LSAT 84, Logic Game 2, Question 10, LSATHacks

This question asks which architect can have x. You can look at past questions to narrow down answers. For instance, question 9 placed x with H (since H is last). So the right answer has to have H.

Likewise, the correct answer to question 7 was D, where F had x. So, the right answer has to have F, too. That leaves only C and E.

E has both G and L. So, if you make any scenario where either G or L have x, then E is the right answer. To make an easy scenario, it’s useful to start by fulfilling rules. So, making H has y is easy, for example, as is making F have w:

Notice that I made sure a rule was fulfilled, and left x to go with either L or G. (The rules this fulfills are rule 3, Hy, and rule 2, Fw)

Since F doesn’t have z, L must. Which leaves G to have x:

And this proves E is CORRECT. But, you might be thinking “don’t we have to prove L can have x?”. Not really. We know the right answer has to have F, G and H, and only E has that combo.

But, if you want further proof, then this proves L can have x:

L: x

F: z

G: w

H: y

It’s the scenario from above, except L, F and G moves their letters one to the right. (From G to L, L to F, F to G, etc.)

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