Q11

[jrdn_pearl]

i set the game up properly and i had 4/5 correct. number 11 gave me some trouble, i was having bad vision. i went for N and S choice D. now i see that S can go with T but i cannot see why C is right.
any feedback would be greatly appreciated.

jp

[aileenann]

Thanks for your question. It's always important to understand both why the wrong answers are wrong but also why the right answers are right (at least during practice when you have the luxury of unlimited timed).

The list of people we choose as the correct answer should be the list of people any of whom individually cannot play a game against T. I'll demonstrate for you why the answer is (C).

The easiest way to deal with this problem is probably if you worked through this example with 2 frames, as dictated by the places of the SOS chunk which can either be SOS in 1, 2, 3 respectively or SOS in 2, 3, 4 respectively.

In the first frame, where SOS is in 1,2,3, every game already has someone assigned (1 = S, 2 = O, 3 = S, 4 + L). At this point T, P, and N have not been placed. Since there has been someone assigned to every single game already there can be no game that can pair T and P or T and N because there are no completely empty games to fill.

Now let’s think about the second frame, where S,O,S is in 2, 3, and 4. In that case, three of the games (2, 3, 4) are partly or fully filled, and again, T, P, and N have not been assigned. Now remember that T cannot be placed in 1 and 3. In that case, the only empty game remaining is 2, so T must go there, playing against S. In such a case, T cannot play against P or N.
Since in either of the two frames T cannot play against P or N, that means that overall T cannot play against P or N.