Q10

[pkashani06]

I have a question regarding Question 10... The explanation provided on the PDF posted earlier states:

"If 5 and 6 are on, we immediately know that 1 and 4 are off. (A) and (D) are out. Let’s consider the
possible totals: two, three, and five (we know that six and seven are impossible). For reasons that
should be well-known at this point, we can eliminate two (it’s not part of the "on" crowd, and once it
joins it, like an awkward friend joining in on your romantic date, it ruins it). We can already eliminate _
or at least defer judgment on (B)."

My PowerScore book translated rule 2 as: 4 cannot equal 2 and 4 cannot equal 5. I believe that the book incorrectly wrote this rule out because the contrapositive of rule 2 states: 2on or 5on --> 4 off. I believe that means that if 4 is off, either 2 or 5 or both must be on, but not necessarily both... Is that correct?

If I understand that rule correctly, it would make my books interpretation of the rule incorrect (4 cannot equal 2, 4 cannot equal 5) because if 4 is off then 2 could off and 5 could be on (or 4off, 2 on, 5off).

Am I understanding this correctly? Or is my book actually correct?

[giladedelman]

Here's the rule:

If switch 4 is on, then switch 2 and switch 5 are off.

You're right that the contrapositive is therefore:

If switch 2 OR switch 5 (or both) is on, then switch 4 is off.

That means that 4 and 2, as well as 4 and 5, cannot be ON at the same time. However, they can be OFF at the same time; according to our conditional statements, there's nothing that has to be true if switch 2, 4, or 5 is OFF. (Be careful not to reverse that contrapositive.)

Does that answer your question?

[tzyc]

I'm redoing this problem and I understand why we can eliminate (A) and (D), but not sure why (C) is correct...is it because of the third constraint?
Thanks.

[timmydoeslsat]

I'm redoing this problem and I understand why we can eliminate (A) and (D), but not sure why (C) is correct...is it because of the third constraint?
Thanks.

So we know that 5 and 6 are both on, which forces 1 and 4 out.

5 6 .......... 1 4

We know that we have to have the circuit load of the panel on. This number represents the total number of switches to be on. So we know that none of our conditional rules are applicable at this point. We had information concerning what happens if 1 or if 4 is is on.

So we have three numbers left to place: 2 3 7

We know that 5 could represent the circuit load of the panel and 3 could. Those are the only two options for the circuit load right now. If you choose 5 to be the circuit load, all of 2 3 and 7 must be in. If you choose 3 to be the circuit load, then you must only put in 3. Either way, we must have 3 in at all times.