LSAT 84, Logic Game 4, Question 20, LSATHacks

This question took me a while when I first did it. Looking at it now though, it’s easy. I realized that I missed a key dynamic in the game the first time I did it. I’ll explain; this dynamic eliminates B and D, and also shows the right answer is correct.

But first, the easy answers. A and C are wrong because of rule 1. H and L have to have different bonds, and therefore they can each only have one bond.

Now, the dynamic that I missed is this:

• There are six corporations. Two of them have two bonds, the rest have one.
• If V has two bonds, then S also has two (rule 2, V5 —> S 10). That means that all the other corps have one bond.

The real key insight is the first point. There can only be two with two. But, knowing that insight lets you figure out the deduction about V. And V is present in three answers, so it’s worth thinking about what happens if it has both.

So that’s why C and D are wrong. If V has two, then S does, so G and R can’t. Only two corporations can have two.

E is CORRECT. As I showed above, if V has both then S has both (rule 3). So E works with that rule. The others are all fine with one bond. The only other rules are that H and L have different bonds, and that R has 10 if L does. That’s easy to fulfill.

Counting was the real constrain for question 20. You need to see how all eight of the bonds can be placed: if two corporations have two, then the others each have just one.

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