LSAT Prep question... formal logic game

DrPepper237
Posts: 7
Joined: Tue Jul 26, 2011 3:20 pm

LSAT Prep question... formal logic game

Postby DrPepper237 » Tue Jul 26, 2011 3:35 pm

So... i'm prepping right now with the 2nd edition of the McGraw-Hill's Conqering the LSAT Logic Games book. I am working through formal logic game 3 and I am disputing question 3.

Here are the constraints. There are 7 fish that can be put in the aquarium, Anna, Ben, Chris, Dana, Evan, Frank, and Garry.

If Anna is in, then Frank is out.
If Chris is out, then Anna is in and Ben is in.
If Evan is out, then Chris is out.
If Dana is in, then Evan is out.

The question: What is the maximum number of fih that could be present in the aquarium?
A. 3
B. 4
C. 5
D. 6
E. 7

I answered with A (3 fish). The book says the answer is actually C (5 fish). It say that Five fish can be present in the aquarium: Anna, ben, Chris, Evan, and Garry. You cannot hav Frank because of the contraint that says if Anna is in, then Frank is out. You cannot have Dana due to the contrapositive that says if Dana is in, then Evan is out (therefore, if Evan I in, Dana is out).

So here are the diagrams I have.
D --> not E --> not C --> A (also not C-->B) --> F
F --> not A (also not B) --> C --> E --> not D
So HOW can Chris be in the tank at the same time as Ben under the contraint AND contrapositives concerning “Chris is out, thn Anna is in and Ben is in” which implies that if Ben is out then Chris is in?? What am I missing?

bhan87
Posts: 850
Joined: Mon Mar 12, 2007 8:08 pm

Re: LSAT Prep question... formal logic game

Postby bhan87 » Tue Jul 26, 2011 3:50 pm

Diagrammed:

A -> Not F
Not C -> A
Not C -> B
Not E -> Not C
D -> Not E

Combine these into a conditional chain:

D-> Not E -> Not C -> A -> Not F
(Under not C, draw another conditional arrow with "-> B")

Contrapositive:

F -> Not A -> C -> E -> Not D
(Under C, draw another conditional arrow with "Not B ->")

Under the Contrapositive situation, F, B, C, E, G can all be in at the same time. Note that "Not A" and "Not B" are sufficient conditions for C to be in, but that does not mean that C has to be OUT if you place either A or B in (only that if either A or B are out, then C has to be in). This is because our sufficient condition for B only applies to when he's OUT. There are no rules for when he's IN. Under that situation you can have 5 situations:

1. Just C
2. A and C
3. B and C
4. A, B, and C
5. A and B

The only impossible siutations are: Just A, Just B

To explain the other fish being in, if we Assume F is in, the chain says that A must be out, which forces C in (now we have F and C). If C is in, then E must also be in (F, C, and E). G is a floater, so he can be in or out at any time, let's put him in to maximize the number (F, C, E, and G). Lastly, we must be careful about our conditional rules. The only situation that B triggers anything is if he's out (that forces C to be in), but remember that B being IN doesn't trigger ANYTHING.

Edit: TLS was messing up my chain formatting, so sorry for the ghetto parantheses

DrPepper237
Posts: 7
Joined: Tue Jul 26, 2011 3:20 pm

Re: LSAT Prep question... formal logic game

Postby DrPepper237 » Wed Jul 27, 2011 7:12 pm

Ooooooooooooooooooooooooo

Derp :? Lol

Thank you so much! You've opened a whole new world of logic for me :D




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