This is an example of a specific calculation question about the key points of subnetting calculations, and as preparation for the CCNA exam, you should be able to do the IP address calculation question shown on this page in about one minute, in a flash!
Click to open “Answer” and “Explanation”. If you have not read “Key Points for Subnetting Calculations,” please do so first.
What are the IP addresses of the valid hosts in the subnet 10.1.160.0/20? (Choose three)
10.1.168.0
10.1.176.1
10.1.174.255
10.1.160.255
10.1.160.0
10.1.175.255
Answer
A,C,D
Explanation
From the subnet mask “/20”, the third byte of the network address changes in multiples of 16. The next network address after “10.1.160.0/20” is “10.1.176.0/20”.
From here, -1 (0.0.0.1) gives the broadcast address “10.1.175.255”. And the IP address range is “10.1.160.1-10.1.175.254” between “10.1.160.0” and “10.1.175.255”.
Q2
Which is the address of the subnet for 192.168.1.42 255.255.255.248?
192.168.1.8/29
192.168.1.32/27
192.168.1.40/29
192.168.1.16/28
192.168.1.48/29
Answer
C
Explanation
From the subnet mask 255.255.255.248, the network address changes in multiples of 8 for the fourth byte. So you look at the fourth byte of the IP address “192.168.1.42” in question, which is a multiple of 8 less than 42, which is 40. So the network address for 192.168.1.42 255.255.255.248 is 192.168.1.40 255.255.255.248. The subnet mask in prefix notation is 192.168.1.40/29.
Q3
Which is the last valid IP address for a subnet using subnet mask 255.255.255.224?
192.168.2.63
192.168.2.62
192.168.2.61
192.168.2.60
192.168.2.32
Answer
B
Explanation
With a subnet mask of 255.255.255.224, the network address changes in multiples of 32 for the fourth byte. The -1 from the network address is the broadcast address. Also, the last valid IP address is the network address -2 because it is the upper limit of the IP address range.
To summarize the above For the subnet mask 255.255.255.224, the relationship between network address, broadcast address, and last valid IP address is as follows
Network address : multiple of 32 Broadcast address : multiple of 32 -1 Last valid IP address : multiple of 32 -2
Of the IP addresses in the choices, the IP address whose fourth byte is a multiple of 32 – 2 is 192.168.2.62.
Q4
Which network address is the IP address 172.16.159.159/22?
172.16.0.0
172.16.128.0
172.16.156.0
172.16.159.0
172.16.159.128
172.16.192.0
Answer
C
Explanation
From the subnet mask /22, the network address changes in multiples of 4 for the third byte. So, you look at the third byte of the IP address in question, “172.16.159.159,” which is 159. 156 is a multiple of 4 smaller than 159, so the network address is 172.16.156.0/22.
Q5
Which network address is the IP address 192.168.23.61/28?
192.168.23.0
192.168.23.32
192.168.23.48
192.168.23.56
192.168.23.60
Answer
C
Explanation
From subnet mask /28, the network address changes in multiples of 16 for the fourth byte; the multiple of 16 below 61 for the fourth byte is 48. Therefore, the network address is 192.168.23.48/28.
Q6
Which statement is true about 10.16.3.65/23? (Choose two)
The smallest IP address in the subnet is 10.16.2.1 255.255.254.0
The last valid IP address for the subnet is 10.16.2.254 255.255.254.0
Network is not subnetted.
Broadcast address of subnet is 10.16.3.255 255.255.254.0
Subnet address is 10.16.3.0 255.255.254.0
Answer
A,D
Explanation
From subnet mask /23, the third byte of a network address varies in multiples of 2. Note the “3” in the third byte of the IP address in question, 10.16.3.65, which is a multiple of 2 less than 3. The network address is 10.16.2.0/23. The network address of the next subnet is 10.16.4.0/23, and the broadcast address is 10.16.3.255, which is -1 from 10.16.4.0.
The available IP address range for 10.16.2.0/23 is 10.16.2.1 to 10.16.3.254 since the IP address is between 10.16.2.0 and 10.16.3.255.
Q7
The network 172.16.0.0 is subnetted with a subnet mask of 255.255.255.192. When the IP address 172.16.2.120 is assigned to a host, a conflict was detected. Assign an IP address on the same subnet as 172.16.2.120 to the host. Which IP address is appropriate?
172.16.1.80
172.16.2.80
172.16.1.64
172.16.2.64
172.16.2.127
172.16.2.128
Answer
B
Explanation
For this problem, you only need to consider the IP address of the same network as 172.16.2.120 255.255.255.192. From the subnet mask 255.255.255192, the network address changes in multiples of 64 with the fourth byte. Notice the “120” in the fourth byte of “172.16.2.120”; a multiple of 64 smaller than 120 is 64, so the network address is 172.16.2.64. And the network address of the next subnet is 172.16.2.128. The broadcast address is 172.16.2.127, which is -1 from the next network address, 172.16.2.128. For the network address 172.16.2.64 255.255.255.192, the available IP address range is 172.16.2.65 to 172.16.2.126.
Q8
Statically assign an IP address to the server. Assign the first available IP address to the router with network address 192.168.20.24/29. The server is assigned the last available IP address. Which of the following is the appropriate configuration for the server?
IP address 192.168.20.14 Subnet mask 255.255.255.248 Default gateway 192.168.20.9
IP address 192.168.20.254 Subnet mask 255.255.255.0 Default gateway 192.168.20.1
IP address 192.168.20.30 Subnet mask 255.255.255.248 Default gateway 192.168.20.25
IP address 192.168.20.30 Subnet mask 255.255.255.240 Default gateway 192.168.20.17
IP address 192.168.20.30 Subnet mask 255.255.255.240 Default gateway 192.168.20.25
Answer
C
Explanation
Just think of the IP address range of 192.168.20.24/29. Since the subnet mask is /29, the network address changes in multiples of 8 for the fourth byte. So the network address for the next subnet is 192.168.20.32/29. From here -1 gives us the broadcast address 192.168.20.31. The IP address range is 192.168.20.25 to 192.168.20.30.
The first IP address in this range, 192.168.20.25, would be assigned to the router, and the last IP address, 192.168.20.30, would be assigned to the server. The IP address set on the router will be the default gateway IP address for the server.
Q9
Which host addresses are valid for subnet mask 255.255.248.0? (Choose three)
172.16.9.0
172.16.8.0
172.16.31.0
172.16.20.0
Answer
A,C,D
Explanation
With a subnet mask of 255.255.248.0, the network address changes in multiples of 8 for the third byte. And all of the choices have “0” as the fourth byte. Therefore, if the third byte of the IP address of a choice is a multiple of 8, it is a network address.
In other words, a valid IP address among the choices is an address whose third byte is not a multiple of 8.
Q10
Which IP addresses can be configured for Host B? (Choose two)
192.168.10.32
192.168.10.38
192.168.10.46
192.168.10.47
192.168.10.49
192.168.10.51
Answer
B,C
Explanation
Just consider the range of IP addresses available at the network address of 192.168.10.32/28 to which Host B in the figure is connected. Since the subnet mask is /28, the network address changes in multiples of 16 for the fourth byte. the next network address after 192.168.10.32/28 is 192.168.10.48/28. -1 from 192.168.10.48 gives the broadcast address 192. From 192.168.10.48, you get the broadcast address 192.168.10.47. From the above, the available IP address range for 192.168.10.32/28 is 192.168.10.33 to 192.168.10.46.
