Bit-rate

Question 1
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
Question 1 Explanation: 
Given data,
-- Number of Pages(P)=100 per second
-- Number of Lines(L)= 24
-- Number of characters(C) =80
-- One character(O) =8 bits
-- Bit rate of channel=?
Step-1: Bit rate of channel= P*L*C*O
= 100*24*80*8 bits per second
= 1536000
Step-2: Given options in Megabits per seconds.
= 1.536 Mbps
Question 2
If a file consisting of 50,000 characters takes 40 seconds to send, then the data rate is __________.
A
1 kbps
B
1.25 kbps
C
2 kbps
D
10 kbps
Question 2 Explanation: 
Given data,
-- Total number of characters=50,000
-- One character=8 bits
-- Total time to send 50,000 characters= 40 sec
-- Data rate=?
Step-1: Data rate= (Total number of characters* one character bits)/ Total time
= (50,000*8)/40
= 4,00,000/40
= 10,000 bits per seconds
Step-2: We calculated bits per second but given options in Kbps.
= 10 Mbps
There are 2 questions to complete.

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