Q7

[Acing LSAT]

I guessed on this one and got it right. Later I figured it out, so I want to share.

The main factor here is the constraint that the switch which equals the number of switches on must also be on.

We know that 1, 3 are out. so the max number of in is now 5. If 5 is in then 1 and 4 is out and we are down to three in. We already have 3 out, so that forces that cannot be in.

4 also would force two out and leave 3 in, but three is out. so 4 does not work.

We are left with (A) which gives us 2 and 7, which works because we can have two in.

[ohthatpatrick]

Nicely done.

This game ends up being all about that oddball 3rd rule. I found that it was helpful to write out / talk through a couple examples to understand what the 3rd rule was saying:

"Oh, okay. If there are 5 things IN, then switch 5 must be one of them. If there are 4 things IN, then switch 4 must be one of them ..."

You could actually frame this game somewhat quickly (although with more frames than we would normally attempt)

7 switches IN = impossible, R1 and R2 force at least 2 things to be OUT

6 switches IN = impossible, same as above.

5 switches IN
in: 5 2 3 6 7
out: 1 4

4 switches IN
in: 4 6 7 1/3
out: 2 5 3/1

3 switches IN
in: 3 _ _
out: 1 _ _ _

2 switches IN
in: 2 _
out: 4 _ _ _ _

1 switch IN
in: 1
out: 2 3 4 5 6 7

From these frames, only 2 switches IN allows both 1 and 3 to be OUT.