PT 34 LG #4 Forum

[PeanutsNJam]

In the set-up of this LG, there is a constraint that, if true, wouldn't be possible. It's a binary grouping game. I'll give a simplified example:

Constraint 1: If A is chosen, then B is not.
Constraint 2: If B is not chosen, then C is.
Constraint 3: If C is chosen, then A is not.

Therefore, the conclusion is apparently that A *cannot* be chosen, because it leads to a logical impossibility.

I caught that, but spent all my time thinking I made a mistake because the LSAT would *never* place a constraint that, if all other constraints are true, is impossible. But apparently they do? Has anybody encountered this anywhere else? It's just a red herring?

Here's the Manhattan LSAT diagram because I'm too lazy to explain the game:

http://www.manhattanlsat.com/forums/diagram-t268.html

[03152016]

yeah they do it a couple times
off the top of my head, the nine treatments game and the souderton/randsborough game
if triggering a conditional results in a logical impossibility, it won't trigger

[Colonel_funkadunk]

yeah they do it a couple times
off the top of my head, the nine treatments game and the souderton/randsborough game
if triggering a conditional results in a logical impossibility, it won't trigger

This. The souderton game is the one I thought of too. It just means it won't happen, they do it to trip you up

[03152016]

make sure to eliminate the element
if my elements are: A B C D E F G
and i have conditional A -> D
and i realize it can't trigger
that means A is out of the game
go to the top of your main where you should have written the elements out
and strike out A

[Harry_Pluxen]

In the set-up of this LG, there is a constraint that, if true, wouldn't be possible. It's a binary grouping game. I'll give a simplified example:

Constraint 1: If A is chosen, then B is not.
Constraint 2: If B is not chosen, then C is.
Constraint 3: If C is chosen, then A is not.

Therefore, the conclusion is apparently that A *cannot* be chosen, because it leads to a logical impossibility.

I caught that, but spent all my time thinking I made a mistake because the LSAT would *never* place a constraint that, if all other constraints are true, is impossible. But apparently they do? Has anybody encountered this anywhere else? It's just a red herring?

Here's the Manhattan LSAT diagram because I'm too lazy to explain the game:

http://www.manhattanlsat.com/forums/diagram-t268.html

I think that the difficulty subsides when you realize that L (I think it was L?) has to be out of the game (at the R clinic).

Then its just a matter of using the conditional rules and their contrapositives.

here:http://7sage.com/lsat_explanations/lsat ... -4-game-4/