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Difference between revisions of "1998 AHSME Problems/Problem 9"

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==Solution==
 
==Solution==
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Assume that there are <math>100</math> people in the audience.
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<math>20</math> people heard <math>60</math> minutes of the talk, for a total of <math>20\cdot 60 = 1200</math> minutes heard.
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<math>10</math> people heard <math>0</math> minutes.
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<math>\frac{70}{2} = 35</math> people heard <math>20</math> minutes of the talk, for a total of <math>35\cdot 20 = 700</math> minutes.
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<math>35</math> people heard <math>40</math> minutes of the talk, for a total of <math>1400</math> minutes.
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Altogether, there were <math>1200 + 0 + 700 + 1400 = 3300</math> minutes heard among <math>100</math> people.
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Thus, the average is <math>\frac{3300}{100} = 33</math> minutes, and the answer is <math>\boxed{D}</math>.
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==See Also==
 
==See Also==
 
{{AHSME box|year=1998|num-b=8|num-a=10}}
 
{{AHSME box|year=1998|num-b=8|num-a=10}}
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{{MAA Notice}}

Latest revision as of 13:29, 5 July 2013

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 $\boxed{D}$.

See Also

1998 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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