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

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Pablo, Sofia, and Mia got some candy eggs at a party.  Pablo had three times as many eggs as Sofia, and Sofia had twice as many eggs as Mia.  Pablo decides to give some of his eggs to Sofia and Mia so that all three will have the same number of eggs.  What fraction of his eggs should Pablo give to Sofia?
 
Pablo, Sofia, and Mia got some candy eggs at a party.  Pablo had three times as many eggs as Sofia, and Sofia had twice as many eggs as Mia.  Pablo decides to give some of his eggs to Sofia and Mia so that all three will have the same number of eggs.  What fraction of his eggs should Pablo give to Sofia?
  
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<math> \textbf{(A)}\ \frac{1}{12}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2}</math>
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<math> \textbf{(A)}\ \frac{1}{12}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2}</math>
  
 
==Solution==
 
==Solution==

Revision as of 18:46, 1 March 2015

Problem

Pablo, Sofia, and Mia got some candy eggs at a party. Pablo had three times as many eggs as Sofia, and Sofia had twice as many eggs as Mia. Pablo decides to give some of his eggs to Sofia and Mia so that all three will have the same number of eggs. What fraction of his eggs should Pablo give to Sofia?

$\textbf{(A)}\ \frac{1}{12}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2}$

Solution

Assign a variable to the number of eggs Mia has, say $m$. Then, because we are given that Sofia has twice the number of eggs Mia has, Sofia has $2m$ eggs, and Pablo, having three times the number of eggs as Sofia, has $6m$ eggs.

For them to all have the same number of eggs, they must each have $\frac{m+2m+6m}{3} = 3m$ eggs. This means Pablo must give $2m$ eggs to Mia and a $m$ eggs to Sofia, so the answer is $\frac{m}{6m} = \boxed{\textbf{(B) }\frac{1}{6}}$

See Also

2015 AMC 10A (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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