July 24, 2016 at 8:03 pm #2297
This game has probably been covered ad nauseam, but I’d very much like to see an LSAT Trainer approach to it. BTW, just watched the 7sage explanation. It was good, but I don’t think I’d ever think to map every possible inference… That said, the inference of U/V must be out was genius. Hopefully, I’ll remember to keep that in my back pocket. Two things to point out though, 1) they didn’t cover the scenario with both U and V out, 2) they made a faulty inference with the green iguanadon out… not even sure that they mattered. Anyways, thanks for your time.
July 24, 2016 at 9:35 pm #2298LSAT DanParticipant
One of the most important skills on the Logic Games section is figuring out which rules are more important than other rules. Here, for instance, the first rule (“Exactly two mauve toys are included”) is what I call a “key rule.” How can we recognize it as such? EVERY other rule is designed to prevent us from satisfying that first rule! The stegosaur can’t be mauve; the iguanadon can’t be mauve; the platesaur can’t be mauve; U and V can both be mauve – but they can’t both be included; and finally, L and U can both be mauve, but they can’t both be included, either. So focusing on that key rule – in other words, asking ourselves periodically, “What are my two mauve dinosaurs going to be?” is going to be important and helpful.
Brief digression, with respect to your comment about the U/V rule. That U or V (or both) must be out isn’t really even an inference; it’s just what that next-to-last rule is saying – if one of them is in, the other is out. It’s a little tricky, because of the word “only” and because the word “if” appears in the second half of the sentence, but this type of formal logic rule untangling should be bread & butter, and you might want to brush up on the formal logic drills if you translated this rule incorrectly.
The other thing I’d focus on, besides the mauve dinosaurs, is, as always, the smaller group – in this case, the out group. Only two dinosaurs won’t be included, and (at least) one of them will be U or V. So as soon as we’re told that another dinosaur is in the out group (as in question 13), then everyone else is in.
So, what COULD our mauve dinosaurs be? The only ones that can be mauve are T, U, V, and L. That gives us six pairs:
But two of the pairs aren’t allowed: UL and UV. So there are only four possibilties:
I wouldn’t map out four parallel universes (too many for my taste), but I’d definitely jot down the four options off to the side and keep them in mind – that’s our goto question when we’re stuck – what are the mauve dinosaurs going to be?
This will help untangle that acceptability question, where one of the eliminations (C) depends on the colors, even though they’re not given.
Quick look at the individual questions –
13. If T is out, notice that eliminates 3 of our 4 mauve pairs – we’re left with V and L as the mauve dinosaurs, and that’s enough to know that (D) must be correct; there are only exactly 2, and we know what they are.
14. Again, focus on the possible mauve pairs – TL, TV, TU, and LV. That eliminates the wrong answers:
B) is wrong, because all mauve pairs include L or T.
C), D), and E) are wrong, because in each case, we’d need the other two possible mauves to be included, and they would all result in displays that include both U and V.
A) is correct, and should be intuitive, because it gives us something we need – a mauve dinosaur. The other one can be T, and we can leave out U to avoid the U/V pair that eliminated (C), (D), and (E).
15. Again, what can our mauve dinosaurs be? If T isn’t one of them (because on this question it’s yellow), then the mauves are L and V, so (E) is correct.
16. If U is in, then V is out. If V is out, the mauve pair is either TL or TU; either way, we need a mauve tyrannosaur, so (A) is correct.
17. If there are two pairs of green toys, we know the color distribution: 2 greens, 2 mauves, and the red stegosaur. So (A) is wrong. No room for a yellow toy. What about (B)? If T is not mauve, we’d need to use L and V. Could that work? Test it…(It will be easy; the first 4 are gimmes:
We need another green,and we have the iguanadon. U and P are out. That works, so (B) is right.
Hope this helps; kind of rushed it, so please ask for any clarification.
July 25, 2016 at 6:41 am #2299dannypearlbergParticipant
Yeah, I don’t think this is a good game to split into multiple scenarios (as it was done on 7sage), and I don’t think I’d ever split a game into 6(!) scenarios. In general, a good rule of thumb when thinking about splitting a game into multiple scenarios is that you shouldn’t bother splitting if it’s going to result in more than 4 scenarios. So, besides the U/V “inference” (very important to always make this move for in/out games), I wouldn’t worry about any of the other inferences that they made.
