# Reply To: pt 57 section 3 game 3

July 24, 2016 at 9:35 pm #2298
LSAT Dan
Participant

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:
TU
TV
TL
UV
UL
VL

But two of the pairs aren’t allowed: UL and UV. So there are only four possibilties:

TU
TV
TL
VL

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:

Mauve L
Mauve V
Red S
Green T

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.

-Dan