October 2005 LSAT
Section 4
Question 9

October 2005 LSAT
Section 4
Question 9

Reply

Ravi on July 10, 2019

@Lizzie-Annerino,Happy to help. Let's take a look.

We know that 1, 2, 3, 4, and 5 are left. Let's start with the max and

work our way down.

Can 5 work? No, it can't because of the rules in this game.

1) 1 - >/3

1 - >/5

2) 4 - >/2

4 - >/5

The 'not both' rules ensure that there's no way the remaining 5 pieces

can all be in. Also, remember that if 5 switches are in, then the 5

switch must be on (the last rule of the game says the switch whose

number corresponds to the circuit load of the panel is itself on)

What about 4 switches? Well, we know 4 would have to be in. If 4 is

in, then 2 and 5 are out, so at most 4 could take 1 and 3 with it, so

4 isn't going to work.

What about 3? If 3 is in, then 1 is out with 6 and 7. We could put 2

and 5 in 2 with 3 and have 4 out, and this would work.

In Out

3 6

2 7

5 1

4

This means that 3 is the maximum circuit load if 6 and 7 are out, so

(C) is the correct answer choice.

Does this make sense? Let us know if you have any other questions!

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