Test your basic knowledge |

IP Subnetting VLSMs And Troubleshooting IP

Subject :

Instructions:

Answer 50 questions in 15 minutes.
If you are not ready to take this test, you can study here.
Match each statement with the correct term.
Don't refresh. All questions and answers are randomly picked and ordered every time you load a test.

This is a study tool. The 3 wrong answers for each question are randomly chosen from answers to other questions. So, you might find at times the answers obvious, but you will see it re-enforces your understanding as you take the test each time.

1. Given 192.168.10.0/28 - what is the subnet mask?
0.0 - 64.0 - 128.0 - and 192.0
A /29 (255.255.255.248) - regardless of the class of address - has only three hosts bits. Six hosts is the maximum amount of hosts on this LAN - including the router interface.
255.255.255.128
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25

2. What's the broadcast address of each subnet?
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
All nodes in the network use the same subnet mask
The number just before the next subnet; the broadcast of the last subnet is always 255
IP stack failure; Reinstall TCP/IP

3. 192.168.100.99/25
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
The number just before the next subnet; the broadcast of the last subnet is always 255
/25 is 255.255.255.128. The fourth octet is a block size of 128. 0 - 128. The host is in the 0 subnet - broadcast of 127. Valid host 1-126

4. Classful Routing
Create many networks using subnet masks of different lengths from one network
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination
/27 is 255.255.255.224. The fourth octet is a block size of 32. Count by 32s until you pass the host address of 66. 0 - 32 - 64. The host is in the 32 subnet - broadcast address of 63. Valid host range of 33-62.
All interfaces within the classful address space have the same subnet mask

5. What subnet and broadcast address is the IP address 172.16.66.10 255.255.192.0 (/18) a member of?
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
11110000 = 240
63.255 - 127.255 - 191.255 - 255.255
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255

6. 192.168.100.99/26
11000000 = 192
The 10.32 subnet; The broadcast is 10.63
/26 is 255.255.255.192. The fourth octet has a block size of 64. 0 - 64 - 128. The host is in the 64 subnet - broadcast of 127. Valid host 65-126
The mask 255.255.254.0 (/23) used with a Class A means that there are 15 subnet bits and 9 hosts bits. The block size in the third octet is 2 (256 - 254). So this makes the subnets in the interesting octet 0 - 2 - 4 - 6 - etc. - all the way to 254. T

7. What subnet and broadcast address is the IP address 172.16.10.33 255.255.255.224 (/27) a member of?
A /29 (255.255.255.248) has a block size of 8 in the fourth octet. This means the subnets are 0 - 8 - 16 - 24 - etc. 10 is in the 8 subnet. The next subnet is 16 - so 15 is the broadcast address.
A Class C subnet mask of 255.255.255.224 is 3 bits on and 5 bits off (11100000) and provides 8 subnets - each with 30 hosts. However - if the command ip subnet-zero is not used - then only 6 subnets would be available for use.
The 10.32 subnet; The broadcast is 10.63
11110000 = 240

8. /30
/29 is 255.255.255.248. The fourth octet is a block size of 8. 0 - 8 - 16 - 24. The host is in the 16 subnet - broadcast of 23. Valid host 17-22
11111100 = 252
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
A CIDR address of /19 is 255.255.224.0. This is a Class B address - so that is only 3 subnet bits but provides 13 host bits - or 8 subnets - each with 8 -190 hosts.

9. What subnet and broadcast address is the IP address 172.16.50.10 255.255.224.0 (/19) a member of?
A /28 is a 255.255.255.240 mask. We need to count to the eighth subnet - not starting at subnet-zero. 16 - 32 - 48 - 64 - 80 - 96 - 112 - 128. The ninth subnet is 144 (we need this to help us find the 128 subnet broadcast address - which is 143).The
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
0.0 - 64.0 - 128.0 - and 192.0
A /29 (255.255.255.248) - regardless of the class of address - has only three hosts bits. Six hosts is the maximum amount of hosts on this LAN - including the router interface.

