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- Mon Jan 20, 2014 12:00 am
#43583
Setup and Rule Diagram Explanation
This is a Grouping: Defined-Fixed, Unbalanced: Overloaded game.
L is the only random fish in the game, and Z is the only random plant in the game.
Exactly three of the five fish must be selected. This means that only two of the five fish will not be selected. And, because exactly two plant species will be selected, exactly two will not.
Because X is a necessary condition for two other variables, you should examine the effects of the contrapositive. If X is not selected, neither H nor K can be selected. Therefore, if X is not selected, G, J, and L must be the three fish selected. When J is selected, then W is selected, and when G is selected Y cannot be selected, leaving Z to be selected. Hence, when X is not selected, only one solution to the game exists:
A very tricky inference involves Y and X. When Y is selected, G cannot be selected. This leaves only four fish in the selection pool (no pun intended). Because there must be exactly three fish selected, we cannot eliminate X from the selection list because removing X would remove H and K as well, which would leave an insufficient number of fish. Thus, if Y is selected, the other plant that must be selected is X. Consequently, if Y is selected, J cannot be selected, since J requires W, and from the first rule G cannot be selected, meaning that when Y is selected there is only one possible solution to the game.
This inference is tested on questions #16 and #17.
This is a Grouping: Defined-Fixed, Unbalanced: Overloaded game.
L is the only random fish in the game, and Z is the only random plant in the game.
Exactly three of the five fish must be selected. This means that only two of the five fish will not be selected. And, because exactly two plant species will be selected, exactly two will not.
Because X is a necessary condition for two other variables, you should examine the effects of the contrapositive. If X is not selected, neither H nor K can be selected. Therefore, if X is not selected, G, J, and L must be the three fish selected. When J is selected, then W is selected, and when G is selected Y cannot be selected, leaving Z to be selected. Hence, when X is not selected, only one solution to the game exists:
A very tricky inference involves Y and X. When Y is selected, G cannot be selected. This leaves only four fish in the selection pool (no pun intended). Because there must be exactly three fish selected, we cannot eliminate X from the selection list because removing X would remove H and K as well, which would leave an insufficient number of fish. Thus, if Y is selected, the other plant that must be selected is X. Consequently, if Y is selected, J cannot be selected, since J requires W, and from the first rule G cannot be selected, meaning that when Y is selected there is only one possible solution to the game.
This inference is tested on questions #16 and #17.
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Dave Killoran
PowerScore Test Preparation
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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/