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Difference between revisions of "2002 AMC 8 Problems/Problem 25"

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<math> \text{(A)}\ \frac{1}{10}\qquad\text{(B)}\ \frac{1}{4}\qquad\text{(C)}\ \frac{1}{3}\qquad\text{(D)}\ \frac{2}{5}\qquad\text{(E)}\ \frac{1}{2} </math>
 
<math> \text{(A)}\ \frac{1}{10}\qquad\text{(B)}\ \frac{1}{4}\qquad\text{(C)}\ \frac{1}{3}\qquad\text{(D)}\ \frac{2}{5}\qquad\text{(E)}\ \frac{1}{2} </math>
  
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==Solution==
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==Solution 1==
  
 
Since Ott gets equal amounts of money from each friend, we can say that he gets <math>x</math> dollars from each friend. This means that Moe has <math>5x</math> dollars, Loki has <math>4x</math> dollars, and Nick has <math>3x</math> dollars. The total amount is <math>12x</math> dollars, and since Ott gets <math>3x</math> dollars total, <math>\frac{3x}{12x}= \frac{3}{12} = \boxed{\text{(B)}\ \frac14}</math>.
 
Since Ott gets equal amounts of money from each friend, we can say that he gets <math>x</math> dollars from each friend. This means that Moe has <math>5x</math> dollars, Loki has <math>4x</math> dollars, and Nick has <math>3x</math> dollars. The total amount is <math>12x</math> dollars, and since Ott gets <math>3x</math> dollars total, <math>\frac{3x}{12x}= \frac{3}{12} = \boxed{\text{(B)}\ \frac14}</math>.
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<math>\blacksquare</math>
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==Solution 2==
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We can assign any natural number to the price that Ott's friends gave him. For this example, we will use 5.
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Moe gave Ott a fifth of his money, and also five dollars. So in total Moe has 25 dollars.
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Loki gave Ott a fourth of his money, and also five dollars. So in total Loki has 20 dollars.
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Nick gave Ott a third of his money, and also five dollars. So in total Nick has 15 dollars.
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<math>5\cdot3=15</math> which is the total money the group gave Ott.
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<math>25+20+15=60</math> is the group's balance. Therefore, the fraction of the group's money that Ott now has is <math>\frac{15}{60} = \boxed{\text{(B)}\ \frac14}</math>.
  
 
==Video Solution==
 
==Video Solution==
  
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https://youtu.be/ysNxyATCxzg - Happytwin
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https://www.youtube.com/watch?v=F-ZvPoJdnfk  ~David
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2002|num-b=24|after=Last <br /> Problem}}
 
{{AMC8 box|year=2002|num-b=24|after=Last <br /> Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 15:14, 30 December 2024

Problem

Loki, Moe, Nick and Ott are good friends. Ott had no money, but the others did. Moe gave Ott one-fifth of his money, Loki gave Ott one-fourth of his money and Nick gave Ott one-third of his money. Each gave Ott the same amount of money. What fractional part of the group's money does Ott now have?

$\text{(A)}\ \frac{1}{10}\qquad\text{(B)}\ \frac{1}{4}\qquad\text{(C)}\ \frac{1}{3}\qquad\text{(D)}\ \frac{2}{5}\qquad\text{(E)}\ \frac{1}{2}$

Solution 1

Since Ott gets equal amounts of money from each friend, we can say that he gets $x$ dollars from each friend. This means that Moe has $5x$ dollars, Loki has $4x$ dollars, and Nick has $3x$ dollars. The total amount is $12x$ dollars, and since Ott gets $3x$ dollars total, $\frac{3x}{12x}= \frac{3}{12} = \boxed{\text{(B)}\ \frac14}$.

Solution 2

We can assign any natural number to the price that Ott's friends gave him. For this example, we will use 5.

Moe gave Ott a fifth of his money, and also five dollars. So in total Moe has 25 dollars.

Loki gave Ott a fourth of his money, and also five dollars. So in total Loki has 20 dollars.

Nick gave Ott a third of his money, and also five dollars. So in total Nick has 15 dollars.

$5\cdot3=15$ which is the total money the group gave Ott.

$25+20+15=60$ is the group's balance. Therefore, the fraction of the group's money that Ott now has is $\frac{15}{60} = \boxed{\text{(B)}\ \frac14}$.

Video Solution

https://www.youtube.com/watch?v=F-ZvPoJdnfk ~David

See Also

2002 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last
Problem
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 AJHSME/AMC 8 Problems and Solutions

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