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

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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 leftovers. What is the sequence of the siblings in decreasing order of the part of pizza they consumed?
 
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 leftovers. What is the sequence of the siblings in decreasing order of the part of pizza they consumed?
  
<math>\textbf{(A) } \text{Alex, Beth, Cyril, Dan}</math>
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<math>\textbf{(A) } \text{Alex, Beth, Cyril, Dan}</math> <br>
<math>\textbf{(B) } \text{Beth, Cyril, Alex, Dan}</math>
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<math>\textbf{(B) } \text{Beth, Cyril, Alex, Dan}</math> <br>
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<math>\textbf{(C) } \text{Beth, Cyril, Dan, Alex}</math> <br>
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<math>\textbf{(D) } \text{Beth, Dan, Cyril, Alex}</math> <br>
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<math>\textbf{(E) } \text{Dan, Beth, Cyril, Alex}</math> <br>
  
<math>\textbf{(C) } \text{Beth, Cyril, Dan, Alex}</math>
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==Solution==
<math>\textbf{(D) } \text{Beth, Dan, Cyril, Alex}</math>
 
  
<math>\textbf{(E) } \text{Dan, Beth, Cyril, 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>.
  
==Solution==
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==Video Solution==
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https://youtu.be/PW6_dcQbhyE
  
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>.
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~savannahsolver
  
 
==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}}

Latest revision as of 13:09, 18 June 2020

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

Video Solution

https://youtu.be/PW6_dcQbhyE

~savannahsolver

See Also

2015 AMC 10B (ProblemsAnswer KeyResources)
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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