Howdy, Stranger!

It looks like you're new here. If you want to get involved, click one of these buttons!

PT 28 SECTION 2 GAME 2

Ashley MonroeAshley Monroe Member
edited June 2020 in Logic Games 34 karma

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://7sage.com/lsat_explanations/lsat-28-section-2-game-2/

Comments

  • xicedcoffeexxicedcoffeex Free Trial Member
    edited June 2020 12 karma

    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.

Sign In or Register to comment.