Q 17

[noah]

Let me re-post a discussion about this question here, to keep things nice and tidy!

Hi,

I can't understand why (E) is not right for the Q17. Because T can't be more than two in this set-up. There's no explanation about (E) in your diagram.

Can anybody explain it?

Thanks!

I think I can help with your question. The easiest way to figure out this question is to look back at diagrams that you did in figuring out the other questions.

Why D is right:
We know we have to have two M's, and at least one G, J, and T. For each J, we must have a G right before it. So if we have 2 J's, 2 G's, and 2 M's, that's already six and forces us to make the last destination T. If we had THREE J's, that would force us to have THREE G's, and our TWO required M's, which already makes seven and we didn't even include T.

Why E is wrong:
However, if we had three T's, that is, Freedom made MORE than 2 voyages to Trinidad, then we could have the following scenarios:

From 1-7: T M G J T M T, for example

Or you could have: M T G J T M T

There are many other possibilities. It's really easiest to just try it out and see if you can.

Basically, the point in these kinds of questions are to look for the variables that have constraints on them. Example, J has a huge constraint on it because it's always attached with G. M has a constraint on it because we know there are only 2 and they are separated by at least one G.

T is the only variable without constraint--we know it has to be in the 7th spot, but that's it. For questions that are must be true/false, etc the answer tends to be variables with constraints on them, because those usually have to be one way or another (there aren't as many possibilities about where to put them).

I am also questioning the difference between answers D and E. I can see why D must be true. However, since J is never at week 4, T cannot be visited more than two times within the seven day range. Am I missing something here?

[noah]

I am also questioning the difference between answers D and E. I can see why D must be true. However, since J is never at week 4, T cannot be visited more than two times within the seven day range. Am I missing something here?

I'll "break" (E): With 3 Ts, we can do it! We could have

T M T G J M T

So, (E) is not something that must be true.

Let's try to "break" (D) - what happens if we have 3 voyages to J. Can that work? Why not?

[JanetS46]

I'll "break" (E): With 3 Ts, we can do it! We could have

T M T G J M T

So, (E) is not something that must be true.

Let's try to "break" (D) - what happens if we have 3 voyages to J. Can that work? Why not?

D cannot be broken, and is therefore the correct answer. The reason for this is because G must always come before J. Since T takes up slot 7, we only have 6 slots left. If we went to J 3x (more than twice), we would also go to G 3x. That means the rest of the 6 slots are taken- no room for M, which we must go to exactly twice according to the rules. Therefore, we can only go to J at most 2x.

[LawrenceG622]

My question is in regards to the wording of the answers here. If the answer choice says we can have "at most two voyages to Jamaica" does this mean we must verify that the cruise can actually make two voyages? Let's suppose that given the games rules, the cruise can only make one trip to Jamaica (I know this is not true). I see the general idea here is that we clearly cannot fit Jamaica three times in the schedule, but why isn't it necessary to verify that we can fit Jamaica in twice?

[ohthatpatrick]

I see what you're saying, but I don't think the phrasing "at most" implies that the maximum stated is a proven possibility.

If you were at my house and said, "Ya got any beers in the fridge?"
and I said, "We've got at most 2", I would only be wrong if you went to the fridge and saw 3 or more.

If you see 0, 1, or 2 beers, I was still saying something true when I said "We've got at most 2".

The phrasing "at most" is a fuzzy basement, but a hard ceiling.

I'm not committing to whether we have 0, 1, or 2.
I'm only committing to the idea that we do NOT have 3 or more.

At any rate, if it makes you feel better, we COULD do two trips to J:

G J M G J M T

Hope this helps.

[AbhistD667]

I did understand why D is right but my concern is how can we deal with this question in an efficient way? I eliminated A and B using diagrams of previous question but I had to try out the rest of answer choices which took time.

[Misti Duvall]

I did understand why D is right but my concern is how can we deal with this question in an efficient way? I eliminated A and B using diagrams of previous question but I had to try out the rest of answer choices which took time.

I'd initially framed this game based on the possibilities for the GJ chunk (since we know J has be used at least once, and G has to come immediately before it, there has to be at least one GJ chunk). And, since J can't go in 4 and T has to be in 7, there are only three possibilities:

1) GJ is somewhere in 1-3, either 1-2 or 2-3
2) GJ is in 4-5
3) GJ is in both two of the 1-3 spots and 4-5

Which means that there are at most two Js, cause there's no more room for the GJ chunk.

Hope this helps.