2002 AMC 8 Problems/Problem 25

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

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

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

Page

Toolbox

Search

2002 AMC 8 Problems/Problem 25

Contents

Problem

Solution

Solution 2 (easiest)

Video Solution

See Also