Difference between revisions of "2002 AMC 8 Problems/Problem 25"
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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>. | 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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| + | ==Video Solution== | ||
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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 09:18, 14 June 2024
Contents
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?
Solution
Since Ott gets equal amounts of money from each friend, we can say that he gets 
dollars from each friend. This means that Moe has 
dollars, Loki has 
dollars, and Nick has 
dollars. The total amount is 
dollars, and since Ott gets dollars total, 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
