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UGC NET CSE | November 2017 | Part 3 | Question: 25

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Suppose we want to download text documents at the rate of $100$ pages per second. Assume that a page consists of an average of $24$ lines with $80$ characters in each line. What is the required bit rate of the channel?

  1. $192$ kbps
  2. $512$ kbps
  3. $1.248$ Mbps
  4. $1.536$ Mbps
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We have 100 pages, each page is having 24 line and in each line there are 80 character and each character is of 8 bits.
Now we have to calculate no of bits can be downloaded:
ie.

 Downloading rate = 100 pages 
= 100 pages *24 line * 80 character * 8 bits
= 1536000 bits per second
ie 1.536 Mbps.

So, option (D) is correct.

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