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

Question

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?

• A. $192$ kbps
• B. $512$ kbps
• C. $1.248$ Mbps
• D. $1.536$ Mbps

Answer 1

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.