Difference between revisions of "1999 AMC 8 Problems/Problem 6"
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==Problem== | ==Problem== | ||
| − | Bo, Coe, Flo, | + | Bo, Coe, Flo, Jo, and Moe have different amounts of money. Neither Jo nor Bo has as much money as Flo. Both Bo and Coe have more than Moe. Jo has more than Moe, but less than Bo. Who has the least amount of money? |
<math> \text{(A)}\ \text{Bo}\qquad\text{(B)}\ \text{Coe}\qquad\text{(C)}\ \text{Flo}\qquad\text{(D)}\ \text{Jo}\qquad\text{(E)}\ \text{Moe} </math> | <math> \text{(A)}\ \text{Bo}\qquad\text{(B)}\ \text{Coe}\qquad\text{(C)}\ \text{Flo}\qquad\text{(D)}\ \text{Jo}\qquad\text{(E)}\ \text{Moe} </math> | ||
| Line 16: | Line 16: | ||
The only person who has not been ruled out is Moe. So <math>\boxed{\text{(E)}}</math> is the answer. | The only person who has not been ruled out is Moe. So <math>\boxed{\text{(E)}}</math> is the answer. | ||
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| + | ==Video(By YippieMath)== | ||
| + | https://www.youtube.com/watch?v=liJf550775M | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=1999|num-b=5|num-a=7}} | {{AMC8 box|year=1999|num-b=5|num-a=7}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 20:26, 25 May 2025
Problem
Bo, Coe, Flo, Jo, and Moe have different amounts of money. Neither Jo nor Bo has as much money as Flo. Both Bo and Coe have more than Moe. Jo has more than Moe, but less than Bo. Who has the least amount of money?
Solution
Use logic to solve this problem. You don't actually need to use any equations.
Neither Jo nor Bo has as much money as Flo. So Flo clearly does not have the least amount of money. Rule out Flo.
Both Bo and Coe have more than Moe. Rule out Bo and Coe; they clearly do not have the least amount of money.
Jo has more than Moe. Rule out Jo.
The only person who has not been ruled out is Moe. So 
is the answer.
Video(By YippieMath)
https://www.youtube.com/watch?v=liJf550775M
See Also
| 1999 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
