LSAT 60, Logic Game 3, Question 17, LSATHacks

If only two loads are consecutive, that means we’re in one of the “Two mulch together, with mulch on the end” scenarios. We also must space out the stones so that there aren’t two in a row. This is tricky. Here are our parameters:

• We must put two mulch together
• We must put at least one mulch on one of the ends
• We can’t have more than two stone together (or mulch, of course.)

We already start with a mulch in five. There are two spaces to the right of it. We could make those stone without consequence:

We can then put our end-mulch in spot 1:

The reverse of this (mulch 7, stone 1) wouldn’t have worked. Try it: you would have ended up with three stones in a row at the front, because you’d have to put your other mulch in spot 4 in order to avoid having too many transfers.

Next, we need to put two mulches together. There are two ways to do this:

Both scenarios only have three transfers, and no more than two loads of the same material in a row. I believe these are the only two possible scenarios, though it’s not impossible that I missed one.

These scenarios prove that B is CORRECT. The fourth load could be stone.

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