pt 33 game 3 (grouping game)
 flash21
 Posts: 1536
 Joined: Fri Apr 19, 2013 8:56 pm
pt 33 game 3 (grouping game)
would you guys consider this to be a hard grouping game? I got 4/6 on it, and it took me like 14 minutes to get those.. questions 14 and 17 screwed me up.

 Posts: 21
 Joined: Wed Jul 31, 2013 7:31 pm
Re: pt 33 game 3 (grouping game)
I was about to write an explanation of this game (which I really like) but 33: Game 3 (Jeweler's Stones) is a In/Out game, is this the one you are talking about?
EDIT: I guess that in/out is a type of grouping game, hadn't really thought about it in that way.
This game is easier when you figure out the possible distributions before hand. In particular this answers questions 14 for you, and makes some other questions easier.
Because you must have at least 2 Topazes (rule #1), and you can't have 4 (because w > not z), you must have 23 Topazes.
So possible distribution with 2 Topazes would be: 2T/3S/1R or 2T/1S/3R (you can't have 2/2/2 because 2S > 1 R from rule #2). Possible distribution with 3 Topazes would be 3T/1S/2R, 3T/2S/1R, 3T/0S/3R, or 3T/3S/0R. I think that those six possibilities are the only possible distributions of stones. This answers question 14 because they all have 3 of at least one type of stone. This also helps with 17 because you can eliminate A and B, and then you can just check the other three. You can also get 17 by realizing that since you must have 2 T, and can't have W AND Z, you must have at least one of X and Y.
EDIT: I guess that in/out is a type of grouping game, hadn't really thought about it in that way.
This game is easier when you figure out the possible distributions before hand. In particular this answers questions 14 for you, and makes some other questions easier.
Because you must have at least 2 Topazes (rule #1), and you can't have 4 (because w > not z), you must have 23 Topazes.
So possible distribution with 2 Topazes would be: 2T/3S/1R or 2T/1S/3R (you can't have 2/2/2 because 2S > 1 R from rule #2). Possible distribution with 3 Topazes would be 3T/1S/2R, 3T/2S/1R, 3T/0S/3R, or 3T/3S/0R. I think that those six possibilities are the only possible distributions of stones. This answers question 14 because they all have 3 of at least one type of stone. This also helps with 17 because you can eliminate A and B, and then you can just check the other three. You can also get 17 by realizing that since you must have 2 T, and can't have W AND Z, you must have at least one of X and Y.
 flash21
 Posts: 1536
 Joined: Fri Apr 19, 2013 8:56 pm
Re: pt 33 game 3 (grouping game)
duallys21 wrote:I was about to write an explanation of this game (which I really like) but 33: Game 3 (Jeweler's Stones) is a In/Out game, is this the one you are talking about?
EDIT: I guess that in/out is a type of grouping game, hadn't really thought about it in that way.
This game is easier when you figure out the possible distributions before hand. In particular this answers questions 14 for you, and makes some other questions easier.
Because you must have at least 2 Topazes (rule #1), and you can't have 4 (because w > not z), you must have 23 Topazes.
So possible distribution with 2 Topazes would be: 2T/3S/1R or 2T/1S/3R (you can't have 2/2/2 because 2S > 1 R from rule #2). Possible distribution with 3 Topazes would be 3T/1S/2R, 3T/2S/1R, 3T/0S/3R, or 3T/3S/0R. I think that those six possibilities are the only possible distributions of stones. This answers question 14 because they all have 3 of at least one type of stone. This also helps with 17 because you can eliminate A and B, and then you can just check the other three. You can also get 17 by realizing that since you must have 2 T, and can't have W AND Z, you must have at least one of X and Y.
Hey thanks  I missed that inference limiting the distributions which made this a lot harder than it should have been probably. Thanks for the help!
Return to “LSAT Prep and Discussion Forum”
Who is online
Users browsing this forum: No registered users and 6 guests