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

(Solution 2 (easiest))
(Solution 2 (easiest))
Line 11: Line 11:
  
 
==Solution 2 (easiest)==
 
==Solution 2 (easiest)==
Assume Moe has <math>\$ 5</math>. Using the information given in the problem, we can conclude that Loki has <math>\$ 4</math> and Nick has <math>\$ 3</math>. After everyone gives Ott some fraction of their money, he gets <math>\$ 1</math> <math>+</math> <math>\$ 1</math> <math>+</math> <math>\$ 1</math> <math>=</math> <math>\$ 3</math>. reviewing the situation, Moe now has <math>\$ 5</math> <math>-</math> <math>\$ 1</math> <math>=</math> <math>\$ 4</math>, Loki has <math>\$ 4</math> <math>-</math> <math>\$ 1</math> <math>=</math> <math>\$ 3</math>, and Nick has <math>\$ 3</math> <math>-</math> <math>\$ 1</math> <math>=</math> <math>\$ 2</math>. Ott has <math>\frac{3}{4+3+2+3}</math> fraction of the group's money. Thus, the answer is <math>\frac{3}{12}=\boxed{\text{(B)}\ \frac14}</math>.  
+
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, he gets <math>\$ 1</math> <math>+</math> <math>\$ 1</math> <math>+</math> <math>\$ 1</math> <math>=</math> <math>\$ 3</math>. reviewing the situation, Moe now has <math>\$ 5</math> <math>-</math> <math>\$ 1</math> <math>=</math> <math>\$ 4</math>, Loki has <math>\$ 4</math> <math>-</math> <math>\$ 1</math> <math>=</math> <math>\$ 3</math>, and Nick has <math>\$ 3</math> <math>-</math> <math>\$ 1</math> <math>=</math> <math>\$ 2</math>. Ott has <math>\frac{3}{4+3+2+3}</math> fraction of the group's money. Thus, the answer is <math>\frac{3}{12}=\boxed{\text{(B)}\ \frac14}</math>.  
  
 
~sakshamsethi
 
~sakshamsethi

Revision as of 18:53, 14 July 2020

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, he gets $$ 1$ $+$ $$ 1$ $+$ $$ 1$ $=$ $$ 3$. reviewing the situation, Moe now has $$ 5$ $-$ $$ 1$ $=$ $$ 4$, Loki has $$ 4$ $-$ $$ 1$ $=$ $$ 3$, and Nick has $$ 3$ $-$ $$ 1$ $=$ $$ 2$. Ott has $\frac{3}{4+3+2+3}$ fraction of the group's money. Thus, the answer is $\frac{3}{12}=\boxed{\text{(B)}\ \frac14}$.

~sakshamsethi

Video Solution

https://youtu.be/ysNxyATCxzg - Happytwin

See Also

2002 AMC 8 (ProblemsAnswer KeyResources)
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

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