October 1996 LSAT
Section 3
Question 17
If on a nine–bead strand the first and fourth beads from me clasp are purple, and the second and fifth beads are yell...
Replies
Naz on October 27, 2015
Don't forget rule # 4.Because we have P in 1 and 4 and Y in 2 and 5, we must have R in 3 and 6 to fulfill rule #1. So that means we must have one G and one O in 7 and 8 to fulfill rule #4. Since rule # 3 tell us that R and O cannot be next to one another, O must be in 8 and that forces G to be in 7 (again because of rule #4). Then other than R, anything can be in 9 (R cannot be because of rule #3). Thus, we cannot have the 8th bead be G. It MUST BE O. If you re-watch the video for the question explanation, this is pointed out.
Hope that helps! Please let us know if you have any other questions.
dallman0303 on October 27, 2015
Completely understand rule #4 and that anything could be in 9. Why can't O be in 9 and 7/8 be G? R and O are still separated and all colors are represented.AndrewArabie on September 13, 2022
This is my problem with the question as well. "O" could be in 9, it doesn't have to be green. Am I missing something?AndrewArabie on March 20, 2023
I just want to reiterate this question. I don't see anything wrong with B. PYRPYRGGO seems like perfectly plausible scenario to me and doesn't violate any rulesAndrewArabie on March 20, 2023
Oh wait never mind. G&O beads have to be represented within the first 8. I interpreted this rule just to mean whenever there are 8 or more beads all must be represented