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Difference between revisions of "2002 AMC 8 Problems/Problem 25"
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==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?
 
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?
  
 
<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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Answer:
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==Solution==
  
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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, 3x/12x=3/12=1/4=B.
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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 (easiest)==
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Assume Moe, Loki, and Nick each give Ott <math>\$ 1</math>. Therefore, Moe has <math>\$ 5</math>, Loki has <math>\$ 4</math>, and Nick has <math>\$ 3</math>. After everyone gives Ott some fraction of their money, the total money at the end situation will be the same as the original; which is <math>\$ 12</math>. Ott gets <math>\$ 1</math> <math>+</math> <math>\$ 1</math> <math>+</math> <math>\$ 1</math> <math>=</math> <math>\$ 3</math>. Thus, the answer is <math>\frac{3}{12}=\boxed{\text{(B)}\ \frac14}</math>.
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~sakshamsethi
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==Video Solution==
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https://youtu.be/ysNxyATCxzg - Happytwin
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== Video Solution ==
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https://youtu.be/HISL2-N5NVg?t=1267
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~ pi_is_3.14
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==See Also==
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{{AMC8 box|year=2002|num-b=24|after=Last <br /> Problem}}
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{{MAA Notice}}

Revision as of 23:19, 27 January 2021

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

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, $\frac{3x}{12x}= \frac{3}{12} = \boxed{\text{(B)}\ \frac14}$. $\blacksquare$


Solution 2 (easiest)

Assume Moe, Loki, and Nick each give Ott $$ 1$. Therefore, Moe has $$ 5$, Loki has $$ 4$, and Nick has $$ 3$. After everyone gives Ott some fraction of their money, the total money at the end situation will be the same as the original; which is $$ 12$. Ott gets $$ 1$ $+$ $$ 1$ $+$ $$ 1$ $=$ $$ 3$. Thus, the answer is $\frac{3}{12}=\boxed{\text{(B)}\ \frac14}$.

~sakshamsethi

Video Solution

https://youtu.be/ysNxyATCxzg - Happytwin

Video Solution

https://youtu.be/HISL2-N5NVg?t=1267

~ pi_is_3.14

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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