2015 AMC 10A Problems/Problem 4

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 1

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 $m$ eggs to Sofia, so the answer is $\frac{m}{6m} = \boxed{\textbf{(B) }\frac{1}{6}}$

Solution 2

Let Mia have $1$ egg. Sofia and Pablo would then have $2$ and $6$ eggs, respectively. If we split the eggs evenly, each would have $\frac{1 + 2 + 6}{3} = \frac{9}{3} = 3$ eggs. Since Sofia needs $3 - 2 = 1$ more egg, Pablo must give her $1$ of his $6$ eggs, or $\boxed{\textbf{(B) }\frac{1}{6}}$

Video Solution

https://youtu.be/uJ68vr4IA2Q

~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/tKsYSBdeVuw?t=3510

~ pi_is_3.14

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search