#5 - Local, Must Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?f=229&p=88682)

The correct answer choice is (A)

If both G and H (two botanists) are selected, then from the second rule exactly one zoologist is selected. Thus, the fifth member of the panel must be a botanist or chemist:

G1-Q5-d1.png

When only one zoologist can be selected, from the contrapositive of the fifth rule M cannot be selected, leaving only K or L to be selected as a chemist:

G1-Q5-d2.png

Note that with M eliminated, answer choices (B), (C), and (D)—each of which include M—can be eliminated (in each instance, the other variable would have to be selected, which is not true in each case). With only answer choices (A) and (E) remaining, (E) can be eliminated because R could be the zoologist selected.

Answer choice (A) is thus correct, and consider what occurs if neither F or K is selected:

L must be selected as the chemist meeting the conditions in the first rule. But, no other scientist remains to be chosen as the fifth and final member of the panel. Hence, one of F or K must always be selected.

[Xantippe]

For Q5: If both G and H are among the scientists selected, the panel must include either...

I based my answer on a distribution of 2 (G and H), 2 (KL), and 1 (P/Q). I chose incorrectly that the panel must include (E) P or else Q. (Note: this did not feel right either because I was thinking a more complete albeit wrong answer would have been P/R and Q). Clearly wrong because of the first rule; If more than one botanist is selected, than at most one zoologist is selected.

The correct answer is (A) F or else K.


My reasoning was that the question was based on a 2-2-1 Distribution. Should I have chose a 3-1-1? The reason I did not was because I thought the question was specifically asking for a G and H (2) in the first distribution.


Thanks in advance!

[James Finch]

Hi Xantippe,

It looks like you assumed that it could only have been one of the two distributions, when it fact it could be either, as there must only be one zoologist but there could be all three botanists, depending on whether F is put in or not. The key to this question is diagramming and making sure to Hurdle the Uncertainty by using both the In and the Out group to figure out which possibilities must be in. Working through this with my diagram, I got that:

In (5 slots): G, H, P/R/Q (only 1!), L (Must have at least one chemist), K/F

Out (4 slots): M (either P or Q must be out, so M must be out), P/R/Q, P/R/Q, K/F (at least one is always out)

Because my out group filled up quickly, it forced either K or F into the in group. Since this question is looking for an either/or pair that must go in, the only pair on the board is K/F, given in answer choice (A).

Hope this clears things up!

[Xantippe]

Yes, I assumed only one distribution! I re-did the question with a 3-1-1 and a 2-2-1; and clearly saw that either F or K are required.


Thanks so much!

[vangorgc]

Hi PowerScore -
I ended up getting this one right, but it took me a while before I just put what I *thought* was right and moved on. Going over it, I can't figure out why my answer is right.
I'm confused why a) the distribution has to be 5; b) why the chemist that we need at least one of needs to be L. Why can't it be, for example, that F is included, and K is excluded?
Thanks,
Grace

[KelseyWoods]

Hi Grace!

The very first line of the stimulus tells us that we are selecting a panel of 5 people. Since we have to have at least one of each of our 3 types of scientist (first rule), that means that our possible distributions of scientists selected could be 3-1-1 or 2-2-1.

If G and H are selected, that means that we have at least 2 botanists, which kicks in that 2nd rule: if more than one botanist is selected then exactly 1 zoologist is selected. That means our distributions could be 3B-1C-1Z or 2B-2C-1Z.

If we have 3B-1C-1Z:
We'd have all 3 botanists (F G H). Since we have F, we cannot have K (3rd rule). If we had M, we'd have to have 2 zoologists: P and R (5th rule). So our 1 chemist must be L. And then we could have any 1 of our zoologists. So the group of 5 would be:
F G H L P/Q/R

If we have 2B-2C-1Z:
Again, we cannot have M because if we had M we'd have to have 2 zoologists (both P and R). So our 2 chemists must be K and L. Since we have K, that would mean that we cannot have F. So our 2 botanists would be G and H. And again we could have any 1 or our zoologists. So the group of 5 would be:
G H K L P/Q/R

The question stem states: "If both G and H are among the scientists selected, then the panel must include either." This means that in the answer choices we are looking for a pair of variables where we have to have one or the other. We always have to have either F (the 3B-1C-1Z possibility) or K (the 2B-2C-1Z possibility). Thus, answer choice (A) is correct. Again, it's not that we have to have both F and K. In fact we can't have them both. We could have F without K or K without F. But we can't make a group of 5 with neither K nor F. We always have to have either one of them. For each of the other answer choices, you could create a group of 5 without either of the options in the answer choice.

Hope this helps!

Best,
Kelsey