Both numbers in “1, 2” are smaller than 3, which goes against the condition a ＞ 3. If a were 1 or 2, machine A's evaluation value would fall below the 3 of machine B and machine C, so machine A would be the inferior one in performance — the opposite of machine A ＞ machine B and machine A ＞ machine C in Table 4. This is the wrong answer that is easy to pick when the direction of the inequality signs is taken the wrong way round (misreading it as a smaller evaluation value being better).
×Incorrectイ
What is wrong with “3, 4, 5, 6” is that it includes both endpoints, 3 and 6. If a ＝ 3, machine A would have the same evaluation value as machine B and machine C, which does not satisfy the comparison result machine A ＞ machine B (and machine A ＞ machine C) that says one is superior to the other. If a ＝ 6, machine A and machine D would have equal values, which does not satisfy machine A ＜ machine D. The condition is not 3 ≦ a ≦ 6 but 3 ＜ a ＜ 6, and equality is not allowed.
○ウ
正解The relative comparison table in Table 4 gives three relations regarding performance: machine A ＞ machine B, machine A ＞ machine C, and machine A ＜ machine D. As item (4) of the selection requirements in question set A states, “the machine with the largest overall evaluation value is selected”, so a larger evaluation value means a better machine, and the direction of each inequality sign in the relative comparison table corresponds directly to the size relation between the evaluation values. In Table 5, machine B ＝ 3, machine C ＝ 3, and machine D ＝ 6 are already filled in, so the following conditions are imposed on a. From machine A ＞ machine B, a ＞ 3. From machine A ＞ machine C, a ＞ 3. From machine A ＜ machine D, a ＜ 6. Putting these together gives 3 ＜ a ＜ 6. In addition, item (3) of the selection requirements lays down that “an evaluation value is one of 1 to 9”, so an evaluation value can only be an integer from 1 to 9. The only integers from 1 to 9 that satisfy 3 ＜ a ＜ 6 are 4 and 5, so the sequence of values a can take is “4, 5”, and ウ is appropriate.
×Incorrectエ
All the numbers in “7, 8, 9” are larger than 6, which goes against the condition a ＜ 6. Machine D's evaluation value is 6, so if a were 7 or more, machine A would be the better one in performance, contradicting machine A ＜ machine D in Table 4. This is the wrong answer that is easy to pick when you look only at two of the three rows — machine A ＞ machine B and machine A ＞ machine C — and drop the upper-bound condition; the comparison results have to be taken all together as a simultaneous set of every row.
ここがポイント
Read the inequality signs in a relative comparison table (a pairwise comparison table) as corresponding directly to the size relation between evaluation values. In this question, item (4) of the selection requirements, “the machine with the largest overall evaluation value is selected”, lets you judge that a larger evaluation value means better. For an unknown evaluation value, the key point is to take all the related comparison results together as a simultaneous set, pin down both the upper and the lower bound, and narrow down the range of values it can take.
Once the range has been narrowed, apply the constraint on the domain of the evaluation values and list the candidates. In this question, item (3) of the selection requirements, “an evaluation value is one of 1 to 9”, limits evaluation values to integers from 1 to 9, so the candidates satisfying 3 ＜ a ＜ 6 are fixed at just two, 4 and 5. Note also that ＞ and ＜ are strict inequality signs that do not include equality, so the boundary values themselves (3 and 6 here) do not enter the candidate list.