ITPEC IP April 2025 Question 19

Topic: SoftwareAlso asked in October 2018, October 2020 and April 2022

Multiple files stored on a hard disk are used in work, from Monday to Friday. In order to handle failure of the hard disk, the data is backed up to another hard disk after the end of each day’s work. When the backup has following conditions, what is the total time required (in minutes) to perform backups of data from Monday to Friday?

[Conditions for backups](1)There are 6,000 files used in work, each with a size of 3 Mbytes.(2)In each day’s work, 1,000 files are modified. Modifications do not change the size of the files.(3)Files are copied to the other hard disk at a speed of 10 Mbytes/second. One (1) file is backed up at a time; backup continues without interruption.(4)From Monday to Thursday, only files modified that day are backed up. On Friday, all files are backed up, whether or not they have been modified.

Explanation

50 minutes — add four daily backups of the modified files to Friday's full backup, then divide by the copy speed.

  • •Monday to Thursday: 4 × 1,000 × 3 = 12,000 Mbytes
  • •Friday, all files: 6,000 × 3 = 18,000 Mbytes
  • •Total: 12,000 + 18,000 = 30,000 Mbytes
  • •Time: 30,000 ÷ 10 = 3,000 seconds = 3,000 ÷ 60 = 50 minutes

Each daily backup takes 300 seconds (5 minutes) and Friday's takes 1,800 seconds (30 minutes).

Answer (c)

Why not others:

  • •(a) 25 — backs up only the 1,000 modified files on Friday too, 5 × 5 minutes
  • •(b) 35 — counts Friday's 30 minutes plus only one 5-minute daily backup instead of four
  • •(d) 150 — backs up all 6,000 files every day, 5 × 30 minutes

Key rule: Total the data copied by every backup first, divide by the transfer rate for seconds, then convert the units; Mbytes ÷ Mbytes/second gives seconds.

Software · AI-written, may contain errors

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