Difference between revisions of "2015 AMC 10B Problems/Problem 4"
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==Solution== | ==Solution== | ||
| − | Let the pizza have <math>60</math> slices, since | + | Let the pizza have <math>60</math> slices, since <math>\lcm(5,3,4)=60</math>. Therefore, Alex ate <math>\frac{1}{5}\times60=12</math> slices, Beth ate <math>\frac{1}{3}\times60=20</math> slices, and Cyril ate <math>\frac{1}{4}\times60=15</math> slices. Dan must have eaten <math>60-(12+20+15)=13</math> slices. In decreasing order, we see the answer is <math>\boxed{\textbf{(C) }\text{Beth, Cyril, Dan, Alex}}</math>. |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2015|ab=B|num-b=3|num-a=5}} | {{AMC10 box|year=2015|ab=B|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 11:06, 1 January 2016
Problem 4
Four siblings ordered an extra large pizza. Alex ate 
, Beth 
, and Cyril 
of the pizza. Dan got the leftovers. What is the sequence of the siblings in decreasing order of the part of pizza they consumed?
Solution
Let the pizza have 
slices, since $\lcm(5,3,4)=60$ (Error compiling LaTeX. Unknown error_msg). Therefore, Alex ate 
slices, Beth ate 
slices, and Cyril ate 
slices. Dan must have eaten 
slices. In decreasing order, we see the answer is 
.
See Also
| 2015 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
