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A classless address is given as 167.199.170.82/27. Find the last address.
- a)167.199.170.95
- b)167.199.170.256
- c)167.199.170.128
- d)167.199.170.88
Correct answer is option 'A'. Can you explain this answer?
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A classless address is given as 167.199.170.82/27. Find the last addre...
Given information: Classless address is 167.199.170.82/27.
To find: The last address.
Steps to find the last address:
Step 1: Convert the subnet mask to its decimal form.
The given subnet mask is /27, which means there are 27 1's in the subnet mask. The remaining bits are 0's. The decimal form of the subnet mask is 255.255.255.224.
Step 2: Identify the network address.
To identify the network address, we need to perform a bitwise AND operation between the IP address and the subnet mask.
167.199.170.82 => 10100111.11000111.10101010.01010010
255.255.255.224 => 11111111.11111111.11111111.11100000
------------------------------------------------------------
Network Address => 10100111.11000111.10101010.01000000
The network address is 167.199.170.64.
Step 3: Identify the broadcast address.
To identify the broadcast address, we need to flip all the bits in the subnet mask and perform a bitwise OR operation with the network address.
255.255.255.224 => 11111111.11111111.11111111.11100000
Inverted subnet mask => 00000000.00000000.00000000.00011111
------------------------------------------------------------
Broadcast Address => 10100111.11000111.10101010.01011111
The broadcast address is 167.199.170.95.
Therefore, the last address is option A, 167.199.170.95.
To find: The last address.
Steps to find the last address:
Step 1: Convert the subnet mask to its decimal form.
The given subnet mask is /27, which means there are 27 1's in the subnet mask. The remaining bits are 0's. The decimal form of the subnet mask is 255.255.255.224.
Step 2: Identify the network address.
To identify the network address, we need to perform a bitwise AND operation between the IP address and the subnet mask.
167.199.170.82 => 10100111.11000111.10101010.01010010
255.255.255.224 => 11111111.11111111.11111111.11100000
------------------------------------------------------------
Network Address => 10100111.11000111.10101010.01000000
The network address is 167.199.170.64.
Step 3: Identify the broadcast address.
To identify the broadcast address, we need to flip all the bits in the subnet mask and perform a bitwise OR operation with the network address.
255.255.255.224 => 11111111.11111111.11111111.11100000
Inverted subnet mask => 00000000.00000000.00000000.00011111
------------------------------------------------------------
Broadcast Address => 10100111.11000111.10101010.01011111
The broadcast address is 167.199.170.95.
Therefore, the last address is option A, 167.199.170.95.
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Community Answer
A classless address is given as 167.199.170.82/27. Find the last addre...
Last address = (any address) OR [NOT (network mask)].
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A classless address is given as 167.199.170.82/27. Find the last address.a)167.199.170.95b)167.199.170.256c)167.199.170.128d)167.199.170.88Correct answer is option 'A'. Can you explain this answer?
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A classless address is given as 167.199.170.82/27. Find the last address.a)167.199.170.95b)167.199.170.256c)167.199.170.128d)167.199.170.88Correct answer is option 'A'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about A classless address is given as 167.199.170.82/27. Find the last address.a)167.199.170.95b)167.199.170.256c)167.199.170.128d)167.199.170.88Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A classless address is given as 167.199.170.82/27. Find the last address.a)167.199.170.95b)167.199.170.256c)167.199.170.128d)167.199.170.88Correct answer is option 'A'. Can you explain this answer?.
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