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

(Video Solution)
(13 intermediate revisions by 7 users not shown)
Line 6: Line 6:
 
==Solution==
 
==Solution==
  
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}{12}= \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>.
 +
<math>\blacksquare</math>
 +
 
 +
 
 +
==Solution 2 (easiest)==
 +
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>.  
 +
 
 +
~sakshamsethi
 +
 
 +
==Video Solution==
 +
 
 +
https://youtu.be/ysNxyATCxzg - Happytwin
 +
 
 +
https://www.youtube.com/watch?v=F-ZvPoJdnfk
 +
 
 +
== Video Solution by OmegaLearn==
 +
https://youtu.be/HISL2-N5NVg?t=1267
 +
 
 +
~ pi_is_3.14
 +
 
 +
==Video Solution==
 +
https://youtu.be/neLz3VW2FYQ Soo, DRMS, NM
  
 
==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}}

Revision as of 16:53, 3 March 2023

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

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

Video Solution by OmegaLearn

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

~ pi_is_3.14

Video Solution

https://youtu.be/neLz3VW2FYQ Soo, DRMS, NM

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