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- Wed Nov 20, 2019 1:17 pm
#72144
This game is discussed in our Podcast: LSAT Podcast Episode 31: The September 2019 LSAT Logic Games Section
Setup and Rule Diagram Explanation
This is a Advanced Linear: Unbalanced: Overloaded, Numerical Distribution game.
The scenario establishes that there are five animals in numbered pens, 1-5. This creates a Linear base for the game. Then, each pen contains an animal (puppy or kitten) and a name. This creates two rows above the five pens, for our Advanced Linear base:
We'll reserve further comment on this aspect until the rules are examined.
Rule #1: This is an extremely helpful rule, with two strong effects:
1. It limits the numerical combinations of kittens and puppies. Now the 1 kitten - 4 puppies option is eliminated:
2. It establishes the type of animal in the first and fifth pens:
Rule #2: Taffy—which is a ridiculous name for a dog—cannot be in a pen next to a kitten. When combined with the first rule, this rule turns out to be extremely limiting:
Rule #3: This rule states that Garnet or Honey, but not both, must be used. G and H are both kittens, and so now the maximum number of kittens that can be used is three:
Thus, another numerical option can be eliminated:
Note that if only two kittens were displayed, one of those kittens would be G or H, and the other would have to be F or J.
Rule #4: The final rule connects the kitten and puppy groups, and states that if Wags is displayed, then Garnet is displayed in the second pen:
Since Garnet is a kitten, you can also connect this rule to the second rule:
Note that, the game has only two remaining numerical options for the distribution of kitten and puppies into the five pens:
Setup and Rule Diagram Explanation
This is a Advanced Linear: Unbalanced: Overloaded, Numerical Distribution game.
The scenario establishes that there are five animals in numbered pens, 1-5. This creates a Linear base for the game. Then, each pen contains an animal (puppy or kitten) and a name. This creates two rows above the five pens, for our Advanced Linear base:
- K: F G H J 4
P: R S T W 4
Name: ___ ___ ___ ___ ___
P/K: ___ ___ ___ ___ ___
1 2 3 4 5
- 4 kittens - 1 puppy
3 kittens - 2 puppies
2 kittens - 3 puppies
1 kitten - 4 puppies
We'll reserve further comment on this aspect until the rules are examined.
Rule #1: This is an extremely helpful rule, with two strong effects:
1. It limits the numerical combinations of kittens and puppies. Now the 1 kitten - 4 puppies option is eliminated:
- 4 kittens - 1 puppy
3 kittens - 2 puppies
2 kittens - 3 puppies
1 kitten - 4 puppies
2. It establishes the type of animal in the first and fifth pens:
- Name: ___ ___ ___ ___ ___
P/K: _K_ ___ ___ ___ _K_
1 2 3 4 5
Rule #2: Taffy—which is a ridiculous name for a dog—cannot be in a pen next to a kitten. When combined with the first rule, this rule turns out to be extremely limiting:
- Taffy is a puppy, and so cannot be in the first or fifth pens.
Since Taffy cannot be next to a pen with a kitten, Taffy cannot be in the second or fourth pens.
Thus, if Taffy is to be used, Taffy would have to be in the third pen, with puppies in the second and fourth pens.
This leads to two basic inferences:
- 1. If Taffy is displayed, it's in the third pen and the numerical distribution is 2 kittens - 3 puppies:
T T3 2K/3P
2. Taffy, if displayed, would have to appear in the middle of a giant puppy block:
- Name: ___ ___ _T_ ___ ___
P/K: _K_ _P_ _P_ _P_ _K_
1 2 3 4 5
- Name: ___ ___ _T_ ___ ___
- 1. If Taffy is displayed, it's in the third pen and the numerical distribution is 2 kittens - 3 puppies:
Rule #3: This rule states that Garnet or Honey, but not both, must be used. G and H are both kittens, and so now the maximum number of kittens that can be used is three:
- Kittens: F, G/H, J
Thus, another numerical option can be eliminated:
- 4 kittens - 1 puppy
3 kittens - 2 puppies
2 kittens - 3 puppies
1 kitten - 4 puppies
Note that if only two kittens were displayed, one of those kittens would be G or H, and the other would have to be F or J.
Rule #4: The final rule connects the kitten and puppy groups, and states that if Wags is displayed, then Garnet is displayed in the second pen:
- W G2
- If W is displayed, then G must be displayed, and consequently H cannot be displayed, resulting in:
W H
Since Garnet is a kitten, you can also connect this rule to the second rule:
- If W is displayed, then G (a kitten) must be displayed second. This means T could not be displayed, resulting in the inference that W and T cannot be displayed together:
W T
Since W and T are both puppies, this means the puppy selection options are: R, S, T/W
- When W is displayed, G is displayed second. This means kittens are in the first, second, and fifth pens (and those three kittens would be G in 2, and F and J in the other two pens). Since you can only have three kittens maximum, the fourth and fifth pens would have to be puppies (one of which is W and then the choice of R or S).
Note that, the game has only two remaining numerical options for the distribution of kitten and puppies into the five pens:
- 3 kittens - 2 puppies — the kitten selection would be F, G/H, J
2 kittens - 3 puppies — the puppy selection would be R, S, T/W; the two kittens would be one each from the G/H and F/J pairs
Dave Killoran
PowerScore Test Preparation
Follow me on X/Twitter at http://twitter.com/DaveKilloran
My LSAT Articles: http://blog.powerscore.com/lsat/author/dave-killoran
PowerScore Podcast: http://www.powerscore.com/lsat/podcast/
PowerScore Test Preparation
Follow me on X/Twitter at http://twitter.com/DaveKilloran
My LSAT Articles: http://blog.powerscore.com/lsat/author/dave-killoran
PowerScore Podcast: http://www.powerscore.com/lsat/podcast/