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									 -  Sat Jan 21, 2012 12:00 am
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							Setup and Rule Diagram Explanation
This is an Advanced Linear Game: Balanced, Identify the Templates.
This is an Advanced Linear game with four variables sets: the three bills, R, S, and T; the three votes of Fu; the three votes of Gianola; and the three votes of Herstein. Notably, either the three bills or the three voters could be chosen as the base. Operationally they will produce no difference. We have chosen to use the three bills as the base and create stacks for Fu, Gianola, and Herstein:
The choice of voting for (F) or against (A) will fill each space. Applying the rules creates the following basic diagram:
The rules provide a considerable amount of specific information: the number of “for” and “against” votes each bill receives; the minimum “for” and “against” votes by Fu, Gianola, and Herstein; and certain votes each voter casts. From the supplied information several inferences can be made. First, since there are two votes for the Recreation bill and one vote against the Recreation bill, and it has already been established that Fu votes for the bill and Gianola votes against the bill, it can be inferred that Herstein votes for the Recreation bill:
Furthermore, since only two voting options exist (F or A), dual-options can be placed on the remaining open spaces:
Of course, further information about some of the dual-options would affect the choices in other dual-options. Regardless, examining the diagram makes it apparent that the voting possibilities are limited. Since there are only four uncertain spaces and even those have restrictions, why not try to show every possibility? Although there are several ways to identify each possibility, the first step we will take is to look at the votes for the school bill. If Gianola votes against the school bill and Herstein votes for the school bill, only one solution exists:
In the diagram above, Gianola must vote for the tax bill since each council member votes for at least one bill; since there must be two votes against the tax bill, it can then be inferred that Fu votes against the tax bill.
The other scenario with the school bill switches the votes of Gianola and Herstein:
Unfortunately, this information does not completely determine the votes of Fu and Gianola on the tax bill. One must vote for the bill and the other must vote against. Since this produces only two scenarios, show each one:
Thus, since all the options for the school bill have been explored, it follows that all the options for the entire voting record have been explored. These three solutions comprise the final setup to the game:
With all of the possibilities fully realized, the questions can be destroyed at light speed.
					
										
					  					This is an Advanced Linear Game: Balanced, Identify the Templates.
This is an Advanced Linear game with four variables sets: the three bills, R, S, and T; the three votes of Fu; the three votes of Gianola; and the three votes of Herstein. Notably, either the three bills or the three voters could be chosen as the base. Operationally they will produce no difference. We have chosen to use the three bills as the base and create stacks for Fu, Gianola, and Herstein:
The choice of voting for (F) or against (A) will fill each space. Applying the rules creates the following basic diagram:
The rules provide a considerable amount of specific information: the number of “for” and “against” votes each bill receives; the minimum “for” and “against” votes by Fu, Gianola, and Herstein; and certain votes each voter casts. From the supplied information several inferences can be made. First, since there are two votes for the Recreation bill and one vote against the Recreation bill, and it has already been established that Fu votes for the bill and Gianola votes against the bill, it can be inferred that Herstein votes for the Recreation bill:
Furthermore, since only two voting options exist (F or A), dual-options can be placed on the remaining open spaces:
Of course, further information about some of the dual-options would affect the choices in other dual-options. Regardless, examining the diagram makes it apparent that the voting possibilities are limited. Since there are only four uncertain spaces and even those have restrictions, why not try to show every possibility? Although there are several ways to identify each possibility, the first step we will take is to look at the votes for the school bill. If Gianola votes against the school bill and Herstein votes for the school bill, only one solution exists:
In the diagram above, Gianola must vote for the tax bill since each council member votes for at least one bill; since there must be two votes against the tax bill, it can then be inferred that Fu votes against the tax bill.
The other scenario with the school bill switches the votes of Gianola and Herstein:
Unfortunately, this information does not completely determine the votes of Fu and Gianola on the tax bill. One must vote for the bill and the other must vote against. Since this produces only two scenarios, show each one:
Thus, since all the options for the school bill have been explored, it follows that all the options for the entire voting record have been explored. These three solutions comprise the final setup to the game:
With all of the possibilities fully realized, the questions can be destroyed at light speed.
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					 Dave Killoran
PowerScore Test Preparation
Follow me on X/Twitter at http://twitter.com/DaveKilloran
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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/



 . So, since you know how many people will vote for and against each group, you can visualize a defined-fixed grouping game, with R,S, and T being the groups, and respectively having 2-1-1 elements. We only have three variables (F,H,G), which must be used at least once (one for), but not more than twice (each has to be against some bill). Given the other rules, we do not have to worry about the latter, as G is not in R, F is not in S, and H is not in T. Recreation group is already filled with H and F, so we do not have to worry about it. S group must have either a G or a H, and T group must have either a F or a G. The only thing that we have to keep in mind is that G must be at least in one of these groups (as we need an affirmative for each variable). In the end, you will still have only three possibilities left (like in the book), but you could avoid all those F and A.
. So, since you know how many people will vote for and against each group, you can visualize a defined-fixed grouping game, with R,S, and T being the groups, and respectively having 2-1-1 elements. We only have three variables (F,H,G), which must be used at least once (one for), but not more than twice (each has to be against some bill). Given the other rules, we do not have to worry about the latter, as G is not in R, F is not in S, and H is not in T. Recreation group is already filled with H and F, so we do not have to worry about it. S group must have either a G or a H, and T group must have either a F or a G. The only thing that we have to keep in mind is that G must be at least in one of these groups (as we need an affirmative for each variable). In the end, you will still have only three possibilities left (like in the book), but you could avoid all those F and A.