 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1998 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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

27

30

33

36

Answer: D

Concepts:weighted averages percentages

Difficulty rating: 1180

Small hint:

The two halves of the remainder are each

35\%

of the audience

Big hint:

Weight the heard times

60,0,20,

and

40

by their audience fractions

Solution:

The average is

0.20(60)+0.10(0)

+0.35(20)+0.35(40)

=12+7+14=33

minutes. Thus D is correct.