Difference between revisions of "2015 AMC 10B Problems/Problem 4"
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==Solution== | ==Solution== | ||
| − | Let the | + | |
| + | Let the pizza have <math>60</math> slices, since the least common multiple of <math>(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 11:54, 5 March 2015
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 the least common multiple of 
. 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 | ||
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