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Difference between revisions of "2015 AMC 10B Problems/Problem 4"

(Created page with "==Problem 4== Four siblings ordered an extra large pizza. Alex ate <math>\frac15</math>, Beth <math>\frac13</math>, and Cyril <math>\frac14</math> of the pizza. Dan got the l...")
 
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<math>\textbf{(A) } \text{Alex, Beth, Cyril, Dan}</math>
 
<math>\textbf{(A) } \text{Alex, Beth, Cyril, Dan}</math>
 
<math>\textbf{(B) } \text{Beth, Cyril, Alex, Dan}</math>
 
<math>\textbf{(B) } \text{Beth, Cyril, Alex, Dan}</math>
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<math>\textbf{(C) } \text{Beth, Cyril, Dan, Alex}</math>
 
<math>\textbf{(C) } \text{Beth, Cyril, Dan, Alex}</math>
 
<math>\textbf{(D) } \text{Beth, Dan, Cyril, Alex}</math>
 
<math>\textbf{(D) } \text{Beth, Dan, Cyril, Alex}</math>
 +
 
<math>\textbf{(E) } \text{Dan, Beth, Cyril, Alex}</math>
 
<math>\textbf{(E) } \text{Dan, Beth, Cyril, Alex}</math>
  
 
==Solution==
 
==Solution==
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Let the total amount of slices be <math>60</math>. Then we know Alex ate <math>12</math>, Beth ate <math>20</math>, Cyril ate <math>15</math>, and Dan ate the remaining, or <math>13</math>. Then the sequence in decreasing order is <math>\boxed{\textbf{(C)} \text{Beth, Cyril, Dan, Alex}}</math>.
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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==
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{{AMC10 box|year=2015|ab=B|before=Problem 3|num-a=5}}
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{{AMC10 box|year=2015|ab=B|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:09, 1 January 2016

Problem 4

Four siblings ordered an extra large pizza. Alex ate $\frac15$, Beth $\frac13$, and Cyril $\frac14$ of the pizza. Dan got the leftovers. What is the sequence of the siblings in decreasing order of the part of pizza they consumed?

$\textbf{(A) } \text{Alex, Beth, Cyril, Dan}$ $\textbf{(B) } \text{Beth, Cyril, Alex, Dan}$

$\textbf{(C) } \text{Beth, Cyril, Dan, Alex}$ $\textbf{(D) } \text{Beth, Dan, Cyril, Alex}$

$\textbf{(E) } \text{Dan, Beth, Cyril, Alex}$

Solution

Let the pizza have $60$ slices, since the least common multiple of $(5,3,4)=60$. Therefore, Alex ate $\frac{1}{5}\times60=12$ slices, Beth ate $\frac{1}{3}\times60=20$ slices, and Cyril ate $\frac{1}{4}\times60=15$ slices. Dan must have eaten $60-(12+20+15)=13$ slices. In decreasing order, we see the answer is $\boxed{\textbf{(C) }\text{Beth, Cyril, Dan, Alex}}$.

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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