Difference between revisions of "2015 AMC 10A Problems/Problem 4"
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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> | <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 1== |
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. | ||
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> | 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> | ||
| + | |||
| + | ==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> | ||
==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}} | ||
Revision as of 13:45, 18 February 2018
Contents
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?
Solution 1
Assign a variable to the number of eggs Mia has, say 
. Then, because we are given that Sofia has twice the number of eggs Mia has, Sofia has 
eggs, and Pablo, having three times the number of eggs as Sofia, has 
eggs.
For them to all have the same number of eggs, they must each have 
eggs. This means Pablo must give eggs to Mia and a eggs to Sofia, so the answer is 
Solution 2
Let Mia have 
egg. Sofia and Pablo would then have 
and 
eggs, respectively. If we split the eggs evenly, each would have 
eggs. Since Sofia needs 
more egg, Pablo must give her of his eggs, or 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
