# 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 couldnâ€™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. Wouldnâ€™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 itâ€™s only M in.

Andrea on January 3 at 08:54PM

Hi @farnoushsalimian

Letâ€™s see if we can break these rules down so weâ€™re on the right track about what is and isnâ€™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, weâ€™re looking for a combination that could be true. That doesnâ€™t mean is has to be trueâ€”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 canâ€™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 weâ€™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 canâ€™t possibly be alone in G now, so M is not going to be in G (rule 2).

So, if M canâ€™t be in G, where can M go? We know from rule 4 that H canâ€™t take M. So, thereâ€™s only one remaining place M can go, and thatâ€™s F.

Now, we want to think about our principals of distribution. Iâ€™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, thatâ€™s how we get our next deduction. If P and S must go together (rule 5), we canâ€™t put them in F because then, there wonâ€™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...itâ€™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.

Letâ€™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 doesnâ€™t leave enough players to fill the two slots we must fill for H (rule 3).

B) Correct, what we broke down above. â€˜

C) It only takes rule 2 to knock this out quickly. If we know from the question stem G has Z, G canâ€™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, thatâ€™s how we get our next deduction. If P and S must go together (rule 5), we canâ€™t put them in F because then, there wonâ€™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 canâ€™t edit R because there would be know one else to fill the remaining spot (rule 3). Remember, H canâ€™t edit less than or more than two textbooks (so P and S canâ€™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.