Three houses are being sold through a

John Robertson

Given:
Mike draws quick diagrams usually, and I found that helps w/o sacrificing a lot of time:

House 1 House 2 House 3
P1 P2 P3

-> (using greater/less than or equal to signs, can't do it on my mac) P1 < P2 < P3 . I think you can just assign House 2 as the mid-priced house and so on without making any math mistakes, so I did.

1) P3- P1 = $130,000. NS for a few reasons. P3 and P1 could be a ton of #s that subtract to $130,000, also tells you nothing about P2. Cross off A & D

2) P3-P2=$85,000. NS. introduces P2, but P3 and P2 could be a ton of different test variables. Cross of B

1&2) P3 - P1 =$130,000 and P3-P2= $85,000. I think its possible to see, "ok two equations, I could solve for P2 bc there's probably a combination that works out to that." But, that's wrong. Strictly speaking, 3 variables, 2 equations, can't sub in and solve. NS. I think the OG provided a few different numbers that would satisfy Eq 1 and Eq 2 in this case.

ANS: E

Jan 15, 2017 • Comment