Difference between revisions of "1998 AHSME Problems/Problem 9"
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| − | == Problem | + | == 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? | 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? | ||
<math> \mathrm{(A) \ } 24 \qquad \mathrm{(B) \ } 27\qquad \mathrm{(C) \ }30 \qquad \mathrm{(D) \ }33 \qquad \mathrm{(E) \ }36 </math> | <math> \mathrm{(A) \ } 24 \qquad \mathrm{(B) \ } 27\qquad \mathrm{(C) \ }30 \qquad \mathrm{(D) \ }33 \qquad \mathrm{(E) \ }36 </math> | ||
| − | + | ==Solution== | |
| + | Assume that there are <math>100</math> people in the audience. | ||
| + | |||
| + | <math>20</math> people heard <math>60</math> minutes of the talk, for a total of <math>20\cdot 60 = 1200</math> minutes heard. | ||
| + | |||
| + | <math>10</math> people heard <math>0</math> minutes. | ||
| + | |||
| + | <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. | ||
| + | |||
| + | <math>35</math> people heard <math>40</math> minutes of the talk, for a total of <math>1400</math> minutes. | ||
| + | |||
| + | Altogether, there were <math>1200 + 0 + 700 + 1400 = 3300</math> minutes heard among <math>100</math> people. | ||
| + | |||
| + | Thus, the average is <math>\frac{3300}{100} = 33</math> minutes, and the answer is <math>\boxed{D}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME box|year=1998|num-b=8|num-a=10}} | ||
| + | {{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?
Solution
Assume that there are 
people in the audience.
people heard 
minutes of the talk, for a total of 
minutes heard.
people heard 
minutes.
people heard minutes of the talk, for a total of 
minutes.
people heard 
minutes of the talk, for a total of 
minutes.
Altogether, there were 
minutes heard among people.
Thus, the average is 
minutes, and the answer is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
