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The answer is in the question.

You provided 192.168.1.10/27. This is a Class C subnet, because the first octet (192) falls in the range of 192-223.

A Class C subnet has a default mask of /24, or 255.255.255.0. In Binary, the first 24 bits are 1, the last 8 bits are 0.

Since the original question used a /27 on a CLass C address, that means 3 bits were borrowed from the CLass C default mask (/24) to get to /27. That is where he "got" the 3 borrowed bits.

The "3 bits" are arbitrary. Had his original question included the address 192.168.1.10/28, then he would be borrowing 4 bits. Had he chosen 192.168.1.10/29, then he would be borrowing 5 bits.

Had he chosen the address 130.168.1.10/27, then he would be borrowing 11 bits. Because this is a Class B address, with a default mask of /16, and to get from /16 to /27 requires 11 bits.

That said, I highly recommend reading the thread linkedrecommend reading the thread linked by @Ron Maupin@Ron Maupin for more detail on the process of Subnetting.

Also, the concept of "classes" in IP address is archaic. No one in the real world uses them anymore. It only persists because of Cisco's insistence of leaving it in all its training and certification material.

The answer is in the question.

You provided 192.168.1.10/27. This is a Class C subnet, because the first octet (192) falls in the range of 192-223.

A Class C subnet has a default mask of /24, or 255.255.255.0. In Binary, the first 24 bits are 1, the last 8 bits are 0.

Since the original question used a /27 on a CLass C address, that means 3 bits were borrowed from the CLass C default mask (/24) to get to /27. That is where he "got" the 3 borrowed bits.

The "3 bits" are arbitrary. Had his original question included the address 192.168.1.10/28, then he would be borrowing 4 bits. Had he chosen 192.168.1.10/29, then he would be borrowing 5 bits.

Had he chosen the address 130.168.1.10/27, then he would be borrowing 11 bits. Because this is a Class B address, with a default mask of /16, and to get from /16 to /27 requires 11 bits.

That said, I highly recommend reading the thread linked by @Ron Maupin for more detail on the process of Subnetting.

Also, the concept of "classes" in IP address is archaic. No one in the real world uses them anymore. It only persists because of Cisco's insistence of leaving it in all its training and certification material.

The answer is in the question.

You provided 192.168.1.10/27. This is a Class C subnet, because the first octet (192) falls in the range of 192-223.

A Class C subnet has a default mask of /24, or 255.255.255.0. In Binary, the first 24 bits are 1, the last 8 bits are 0.

Since the original question used a /27 on a CLass C address, that means 3 bits were borrowed from the CLass C default mask (/24) to get to /27. That is where he "got" the 3 borrowed bits.

The "3 bits" are arbitrary. Had his original question included the address 192.168.1.10/28, then he would be borrowing 4 bits. Had he chosen 192.168.1.10/29, then he would be borrowing 5 bits.

Had he chosen the address 130.168.1.10/27, then he would be borrowing 11 bits. Because this is a Class B address, with a default mask of /16, and to get from /16 to /27 requires 11 bits.

That said, I highly recommend reading the thread linked by @Ron Maupin for more detail on the process of Subnetting.

Also, the concept of "classes" in IP address is archaic. No one in the real world uses them anymore. It only persists because of Cisco's insistence of leaving it in all its training and certification material.

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Eddie
Eddie
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The answer is in the question.

You provided 192.168.1.10/27. This is a Class C subnet, because the first octet (192) falls in the range of 192-223.

A Class C subnet has a default mask of /24, or 255.255.255.0. In Binary, the first 24 bits are 1, the last 8 bits are 0.

Since the original question used a /27 on a CLass C address, that means 3 bits were borrowed from the CLass C default mask (/24) to get to /27. That is where he "got" the 3 borrowed bits.

The "3 bits" are arbitrary. Had his original question included the address 192.168.1.10/28, then he would be borrowing 4 bits. Had he chosen 192.168.1.10/29, then he would be borrowing 5 bits.

Had he chosen the address 130.168.1.10/27, then he would be borrowing 11 bits. Because this is a Class B address, with a default mask of /16, and to get from /16 to /27 requires 11 bits.

That said, I highly recommend reading the thread linked by @Ron Maupin for more detail on the process of Subnetting.

Also, the concept of "classes" in IP address is archaic. No one in the real world uses them anymore. It only persists because of Cisco's insistence of leaving it in all its training and certification material.