Here's my diagram, I'll walk you through it really quickly. First I listed the rules, then I connected them, but what's important is that I connected them in a way to make the maximum number of connections, so the relationships below the rules are not the same as in the rules themselves, there's a couple of slight tweaks. The first logic chain is formatted the same as the rules. But the contrapositive is a little funky. So technically:
if either R ---> ~S and ~S,
But, if ~S or ~S ---> ~New P
So I used ~S and ~S to capture the necessary condition of the presence of R, and juggled the original relationship mentally. The contrapositive is still valid, it just doesn't capture each relationship as well as the rules.
Now, moving on.
To begin with, there's not a lot of inferences up front, so you should expect a bunch of conditional questions. Before we begin, keep in mind that O is the only real variable here. So moving into the questions which I actually did a little work, and/or thought, during the questions:#8, Conditional, If ~New S
So if ~New Soul (or ~Used Soul, or both) ---> ~New Pop
Since Used Pop is already on sale, and this is only the second question, I looked to see if my brief inference would yield the credited response. Hey look, there's E, that's it. Trigger pulled, time for #9#9, Conditional, if J and J, minimum of New CDs on Sale
Whenever I have a minimum question, I start at the smallest possible number, one in this case, and see if it would work:
So if J and J ---> ~R and ~R
But we want the minimum, so let's knock off New P, because that would lead to both S and S:
If New Pop ---> S and S
So now we have New J, ~(New P, New S, New R, and let's throw New O in there), look that works, so the answer is one#10, Absolute, Must be False
We know two things for sure, I expect the credited response to hit on these:
If R ---> ~S and ~S, if S or S ---> ~R and ~R
If ~J and ~J ---> S and S, if ~S and ~S ---> J or J
Boom, the answer hits on the second relationship#11, Conditional, ~J and ~J
If both Js are out, then New P is in, since New P is in, both S and S are in, since both S and S are in both R and R must be out. This eliminates the wrong answers, which is what you should've done, but whenever you have a must be true, except question, with an unrelated variable, the credited answer will likely contain that random variable.#12, Conditional, Must be False, New Soul is the only NEW one on sale
So if New S is the only one, then New (J, O, P, and R) are not on sale. But if both Js are out, then New P is in, and if new P is in, then both S and S are in. But we only have New S in, therefore, we must have one J in, since it can't be new, it must be used.#13, Conditional, four of five are the only ones on sale
Now, if four used ones, of the original five, are the only ones on sale, then all of the new ones, New (J, O, P, R, and S) are not on sale. We know used P is in, we know O must be in, because 4/5, there's nothing on O, the others are restricted so O must be in, now we're down to R, S, and J. Let's look a little closer:
You can immediately eliminate answer choices D and E, via my diagram and the fact that only one is out. Now let's look at answer choice A: If Used J is not on sale, then both J and J would be out, triggering New P to be in, both S and S to be in, and both R and R to be out. Well, that would be two out, that's not right. Let's look at answer choices B: We know used O must be on sale, so this isn't it. Now C: so if both R and R are out, then we can still have J, and look we can still have S, this is it.