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# LSAT Prep Test 9 (October 1993) - S3 - Logic Game 2

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LSAT Prep Test 9 (October 1993) - S3 - Logic Game 2

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I just sent you a PM but I'll put it here too to help.
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Sure, I'll give you my .02. First, we know this is an in/out game. From the first two rules, we know that: 1) J and K cannot both be in, but one of them has to be in and 2) N and P cannot both be in, but one has to be in. We can split this into four worlds based on these first two rules. The four possibilities involving the J/K combo and the N/P combo are as follows:

1) JN in, KP out
2) JP in, KN out
3) KN in, JP out
4) KP in, JN out

Based on the first four rules, these are the only possibilities that we can have. When you have these down, then apply the 2 remaining rules to these worlds:

N--->L
Q--->K

Then, apply each rule. I'll use the corresponding rules from above.

Board 1

JN in, KP out

N will trigger L to be in and K out will trigger Q to be out since you're failing the necessary condition of that rule. Since 4 have to be selected, you're left with M having to go by default. So, you're left with this:

JNLM in, KPQ out

Board 2

JP in, KN out

K being out triggers Q being out. Since we have filled up the out group, everything else has to be thrown in the in. So, you're left with this:

JPLN in, KNQ out

Board 3

KN in, JP out

With K being in, you're affirming the necessary of the rule Q--->K, so the rule falls away. Q can be in or out along with M. They are essentially floaters. N triggers L to be in while Q/M and M/Q fight over the last in spot and out spot.

Board 4

KP in, JN out

Here, everything else becomes a floater. With K being in and affirming the necessary, Q can be in or out. With N being out, you're failing the sufficient which gives L the ability to be in or out. We already know that M is a floater. So, with this board, Q, L and M all fight over the remaining two in spots while the loser is banished to the remaining out spot.

I hope this helped you understand it. Please let me know if I can do anything further to help you.