P.397 Q #6

    • May 22, 2016 at 4:58 pm #1894

      Hi Mike,

      I’m back at Minor Questions on p. 397 and I am not sure about q.6. Answer is C. But I’m not clear why A. isn’t correct. Explanation says that it doesn’t put P&H on opposite shifts. But if P&N are always together, wouldn’t H2 make N go into Shift 1?

      Thanks again!


    • May 23, 2016 at 1:57 pm #1907
      Mike Kim

      Hey Carmen — hope you don’t mind but I moved your Q to a new thread so that if others have the same question they can more easily find this in the future —

      Per answer (A), which says that If N works 1, H works 2 —

      We know that it two things must be true: If N is in 1, H is in 2 and the contrapositive: If H is in 1, N is in 2.

      Per these rules, there are 3 possibilities for how N and H could be arranged: (N1, H2) (H1, N2) (N2, H2).

      It is the third of these possibilities that makes it so that an “if…then” statement has a different consequence than a straight-up “P and H will work different shifts,” and that’s why (A) doesn’t have the same meaning that (C) does — (C), which is a biconditional (discussed on pg. 191), does not allow for them to work either shift together — (C) only allows for two options: (N1, H2) and (H1, N2) just like the original rule only allowed for two possibilities —

      I hope that clears it up — good luck studying this week — Mike


    • May 23, 2016 at 6:59 pm #1909

      Hi Mike, yes, that cleared up the confusion, the biconditional information on p, 191 was exactly on the money for this question. This is starting to come together, slowly. Thanks again!!!

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