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平成29年度 秋期 ITパスポート試験 問40

This question's explanation and your answers so far.

問40
頻出度★★★

Cさんの生産性は, Aさんの1.5倍, Bさんの3倍とする。AさんとBさんの2人で作業すると20日掛かるソフトウェア開発の仕事がある。これをAさんとCさんで担当した場合の作業日数は何日か。

出典 平成29年度 秋期 ITパスポート試験 問40
解説

問 40

  • ア

    正解 Think of productivity (the amount of work done per day) as a ratio. Since C=1.5×A and C=3×B, set C=6, then A=6÷1.5=4 and B=6÷3=2. So A:B:C=4:2:6=2:1:3. The combined productivity of A and B is 2+1=3. Since these two take 20 days to do the work, the total amount of work is 3×20=60. Next, the combined productivity of A and C is 2+3=5. Doing the same total of 60 at a productivity of 5 takes 60÷5=12 days. Therefore the number of work days is 12, so option ア is correct.

  • Incorrectイ

    If you do not correctly set the combined productivity of A and C to 5 but mistakenly use 4 (for example, mistaking C's productivity as 2 times B's instead of 3 times), you get the wrong answer of 60÷4=15 days. Correctly, C is 3 times B, so the combined productivity is 5 and the number of work days is 12.

  • Incorrectウ

    18 cannot be derived from any calculation based on the correct productivity ratio 2:1:3. It is a wrong answer that arises when you mistake the total amount (60) or the combined productivity (5); if you set the ratio correctly and calculate, the result can only be 12 days.

  • Incorrectエ

    Even when A and C handle it, if you forget to reflect C's higher productivity and keep computing with the combined productivity as 3, the same as for A and B, you get the wrong answer of 60÷3=20 days. In reality the combined productivity of A and C is as high as 5, so the work finishes in a shorter 12 days. 20 also equals the corrected number of work days for A and B, making it an easy trap to fall for.

ここがポイント
  • Represent productivity (the amount of work per unit time) as a ratio, and solve it as a work-rate problem using the fact that total work = productivity × days stays constant.
  • From C=1.5A=3B, derive A:B:C=2:1:3, then work backward to the number of work days using the combined productivity of each assigned pair.
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