If the zoology textbook is edited by Gupta, which one of the following could be true?

Farnoush on January 3 at 06:33PM

Why B?

Hi Originally going through this question I couldn’t find an answer choice that was direct because based on the rules If M is in G then it has to b alone so if we have Z in G that means that R must be in F because the first rule states that F-> L/Z. So based on that rule we can infer that L must be in F Based on the rule that P has to have S then they go into H which leaves M and R. Wouldn’t it be then that R has to go into G rather then M based on the rule that if M is in G then it’s only M in.

1 Reply

Andrea on January 3 at 08:54PM

Hi @farnoushsalimian

Let’s see if we can break these rules down so we’re on the right track about what is and isn’t possible. I think you may have missed a combination or two in there and overly restricted your game.

If Z is edited by G, which one of the following could be true?

With this stem, we’re looking for a combination that could be true. That doesn’t mean is has to be true—just that in at lest one scenario, it could be. More importantly, that means that four out of five answer choices here are impossible, or can’t be true in any scenario (or, for a third way of thinking about it, must be false).

So, I like to eliminate based on that sometimes. I start with the premise we’re given in the question stem, and figure out first what must be true based on that.

F: L M (_)...
G: Z (_)...
H: P S l
Floater: R
R can float between F and G.

Based on just that one constraint (that G is edited by Z) we can figure out that all of the above must be true. I can break down for you how I got there.

I started by putting Z in G. Remember, we know from rule 1 that F must edit Z or L. So, since G has Z, F must then get L.

Since Z is in G (the information we are given in the question stem), we also know that M can’t possibly be alone in G now, so M is not going to be in G (rule 2).

So, if M can’t be in G, where can M go? We know from rule 4 that H can’t take M. So, there’s only one remaining place M can go, and that’s F.

Now, we want to think about our principals of distribution. I’ll illustrate this step for you. So far, from the above deductions we have this:

F: L M .....
G: Z ......
H: _ _ l

We have six players. Three are placed, but we have two empty spots left in H (rule 3) that must be filled still. Our remaining players are: PRS

Remember, however, that P and S must go together (rule 5). So, that’s how we get our next deduction. If P and S must go together (rule 5), we can’t put them in F because then, there won’t be enough players to fill H with (rule 3). If we put them in G, we encounter the same problem. So, that means P and S have to take up those two spots in H.

So, just from knowing that Z is in G, we can figure out all of this:

F: L M .....
G: Z ....
H: P S l

Our only remaining player is R. R can float between F and G, and end up in either. R can't be in H, however, because of rule 3, that H must have exactly two textbooks. Since this is a could be true question, I would look first for an answer choice that could be true (as opposed to a must be true, even though a must be true can still be the right answer to a could be true...it’s possible, but less common).

The only thing in our deductions that could be true (as opposed to must be true) is that R could be in F or G. So, I would fist look for an answer choice that gives us one of those.

That gets us to answer choice B.

Let’s go over the rest quickly just to double check our work.

A) No, because like we said above, we know F has to take L and M from rules 1, 2, and 4 working together. If we give F textbook P, we know from rule 5 that S is coming with it. So that means F would get: LMPS, which doesn’t leave enough players to fill the two slots we must fill for H (rule 3).

B) Correct, what we broke down above. ‘

C) It only takes rule 2 to knock this out quickly. If we know from the question stem G has Z, G can’t also have M, because if G has M (rule 2), M must be alone in G which would not be the case with Z there.

D) We know from rule 1 that F must have Z or L. We also know from the question stem that G has Z. So, if G has Z, that means F has to have the remaining of the two, L. That leaves no way for H to have it.

E) My earlier explanation knocks out this answer choice:

F: L M .....
G: Z ......
H: _ _ l

We have six players. Three are placed, but we have two empty spots left in H (rule 3) that must be filled still. Our remaining players are: PRS

Remember, however, that P and S must go together (rule 5). So, that’s how we get our next deduction. If P and S must go together (rule 5), we can’t put them in F because then, there won’t be enough players to fill H with (rule 3). If we put them in G, we encounter the same problem. So, that means P and S have to take up those two spots in H.

...So, H can’t edit R because there would be know one else to fill the remaining spot (rule 3). Remember, H can’t edit less than or more than two textbooks (so P and S can’t go in there together with R). H must edit exactly two textbooks.

Hope this helps! Feel free to follow up if you have any more questions.