Difference between revisions of "2001 AMC 8 Problems/Problem 13"
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There are <math> 36 </math> students in the class: <math> 12 </math> prefer chocolate pie, <math> 8 </math> prefer apple pie, and <math> 6 </math> prefer blueberry pie. Therefore, <math> 36-12-8-6=10 </math> students prefer cherry pie or lemon pie. Half of these prefer each, so <math> 5 </math> students prefer cherry pie. This means that <math> \frac{5}{36} </math> of the students prefer cherry pie, so <math> \frac{5}{36} </math> of the full <math> 360^\circ </math> should be used for cherry pie. This is <math> (\frac{5}{36})(360^\circ)=50^\circ, \boxed{\text{D}} </math> | There are <math> 36 </math> students in the class: <math> 12 </math> prefer chocolate pie, <math> 8 </math> prefer apple pie, and <math> 6 </math> prefer blueberry pie. Therefore, <math> 36-12-8-6=10 </math> students prefer cherry pie or lemon pie. Half of these prefer each, so <math> 5 </math> students prefer cherry pie. This means that <math> \frac{5}{36} </math> of the students prefer cherry pie, so <math> \frac{5}{36} </math> of the full <math> 360^\circ </math> should be used for cherry pie. This is <math> (\frac{5}{36})(360^\circ)=50^\circ, \boxed{\text{D}} </math> | ||
| − | ==Video 1== | + | ==Video Solution 1== |
https://youtu.be/4dhUeOdXvUk | https://youtu.be/4dhUeOdXvUk | ||
Soo, DRMS, NM | Soo, DRMS, NM | ||
Revision as of 23:18, 19 February 2022
Contents
Problem
Of the 36 students in Richelle's class, 12 prefer chocolate pie, 8 prefer apple, and 6 prefer blueberry. Half of the remaining students prefer cherry pie and half prefer lemon. For Richelle's pie graph showing this data, how many degrees should she use for cherry pie?
Solution
There are 
students in the class: 
prefer chocolate pie, 
prefer apple pie, and 
prefer blueberry pie. Therefore, 
students prefer cherry pie or lemon pie. Half of these prefer each, so 
students prefer cherry pie. This means that 
of the students prefer cherry pie, so of the full 
should be used for cherry pie. This is 
Video Solution 1
https://youtu.be/4dhUeOdXvUk Soo, DRMS, NM
See Also
| 2001 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
