I just tried possibilities with the base diagram of rules (with the changed condition as well) but wondering what shortcut there can be -
I got it wrong for my first try as I chose (E) since I didn't think much of 'Must Be True' part of the question.
My second try had me get it right but I just couldn't reduce the work conspicuously.
I diagrammed possibilities centering on Car 5 and 6 since either one would be the purple I (as we need to have purple II among the rest cars than car five and six.
Is it the quickest possible way to get this question right?
Please answer me!
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Re: Q24
Aaah! You picked the worst time to have a "Please answer me!" post! (I was at a family wedding all weekend and then my laptop crashed on Monday, so I can't access all my prep tests).
So sorry for the delay.
The quickest way to think about this problem is to focus on 3 G's, none of which can be next to each other.
If we have to put 3 G's in six spots, and they can't TOUCH each other, that's very limited!
3 non-touching G's requires five spots all by itself: G __ G __ G
So where could the 3 G's go? Normally, in six spots, you could separate your 3 G's by doing
1 3 5
1 3 6
1 4 6
2 4 6
In this case, though, we know that spot 4 is O/P, according to the "no G in 4" rule.
So we really only have to consider the 3 G's in
1 3 5
or
1 3 6
We could frame this, or we could just put a G into our 5/6 cloud where P is (from the rule that there is a P in 5 or 6).
__. __ .__ . O/P. (G, P)
I know that there are G's for sure on 1 and 3.
G. __. G. O/P. (G, P)
So the remaining piece is the other half of the "O/P" placeholder in spot 4.
G . P/O. G. O/P. (G, P)
As it turns out, we could have stopped working a while ago.
They just want to know where there must be G's, so we had that answer from
13 5
or
13 6
There MUST be a G in 1 and 3. So, it's (D).
If you want to practice spacing out 3 copies of the same letter (that can't touch each other), play the "Fuentes Game" from test 44.
https://www.manhattanprep.com/lsat/foru ... -f292.html
So sorry for the delay.
The quickest way to think about this problem is to focus on 3 G's, none of which can be next to each other.
If we have to put 3 G's in six spots, and they can't TOUCH each other, that's very limited!
3 non-touching G's requires five spots all by itself: G __ G __ G
So where could the 3 G's go? Normally, in six spots, you could separate your 3 G's by doing
1 3 5
1 3 6
1 4 6
2 4 6
In this case, though, we know that spot 4 is O/P, according to the "no G in 4" rule.
So we really only have to consider the 3 G's in
1 3 5
or
1 3 6
We could frame this, or we could just put a G into our 5/6 cloud where P is (from the rule that there is a P in 5 or 6).
__. __ .__ . O/P. (G, P)
I know that there are G's for sure on 1 and 3.
G. __. G. O/P. (G, P)
So the remaining piece is the other half of the "O/P" placeholder in spot 4.
G . P/O. G. O/P. (G, P)
As it turns out, we could have stopped working a while ago.
They just want to know where there must be G's, so we had that answer from13 5
or
13 6
There MUST be a G in 1 and 3. So, it's (D).
If you want to practice spacing out 3 copies of the same letter (that can't touch each other), play the "Fuentes Game" from test 44.
https://www.manhattanprep.com/lsat/foru ... -f292.html
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