Bit-rate
Question 1 |
If a file consisting of 50,000 characters takes 40 seconds to send, then the data rate is __________.
1 kbps | |
1.25 kbps | |
2 kbps | |
10 kbps |
Question 1 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
-- 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
Question 2 |
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?
192 kbps | |
512 kbps | |
1.248 Mbps | |
1.536 Mbps |
Question 2 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
-- 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
There are 2 questions to complete.