※ Our check while writing this explanation could not fully confirm the published answer; scoring follows IPA's official key.
×Incorrectア
10 minutes is the value you get if you treat the file size of 2M bytes directly as 2M bits and forget to multiply by 8 to convert bytes into bits. 1,200 × 2M bits = 2,400M bits, divided by the effective speed of 4 Mbps, gives 600 seconds = 10 minutes. Since the transmission speed is given in bps (bits per second), the amount of data must always be converted into bits before dividing.
×Incorrectイ
40 minutes is the value you get if you overlook that the data has to pass through two sections one after another and calculate only one leg. 19,200M bits ÷ 8 Mbps = 2,400 seconds = 40 minutes. In this question the two PCs are not connected directly but go through a router, and the question explicitly states that communication on the old PC side and on the current PC side cannot take place at the same time, so the time must be doubled.
○ウ
正解The correct answer is ウ, 80 (minutes). According to the introduction to question group B, there were 2,000 image files on the current PC when they were deleted, and 60% of them remained on the old PC, so the number of files to transfer is 2,000 × 0.6 = 1,200. Each file is 2M bytes, so the amount of data to transfer is 1,200 × 2M = 2,400M bytes. The unit of communication speed is bps (bits per second), so we convert bytes into bits: 2,400M × 8 = 19,200M bits. Next we find the effective transmission speed. The transmission speed of the line is 40 Mbps, but the transmission efficiency is 20%, so what can actually be carried is 40 × 0.2 = 8 Mbps. This is the key point of the question: copying from the old PC to the current PC first sends the data from the old PC to the router, and then from the router to the current PC, so it passes through two sections. As stated in the question, communication between the old PC and the router and communication between the current PC and the router cannot take place at the same time, so the same data has to be sent twice in a row, and the time required is twice that of one leg (in terms of effective throughput, 8 ÷ 2 = 4 Mbps). Therefore the time required = 19,200M bits ÷ 8 Mbps × 2 = 2,400 seconds × 2 = 4,800 seconds. Converted into minutes, 4,800 ÷ 60 = 80 minutes.
×Incorrectエ
133 minutes is the value you get if you overlook that only 60% of the deleted files remained on the old PC and assume that all 2,000 files have to be transferred. 2,000 × 2M bytes = 4,000M bytes = 32,000M bits, divided by the effective speed of 4 Mbps, gives 8,000 seconds ≈ 133 minutes. What actually has to be transferred is only the 1,200 files that remained on the old PC.
ここがポイント
Transfer time = amount of data ÷ effective transmission speed. Effective transmission speed = nominal transmission speed × transmission efficiency (here 40 Mbps × 20% = 8 Mbps). Because the amount of data is in bytes while the speed is in bps (bits per second), you must multiply by 8 to convert to bits before dividing, so that the units match.
When copying from one PC to another PC via a router over a wireless LAN, if they share the same radio waves and cannot send and receive at the same time, the data flows through two sections one after another, so the time required is twice that of one leg (the effective throughput is halved). Not overlooking the condition that the two communications cannot take place at the same time is what decides whether you get this question right.