#15 - Local, Cannot Be True

[Administrator]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=11115)

The correct answer choice is (A)

The question stem specifies that H is not offered. As the fourth rule lists H as a necessary condition, taking the contrapositive of the fourth rule reveals that G cannot be offered. G is not one of the answer choices, however. But, G is a necessary condition of the second rule, and taking the contrapositive of that rule reveals that L cannot be offered:

June 06_M12_game#3_L6_explanations_game#4_#15_diagram_1.png

Accordingly, answer choice (A) is correct.

[lsatbossintraining]

Hm. Shouldn’t M be kicked out as well?

Or is that because L relates to M in an “or” relationship that M could actually be in while all of H, G and L are out.

Feel like I understand it intuitively but looking for someone to cement the idea for me.

Thanks much.

Kyle

[Jeremy Press]

Hi Kyle,

Yes, you've got the right idea here! If M is in, then one of L or S must be in (though not both). Even if L is out, M could still be in, so long as S is also in.

Here's how we would diagram the rule about M, along with its contrapositive:

Screen Shot 2019-11-04 at 1.17.00 PM.png

We'd have to know that both L and S were out before we could say with certainty that M is out.

I hope this helps!

Jeremy