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Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=17159)

The correct answer choice is (E)

If Z is selected, then from the third rule W is not selected. If W is not selected, then from the contrapositive of the fourth rule M is not selected. Thus, answer choices (A), (B), and (C), each of which contain M, are wrong.

Answer choice (D) is incorrect because it would result in an insufficient number of stones to make the six rings. If no ruby is selected, then all three sapphires must be selected. However, that means that M is selected, which from the fourth rule forces W to be selected. Since Z is already selected according to the question stem, this results in a violation of the third rule.

Answer choice (E) can be confirmed by the following hypothetical: X–Y–Z–F–G–H.
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So, I identified few possible solutions for: If Zt is selected and three sapphires are selected, Fr, Gr, and Hr are selected, at least one of Js or Ks but not both could be selected, and at least Xt or Yt must be selected, or both, if both Xt and Yt are selected, Js and Ks cannot be selected.

Or alternately, if Zt is selected and two sapphires and one ruby are selected, Js and Ks must be selected, since M cannot be selected, one of Fr, Gr, or Hr must be selected in order with the compliance of rule two, and Xt and Yt must be selected for the reason that one of the requirements of the game states that six stones are selected for the placement in six rings.

I have a couple questions regarding this scenario:

Is the language correct if so can it be better communicated in the way of being more concise?
 Jay Donnell
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Hi T.B!

This is a brutal game, and the number of numerical possibilities for how many stones of each type can be included for the rings is frustratingly large. Also, with so many of the players in the game being floaters, there is much open ambiguity that should slow us down before making too many concrete deductions.

Those distribution possibilities limited by rules one and two are:

R S T = 6

3 1 2
1 3 2
1 2 3
2 1 3
3 0 3
0 3 3

Those last two are often overlooked as students assume that at least one stone from each category must be selected, but the game in fact doesn't offer that limitation! With the knowledge that W and Z cannot be selected together ( W <-|-> Z ) the maximum Topazes that can be selected is 3.

For this question, however, like most every Local question, the trick is to input the new information given by the question into your predetermined Global deductions, and know when to stop when those deductions run dry.

By putting Z into the game, we can guarantee that W and (through the combination of rules 3 and 4) M must be unselected.

That leaves the game board looking like this:

Selected ___ ___ ___ ___ Z X/Y

Not selected: W M ___ ___

With Z kicking out W, at least X or Y (both floaters) must be selected, maybe even both of them.

However, this still leaves every potential distribution open but for the 0 3 3. With the huge amount of players in this game without rules, aka floaters, this is the end of the line for forced deductions.

The combination deduction that Z <-|-> M immediately eliminates answer choices A, B and C.

D is eliminated when you realize that if no rubies are selected, all three sapphires are selected which we actually eliminated in the same way as we did with A.

That leaves E as correct as last man standing, which is typical of most Could Be True answers, which are best answered by proving that the other four responses Cannot Be True. The 3 0 3 distribution is certainly still possible, but only for those students who realized wisely that it's possible to select zero members from a particular subgroup.

Hope this helps clear up the right mindset and approach to a Local, Could Be True question, and that this shift can save you lots of time and trouble in the future!
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Thank you, taking the time to review and write the numerical distributions helps with thinking about this question, as I was able to see clearly what could and could not occur, in combination with the grouping inference M :dblline: Z.

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