#15- Local, Must Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=1452)

The correct answer choice is (E)

If T is yellow, then V must be mauve, and from the fifth rule U cannot be mauve and L must be the other mauve toy:

Jun 09__M12_game#3_L11_explanations_game#3_#15_diagram_1.png

Accordingly, answer choice (E) must be true and is correct.

[cpassaro]

I'm lost on this game and I feel like I'm missing an inference. With a lot of time and effort, I was able to answer #12, 13 and 16 correctly, but I'm stumped on the others.

Could you please explain them?

Thanks

[Jon Denning]

Hey cpassaro - thanks for the questions. First, it will be helpful to refer to this previous post regarding the setup: lsat/viewtopic.php?f=156&t=1452.

If T is yellow then V must be mauve, and from the fifth rule U cannot be mauve and L must be the other mauve toy: V and L are mauve, S is red, T is yellow, and the fifth dino is either I or P. So E is correct.

Thanks!

Jon

[karanarnoldrelan]

Hello,

I understand the 5th rule that if T is not mauve then V must be mauve but the rule that P is only displayed if it's yellow. As T is yellow and thus P is not then wouldn't P wouldn't be displayed. So why is answer B wrong?

[Rachael Wilkenfeld]

Hi Karan,

Why can't you have two yellow dinos? Just because T is yellow doesn't mean that P can't be in and yellow.

I found this one easier to do by thinking about the out group. We have only two slots for dinos to be out. One of them must be V/U. In this case, when we consider the M dinos, the RS and the YT, we have one in slot left and one out-slot. Neither P nor I can be M. Therefore, they can't go in the M slots. One can be out, but the other has to be in.

Who else is in here? Everyone else! That means that L has to be in (and M), and either V or U is also in and M. However, we can't have L and U in and M together, so it must be L and V that are in and our M dinos.

Therefore our in group is RS, YT, ML, MV, and one of I/P (appropriately colored, of course). The out group is the other of I/P, and U.

The only answer that must occur is answer choice (E). Answer choice (B) could be true, but doesn't have to be if the I is in instead.

Hope that helps!