1999 AMC 8 Problems/Problem 6

Revision as of 12:30, 23 December 2012 by Gina (talk | contribs) (Solution)

Problem

Bo, Coe, Flo, Joe, 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?

$\text{(A)}\ \text{Bo}\qquad\text{(B)}\ \text{Coe}\qquad\text{(C)}\ \text{Flo}\qquad\text{(D)}\ \text{Jo}\qquad\text{(E)}\ \text{Moe}$

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 $\boxed{\text{(E)}}$ is the answer.

See Also

1999 AMC 8 (ProblemsAnswer KeyResources)
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