Q23

[lizcarney5]

I understand how choices B,C, D and E are incorrect, but what is the most efficient way to approach this problem? Is it by disproving BCDE and accepting A by default?

[timmydoeslsat]

On this exact question, I would be hesitant to select (A) straight up without looking at the other four choices. These other four choices, as you suggested, are simply absolute no-go's.

[giladedelman]

Yep, absolutely. Answer (A) is a head-scratcher. So we defer judgment. In fact, on "could be true" questions, that's typically how we proceed: not by proving that the right answer is possible, but by ruling out the wrong ones.

(B) is out because G is on day 1 and H cannot be.

(C) ditto.

(D) is out because that would force H into day 1.

(E) is out because it would force R into day 1, but G is already in that spot.

So, yeah, now we can just pick (A) and move on! No sense wasting time trying to prove that it works; the other ones were so easy to eliminate. In fact, that makes it a super easy question, provided you don't take the bait and try to prove (A) correct!

[aerialstrong]

For some reason A comes more natural and obvious for me when i'm doing this one. from the rule of "no same pet consecutively", so this comes to my mind:

ABA
CDC

and answer A: 2 days, 4 breeds.
But B.C.D.E took me quite a while to figure out when i was doing the review after the timed test.

It's a bit risky...but works for me.

[zana.nanic]

There is only one thing which is not clear to me.
Answer choice B states: Gs are featured on every day that Hs are.
That translates in If H, then G with the contrapositive If not G, then not H.


Position 1 has a G but cannot have an H. This doesn't go against answer choice B since the presence of G doesn't trigger any conditional. So this could not be the reason. Giladedelman you stated this as a reason, why?How do I translate that conditional?


I thought that B is wrong becuase we don't have space to place three Gs(the ones that go with the three Hs) and the once that goes in slot 1. Infact, it would be:
1g ( H cannot go here)
2(g cannot go here even if H can, but you know that if you place H you must place G and this is something you cannot do)
3gH (is the only slot it can go and then trying not to put the same elements consecutively you fill in the remaing two Hs)
4 skip
6 gH
7 in order to have your three Hs you should place an H here, but this would trigger G here and given the rule "If an element is first then it cannot be on day 7" placing G here would go against the rules!



CDE are clearly wrong:
C) if g, then H. WRONG due to slot 1 where you have a g and you cannot place an H

D)If not r, then H WRONG due to slot 1 where you have a g(not r) and you cannot place H

E)If not H, then R again due to slot 1 this is wrong.


is this correct?

[nbayar1212]

I agree with the above - B is wrong for a different reason than why C is wrong. Simply put, it seems that you can have 3 Gs max (because of the no consecutive breeds rule) and 3 Hs max (which is given) and no matter what, on at least one day that Hs are features, G cannot be which is why B doesn't work.