PT 52 LG 4 - conditional logic Q.

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tuffyjohnson
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PT 52 LG 4 - conditional logic Q.

Postby tuffyjohnson » Tue Mar 26, 2013 11:32 pm

Basic linear LG here. Is there a good way to handle these two rules? They took me a long time to process with question 19.

F-M ---> L-H

Either M's delivery is earlier than H's or it's later than K's but not both.

foggynotion
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Re: PT 52 LG 4 - conditional logic Q.

Postby foggynotion » Wed Mar 27, 2013 9:14 am

"Either M's delivery is earlier than H's or it's later than K's but not both."

I'd do it something like this (actually, I'd probably draw out M with two branches stemming out from it, but I can't seem to type that):

M-H
M-K
or
H-M
K-M

What I did was just take one option, and then decide what happens to the other element, and then I looked at the other option. In other words, first I said, let's say M is earlier than H--if it is, then M can't be later than K, so it must be earlier than K. Then I said, now let's say M is later than K--if it is, M can't also be earlier than H, so it must be later than H. Now I can easily see that M is either before both K and H, or after both of them. I think this makes it alot easier to deal with.

The other you mentioned, I'd do it just the way you did.

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tuffyjohnson
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Re: PT 52 LG 4 - conditional logic Q.

Postby tuffyjohnson » Wed Mar 27, 2013 9:48 am

Thanks!

che3055
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Re: PT 52 LG 4 - conditional logic Q.

Postby che3055 » Wed Mar 27, 2013 3:45 pm

Specifically with question 19, other rules limit the options you have.

If you know that F is 4th, based on other rules (the first rule it lists), G is 5th and K is 6th. Answer A fits this description, and that's all the work you have to do.

With K 6th, you know that it cannot K cannot be before M, so the alternative in the rule you cited must be true - M must be before H, since K cannot be before M. That leaves you with a first three of either MHL or MLH (this eliminates answers B, C, D), or LMH (eliminating E). Of course, all of this is optional really.




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