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Hey anyone down to go through this game. According to the information out there it’s seen as one of the most challenging grouping games. I completed the game untimed and only missed 2; so when I went back to the video explanation on the site, JY set the game up different from mine. I used RSTY for my slots and GPLH for my moving variables. I got question 9 wrong because I flat out could not figure out what it was asking me:( and I got 11 wrong - now for this question I did huddle my variables and realized G was the most restricted, however I didn’t understand why answer B was correct over C :( ... can someone please provide a video and/or feedback for this game.
Admin Note: https://classic.7sage.com/lsat_explanations/lsat-28-section-2-game-2/
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For 9: It's asking which one is not a COMPLETE list of researchers who could learn both S and Y in any given game solution. And, we know that S is restricted to 2 spots, and that Y is occupied by H/L/P (see below). So, let's look at the possibilities.
A) S is occupied by G and H (because G >H), and so H learns both S and Y.
B ) S is occupied by H and another researcher, like L or P. So, H/L and H/P are both possible solutions.
C) S is occupied by L and P. So, L/P is a solution.
This means are possible answers are H, H/L, H/P, and L/P. Answer choice B is not a COMPLETE list, so it's wrong.
For 11:
We know G would have to take H with it, but there's only one spot for R, so G cannot go in R.
We also know that, because there are three spots for Y, at least one of L or P will have to go there (3 spots / 4 researchers). And, since G can't learn anything L or P does, it cannot go there, so Y must be occupied by H/L/P.
So, we know that H must learn Y and that G cannot learn it. And, G must take H along with it on anything it does learn (because G>H). So H has to learn at least one more language than G, and therefore B is correct.
And, C is incorrect because you can imagine a valid solution in which G and L learn equal numbers of languages.