LSAT 84, Logic Game 4, Question 23, LSATHacks

This question gives G and H different types of bonds. That could be either 5/10 or 10/5. You should draw both scenarios.

In both scenarios, you have to give L a different bond from H (rule 1). In the left-hand scenario, that also causes R to have 10 (rule 3).

In the setup, I said this was a counting game. You must place four 5’s, and four 10’s. Notice something funny about the scenarios:

• In scenario 1, there are three 5’s left to place, and three open groups.
• In scenario 2, there are three 10’s left to place, and three open groups.

LSAC leaves little tricks like this all over the place! In both scenarios, you can fill in something definite:

I also added the 2nd rule in scenario 1: if V has 5, S has 5 and 10.

This is a must be true question, so you should look for something both scenarios have in common. The only common points are that R has 10 in both scenarios, and S has 10 in both scenarios.

C is CORRECT. The wrong answers could be true in either scenario, but don’t have to be true.

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