Q2

[christinamattos]

I don't see how to get the answer to #2..what am I missing? My set up & conditionals are right...

I also don't see how to get #4? I don't see how to make inferences when you put P in and nothing comes from P. And I don't see that the other two constraints have anything to do with P being in either...

[tamwaiman]

Hi christinamattos

Q2:
M is the only chemist--> P & R are selected also.(rule5)
M is the only chemist--> K & L are not selected.

According to rule2, we know that only one botanist (one of F/G/H) is selected.

Hitherto, we eliminate (A)(B)(C)(D)

[clare.ess]

I don't know how to eliminate C on this one. C has only one botanist (H), and we know that P must be selected.

Answer E also has one variable we don't know anything about -- Q.

[timmydoeslsat]

Important to remember that this is a must be true question.

If M is the only chemist selected:

M _ _ _ _

So we know that the other two chemists, L and K, are out, and that P and R are in due to our conditional.

M P R _ _ ...LK

With P and R both being in, we know we have two zoologists. This triggers our first rule. We know that if we have more than one zoologist, we have exactly one botanist.

The first rule can be written like this:

If more than 1 B ---> Exactly 1 Z

The reason I did not write at most 1 Z is due to the fact that it must be true for each group to contain at least 1 member.

So for this hypothetical all zoologists will be selected, which means that Q must be selected. We no longer have any rules that are active involving which botanist to pick: FGH

So what must be true? Nothing about FGH. Therefore E must be true.