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

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Assign a variable to the number of eggs Mia has, say <math>m</math>.  Then, because we are given that Sofia has twice the number of eggs Mia has, Sofia has <math>2m</math> eggs, and Pablo, having three times the number of eggs as Sofia, has <math>6m</math> eggs.
 
Assign a variable to the number of eggs Mia has, say <math>m</math>.  Then, because we are given that Sofia has twice the number of eggs Mia has, Sofia has <math>2m</math> eggs, and Pablo, having three times the number of eggs as Sofia, has <math>6m</math> eggs.
  
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For them to all have the same number of eggs, they must each have <math>\frac{m+2m+6m}{3} = 3m</math> eggs.  This means Pablo must give <math>2m</math> eggs to Mia and a <math>m</math> eggs to Sofia, so the answer is <math>\frac{m}{6m} = \boxed{\textbf{(B) }\frac{1}{6}}</math>
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For them to all have the same number of eggs, they must each have <math>\frac{m+2m+6m}{3} = 3m</math> eggs.  This means Pablo must give <math>2m</math> eggs to Mia and <math>m</math> eggs to Sofia, so the answer is <math>\frac{m}{6m} = \boxed{\textbf{(B) }\frac{1}{6}}</math>
  
 
==Solution 2==
 
==Solution 2==
 
Let Mia have <math>1</math> egg. Sofia and Pablo would then have <math>2</math> and <math>6</math> eggs, respectively. If we split the eggs evenly, each would have <math>\frac{1 + 2 + 6}{3} = \frac{9}{3} = 3</math> eggs. Since Sofia needs <math>3 - 2 = 1</math> more egg, Pablo must give her <math>1</math> of his <math>6</math> eggs, or <math>\boxed{\textbf{(B) }\frac{1}{6}}</math>
 
Let Mia have <math>1</math> egg. Sofia and Pablo would then have <math>2</math> and <math>6</math> eggs, respectively. If we split the eggs evenly, each would have <math>\frac{1 + 2 + 6}{3} = \frac{9}{3} = 3</math> eggs. Since Sofia needs <math>3 - 2 = 1</math> more egg, Pablo must give her <math>1</math> of his <math>6</math> eggs, or <math>\boxed{\textbf{(B) }\frac{1}{6}}</math>
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 +
==Video Solution (CREATIVE THINKING)==
 +
https://youtu.be/_TR2cVdIRjw
 +
 +
~Education, the Study of Everything
 +
 +
==Video Solution==
 +
https://youtu.be/uJ68vr4IA2Q
 +
 +
~savannahsolver
 +
 +
== Video Solution by OmegaLearn ==
 +
 https://youtu.be/tKsYSBdeVuw?t=3510
 +
 +
~ pi_is_3.14
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2015|ab=A|num-b=3|num-a=5}}
 
{{AMC10 box|year=2015|ab=A|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:25, 22 September 2024

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 (CREATIVE THINKING)

https://youtu.be/_TR2cVdIRjw

~Education, the Study of Everything

Video Solution

https://youtu.be/uJ68vr4IA2Q

~savannahsolver

Video Solution by OmegaLearn

 https://youtu.be/tKsYSBdeVuw?t=3510

~ pi_is_3.14

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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