LSAT 67, Logic Game 4, Question 21, LSATHacks

LG Game 4 Question 21 Explanation, by LSATHacks

There aren’t many possibilities when we have all nine variables included.

Remember, R can’t go in group 1 (first rule).

We also can’t have three houses together. But we can’t put three houses across all three groups, because H’s and I’s can’t mix.

So we need to put 2H and 1H in two groups. We’d better put at least some of them in group 1, because having H in a group restricts R (rule 3).

R is already blocked from group 1, so putting H there adds no further restriction.

I’m going to build a diagram with 2H in group 1 for now:

Since I’s can’t go with H’s, we have to put all 3I in group 3.

We can put at most 2R with the 3I (rule 4), and at most 1R with any of the H (rule 3).

R can’t go in group 1. So we end up with this diagram:

The only possible alternative would be to put 2H in group 2 rather than group 1:

So the answer will involve H and how we place them.

B is CORRECT. 2H could go in group 2.

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