October 2005 LSAT
Section 4
Question 7

# If switch 1 and switch 3 are both off, then which one of the following could be two switches that are both on?

Replies

Ravi on July 10, 2019

@Lizzie-Annerino,Happy to help. Let's take a look at (A).

Here are the main rules in this game:

1) 1 - >/3

1 - >/5

2) 4 - >/2

4 - >/5

3) circuit load corresponds to one of the switch numbers that's on

We're told that 1 and 3 are both off, and we're looking to figure out

which answer choice could be two switches that are both on.

Starting with (A), we have 2 and 7. Let's try it.

If 2 and 7 are both on, this means that 4 must be off (the

contrapositive of rule 2).

We have 2 and 7 on, and 1, 3, and 4 off with 5 and 6 not yet assigned.

Looking at our rules, we see that with 2 and 7, we already satisfy the

last rule (the switch whose number corresponds to the circuit load of

the panel is itself on). We can just put 5 and 6 in the out/off group,

so we have 2 and 7 in and all of the other switches out. This is why

(A) works and is the correct answer choice.

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

Alexis on September 4 at 10:43PM

What is the best approach to this question? I was stumped when I realized that 1 and 3 being out doesn't tell me anything. Do you jump straight into plugging in answer choices?Alexis on September 4 at 10:44PM

Also do you verify once you think you found one that works? the question stem and the fact that the first answer seemed to work made me think I was doing something wrong.Emil on September 5 at 12:09AM

While sometimes we may need to test out all answer choices, we don't actually need to do that here. We can start by using the third rule. We know that there must be at least three not on based on the rules, since 1, 3, and at least one of two and four must be off. So, the total number on must be between one and four. However, we also know it cannot be one or three, since they are both off. So, there must either be two on or four on.However we cannot have four on. If we put 4 on, that means that 2 and 5 are out, which means that 4 total are out, leaving only three on, which is impossible. So, there must only be two on. Therefore the number 2 must be on, and A is the only one in this case with 2.