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
Explanation: 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:
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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