This question's explanation and your answers so far.
表に示す構成のデータを，流れ図の手順で処理する場合について考える。流れ図中のx，y，zをそれぞれデータ区分A，B，Cと適切に対応させれば，比較（“xか?”，“yか?”，“zか?”）の回数の合計は，最低何回で済むか。
| データ区分 | 件数 |
|---|---|
| A | 10 |
| B | 30 |
| C | 50 |
| その他 | 10 |
流れ図：「開始」→「xか？」（No→「yか？」，Yes→「xの処理」）；「yか？」（No→「zか？」，Yes→「yの処理」）；「zか？」（No→「その他の処理」，Yes→「zの処理」）；各処理後→「終わりか？」（No→「次のデータ」へ戻る，Yes→「終了」）。注記：流れ図中のx，y，zはデータ区分のA，B，Cのいずれかと対応させる。
正解 In the flowchart, each item of data is judged in the order «Is it x?» → (if No) «Is it y?» → (if No) «Is it z?», so the number of comparisons is 1 if it belongs to x, 2 if y, 3 if z, while the «Other» category that matches none of them runs all 3 comparisons. To minimize the total, categories with larger counts should be processed at an earlier stage (fewer comparisons). Since C=50 is the largest count, followed by B=30 and A=10, matching x=C, y=B, z=A is optimal. This gives C: 50×1=50, B: 30×2=60, A: 10×3=30, Other: 10×3=30, for a total of 50+60+30+30=170. Therefore the minimum number is 170, and the correct answer is ア.
190 is the value obtained with a matching such as x=C, y=A, z=B (50×1+10×2+30×3+Other 10×3=190). Because the fairly large count B is assigned to z (3 comparisons), this is not the minimum. Since the question asks «at minimum how many times», it is not appropriate.
230 is the value obtained with a matching such as x=B, y=A, z=C (30×1+10×2+50×3+Other 10×3=230). It places the largest count C at z (3 comparisons), so the number is far from minimal and is actually among the larger ones, hence inappropriate.
250 is the value when x=A, y=B, z=C (10×1+30×2+50×3+Other 10×3=250), the worst matching in which categories with smaller counts are judged earlier. This is the maximum number, not the answer to this question, which asks for the minimum.