July 25, 2016 at 7:29 am #2300
Big thanks to both the Dans. The breakdown of writing out the possible pairs of mauve dinos for sure helps and far more manageable–and effective in my opinion than 7sage’s approach. I guess I wasn’t very direct with my statement, but I when diagramming the game, I wrote out the conditional rule u->~v correctly, but I didn’t take the next step to see that it means that at least one of them must be out. I need to get more comfortable with the rules to see that, I guess.
Thanks again guys.
July 25, 2016 at 11:22 am #2301LSAT DanParticipant
Good thread…glad to see Danny weigh in, too. I strongly agree on the multiple scenarios point (although most of the time, even four is too many for me. I’ll sometimes do four, but my preference is to keep it to 2 or 3).
Thanks for the clarification of your thought process regarding the U/V rule. That’s a potentially very valuable opportunity for a great takeaway on his game, so I’d like to explore it a little further.
This is a very atypical game, in that the “mauve” rule controls so much of the game, that you’ve seen, you don’t really need to worry about much more. But most of the time, that U/V rule and its effect on the out group would be enormous (impossible to overstate, really), and if this game and that explanation brought you to appreciate it, that will have a huge effect on your future results. Here’s how “in & out” games and questions remove, most of the time – the small group fills up, and everyone else is forced into the large group.
For instance, back to question 13, once T is out, and you know that either U or V must also be out, everyone else (S, L, I, P, and other of U or V) must be in, so (A), (C), and (E) can immediately be dismissed (although you have to resort to the mauve rule to eliminate (B)).
The corrolary to that is that rules like the U/V rule are HUGE – they essentially serve to fill up both groups, and that’s especially important in the smaller group (if we knew that T were included, that doesn’t tell us much; the big group has a lot of flexibility. But when T is EXcluded, that’s huge).
Rules that say that two variables can’t be in the same group have an enormous impact in grouping (including in &’out games, especially) games with limited or fixed group sizes, but remember – focus on the smaller group.
A great companion game to look at and work through (or work through again) given your new appreciation for that U/V rule is game 1 from Preptest 63. Judges are being assigned to either the trial court, which has 6 openings, or the appellate court, which has 3 openings. One judge is directly assigned to the appellate court, and we have a rule that says that two of the judges – H and P – can’t be on the same court. That’s going to function similarly to the U/V rule – it’s going to fill up the smaller group. One spot goes to L, and another has to be saved for either H or P.
As soon as you see that setup, you can tell how the questions – most of them – will resolve: The appellate court will fill up as soon as one other judge is assigned to it, and every judge not yet accounted for will be forced onto the trial court. Various answer choices and “if questions” will give you that last appellate court judge.
In fact, EVERY question other than the acceptability question and the rule substitution question resolves this way. If you’re dialed into 1) the effect of the small group filling up; and 2) the way in which rules like the U/V rule and the H/P rule serve to further fill up the smaller groups, it’s going to do wonders not only for your accuracy, but also your timing.
You’ll see similar rules on questions throughout the past LSAT exams. On the “7 bills” game (Preptest 29 or 30, I think), you have the rule at bills 1 and 5 can’t be paid on the same day, and another rule that if Bill 6 is paid on Wednesday, Bill 7 is paid on Thursday. Those two rules are serving to fill up Thursday. Exact same idea. Rules that prevent two variables from being put in the same group always have a subtle effect that extends far beyond merely making sure to separate the variables – they often affect all of the other variables in the game.
Here’s how I explain it to my classes: let’s say 10 of us from LSATters are going to a concert. Mike Kim has (for purposes of this discussion) an awesome 2-seater Lamborghini (that’s the small group), and I have a beat-up pickup truck that seats as many people as we want. Now let’s say that two of the people on the board are an ex-wife and ex-husband who hate each other and can’t ride in the same vehicle (that’s out U/V) rule.
Notice the effect of the exes – you just got screwed out of your chance to ride in the Lamborghini! There’s no more room; no matter what other rules we create, the Lamborghini is going to be Mike and one of the exes. You’ll see this over and over and over, and it’s a great spinoff of the original dinosaur question that extends way beyond that one particular game.
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