This is the LG about tenants and homeowners being selected to go on a committee:
Question 14: I can easily narrow the answer choices down to A (f) and E (r) but cannot figure out which answer it is! I draw my first graph in this order F J M Q R/S which give me answer A. However, you can draw it as Q J M R S and get E as the answer. I'm obviously not catching on to something. Help please.
Question 19: I finally figured this answer out after almost 20 minutes. However, I'm not to sure on my reasoning. I figured that C is right b/c if Q and K aren't selected, P would have to be placed in the graph and wouldn't work with M being in the graph. Thus, C is the answer. It took me forever to make this inference. Is this something I should have seen right off the bat? What method did you guys do in getting this answer?
PT 11 Section 1 Question 14 and 19

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Re: PT 11 Section 1 Question 14 and 19
#14: The committee has to be comprised of J, M, Q, R, and S. Applying the second rule, we know that there will be two tenants and three homeowners. Since J is selected, the fifth rule dictates that M must also be selected. Finally, since M is selected, we know that P is not selected. Since there must be three homeowners on the committee, all three of Q, R, and S must be selected.
#19: You can use your previous work to eliminate some of the choices.
(A) see #17
(B) see #15
(C) Correct
(D) see #15
(E) to eliminate this choice, you could create a hypothetical with both R and S out (F, G, K, P, and Q in this case)
Depending on how linked the variables of the particular game are, you may be able to approach a question like this by assuming one of the two variables in each pair is out, and then checking to see if this condition forces the other variable in the pair to be selected.
#19: You can use your previous work to eliminate some of the choices.
(A) see #17
(B) see #15
(C) Correct
(D) see #15
(E) to eliminate this choice, you could create a hypothetical with both R and S out (F, G, K, P, and Q in this case)
Depending on how linked the variables of the particular game are, you may be able to approach a question like this by assuming one of the two variables in each pair is out, and then checking to see if this condition forces the other variable in the pair to be selected.
Last edited by Cambridge LSAT on Tue Feb 16, 2010 3:27 am, edited 1 time in total.

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 Joined: Tue Jul 14, 2009 7:23 pm
Re: PT 11 Section 1 Question 14 and 19
Cambridge LSAT,
The answer to question 19 is C, not D. I don't see how you can eliminate, for example, B from question 15. In question 14, you still have F and K left over to add too the committee. I don't see how you can eliminate them and come up with J M Q R S.
Thanks for you response
The answer to question 19 is C, not D. I don't see how you can eliminate, for example, B from question 15. In question 14, you still have F and K left over to add too the committee. I don't see how you can eliminate them and come up with J M Q R S.
Thanks for you response

 Posts: 270
 Joined: Mon Aug 24, 2009 3:26 pm
Re: PT 11 Section 1 Question 14 and 19
#14: Since M is the chairperson, the second rule dictates the following slot distribution:
T: __ __
H: __ __ __
The fifth rule ties together J and M, thus filling the two tenant slots. As a result, F and K must both be out.
#15: With F as the chairperson, the second rule dictates the same slot distribution as above. J and M must both be out, since there's only one remaining tenant slot, and the fifth rule ties them together. G must also be out, since the fourth rule would be triggered, and we only have one remaining tenant slot. Thus, only K can fill the second tenant slot. Other than Q being selected (third rule), we're left with a skeletal diagram:
T: F K
H: Q __ __
You can use this diagram to disprove #19, choice B. It can also be used to disprove choice D on #19 since the two open slots can be filled with R and S.
T: __ __
H: __ __ __
The fifth rule ties together J and M, thus filling the two tenant slots. As a result, F and K must both be out.
#15: With F as the chairperson, the second rule dictates the same slot distribution as above. J and M must both be out, since there's only one remaining tenant slot, and the fifth rule ties them together. G must also be out, since the fourth rule would be triggered, and we only have one remaining tenant slot. Thus, only K can fill the second tenant slot. Other than Q being selected (third rule), we're left with a skeletal diagram:
T: F K
H: Q __ __
You can use this diagram to disprove #19, choice B. It can also be used to disprove choice D on #19 since the two open slots can be filled with R and S.

 Posts: 63
 Joined: Tue Jul 14, 2009 7:23 pm
Re: PT 11 Section 1 Question 14 and 19
Ahhh that helps. I comletely misunderstood rule 2. Thanks for clearing that up.
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