#10 - Local, Must Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=16326)

The correct answer choice is (D)

The “if” statement in the question stem specifies that two adults, Q and S, are in groups 1 and 3 respectively. Since the adults are already a restricted group and two of the four adults are now placed, immediately look at the second group and examine the two remaining adult candidates, R and T. According to the Not Laws, T cannot be placed into group 2 and thus R must be the adult in group 2. T must be in either group 1 or group 3:

O94_Game_#2_#10_diagram 1.png

You might expect that making the inference that R is in group 2 would lead immediately to the correct answer, but this question is a bit more subtle. Interestingly, the answer choices come in two distinct types: answer choices (A) and (B) place specific variables into specific groups, and answer choices (C), (D), and (E) specify the number of children in a group. Since the number of children in a group is linked to the adults in a group, and we already have a large amount of information about the adults in this question, answer choices (C), (D), and (E) seem to offer greater prospects for the correct answer. In fact, since R is the only adult in group 2, there must be two children in group 2, and thus answer choice (D) is correct (remember, this is a “cannot be true” question).

[ksandberg]

Hello,

This game took me longer than the other games in the section. I approached this question by placing Q and S in the assigned groups and looking at the rules discussing which people cannot be with each other. So, I focused on the restrictive rules that G cannot be with T and H cannot be with R. I also looked at the fact that H and T cannot be in group 2. Nothing appeared obvious to me, which led me to diagram every possible answer. Is there a faster way to do this question or something that I am missing? Thank you for your time.

[Adam Tyson]

Thanks for asking, ksandberg! A key rule in this game is that there must be at least one adult in each canoe. Those adults are QRST, and that's the key rule for answering this question using our "hurdle the uncertainty" principle.

Once you have placed Q in group 1 and S in group 3, as per the local question's instructions, you've used two of your four adults. Looking at the other rules of the game, you will see that T can never go in group 2. That means R has to be in group 2, and he must be the ONLY adult in group 2. Now, we don't know who else is in that group with him, but we can hurdle that uncertainty and say that we know for sure that there are two children (FGHJK, but we can rule out H since that's another rule about group 2 and H can also never be with R). That also means that those two children are either F and J (because those two are always together) or else they are G and K (the only other two that could be with R).

That rule, requiring an adult in every group, coupled with the restriction against T in group 2, should do the trick for you here. Make sure you have both of those rules accounted for in your setup and you should be good to go!

One mistake I see people making on this question is putting Q in 1 and S in 2, instead of in 3. Read carefully! That ruins this question, because suddenly everything is possible and you cannot solve it. That should be a clue that you need to re-read to make sure you have followed instructions correctly. Ultimately, if things still aren't working, it means there is a problem in your original diagram, and this is your chance to go back and find it and fix it (or, to guess on the rest of the questions and move on to another game, rather than throwing good effort after bad, so to speak).

Let us know if that clears it up for you!