This is an explanation of the fourth logic game from Section IV of LSAT 33, the December 2000 LSAT.
There are ten adjacent stores on Oak Street, five stores on the north side (1, 3, 5, 7, 9) and five on the south (2, 4, 6, 8, 10). The stores are facing each other in pairs: 1 and 2, 3 and 4, 5 and 6, 7 and 8, and 9 is facing 10. Each store has lights in one of these colors: green, red, and yellow (G, R, Y).
This is a fun game. Starting from a couple of facts, you can figure out almost everything.
A couple of ground rules:
- The same color can’t be beside itself, or across from itself.
- You need one and only one yellow light on each side.
It’s easiest to do a horizontal layout for this game:
You can see they use this same layout in the first question.
Next, fill in the rules 4 and 5 on the diagram. 4 is red and 5 is yellow:
Two stores across from or beside each other can’t be the same color. So 3 and 6 have to be green.
1 has to be red, because 3 is green and yellow has already been used in the top row.
Most people would stop here, and try the questions.
But an extremely useful LSAT logic games principle is: “If something can only be drawn one of two ways, try drawing it both ways.”
In this case, store 2 can only be green or yellow.
If it’s yellow, everything falls into place.
Store 8 has to be red, because 6 is green. There are no yellow lights left, since each row has its yellow lights.
That means 7 and 10 are green, and 9 is red.
Now, what if store 2 is green?
Well, not much happens. But it’s very useful to know that everything is decided if store 2 is yellow.
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