10. Given 192.168.10.0/28 - What's the broadcast address for each subnet?
127 & 255
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
RIPv2 - EIGRP - or OSPF

11. Using the following illustration - what would be the IP address of E0 if you were using the eighth subnet? The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range. The zero subnet should not be considered vali
/26 is 255.255.255.192. The fourth octet has a block size of 64. 0 - 64 - 128. The host is in the 64 subnet - broadcast of 127. Valid host 65-126
A 240 mask is 4 subnet bits and provides 16 subnets - each with 14 hosts. We need more subnets - so let's add subnet bits. One more subnet bit would be a 248 mask. This provides 5 subnet bits (32 subnets) with 3 hosts bits (6 host per subnet).
To test the local stack on your host - ping the loopback interface of 127.0.0.1
A /28 is a 255.255.255.240 mask. We need to count to the eighth subnet - not starting at subnet-zero. 16 - 32 - 48 - 64 - 80 - 96 - 112 - 128. The ninth subnet is 144 (we need this to help us find the 128 subnet broadcast address - which is 143).The

12. Which configuration command must be in effect to allow the use of 8 subnets if the Class C subnet mask is 255.255.255.224?
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21
A CIDR address of /19 is 255.255.224.0. This is a Class B address - so that is only 3 subnet bits but provides 13 host bits - or 8 subnets - each with 8 -190 hosts.
A Class C subnet mask of 255.255.255.224 is 3 bits on and 5 bits off (11100000) and provides 8 subnets - each with 30 hosts. However - if the command ip subnet-zero is not used - then only 6 subnets would be available for use.

13. What is the maximum number of IP addresses that can be assigned to hosts on a local subnet that uses the 255.255.255.224 subnet mask?
First - if you have two hosts directly connected - as shown in the graphic - then you need a crossover cable. A straight-through cable won't work. Second - the hosts have different masks - which puts them in different subnets. The easily solution is
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
A /27 (255.255.255.224) is 3 bits on and 5 bits off. This provides 8 subnets - each with 30 hosts. Does it matter if this mask is used with a Class A - B or C network address? Not at all. The amount of hosts bits would never change.

14. Variable Length Subnet Masks (VLSMs)
Create many networks using subnet masks of different lengths from one network
A /28 is a 255.255.255.240 mask. We need to count to the eighth subnet - not starting at subnet-zero. 16 - 32 - 48 - 64 - 80 - 96 - 112 - 128. The ninth subnet is 144 (we need this to help us find the 128 subnet broadcast address - which is 143).The
Local physical network problem between NIC and router
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21

15. What are the valid subnets?
32-bit value that allows the recipient of IP packets to distinguish the network ID portion of the IP address from the host ID portion of the IP address
A /28 is a 255.255.255.240 mask. We need to count to the eighth subnet - not starting at subnet-zero. 16 - 32 - 48 - 64 - 80 - 96 - 112 - 128. The ninth subnet is 144 (we need this to help us find the 128 subnet broadcast address - which is 143).The
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached
A /27 (255.255.255.224) is 3 bits on and 5 bits off. This provides 8 subnets - each with 30 hosts. Does it matter if this mask is used with a Class A - B or C network address? Not at all. The amount of hosts bits would never change.

16. Variable Length Subnet Masks (VLSMs)
11111100 = 252
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached
Each network segment can use a different subnet mask
A CIDR address of /19 is 255.255.224.0. This is a Class B address - so that is only 3 subnet bits but provides 13 host bits - or 8 subnets - each with 8 -190 hosts.

17. What subnet and broadcast address is the IP address 172.16.45.14 255.255.255.252 (/30) a member of?
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
Traceroute - Arp -a - Ipconfig /all

18. /29
Create many networks using subnet masks of different lengths from one network
.1 - .126 & .129 - .254
/30 is 255.255.255.252. The valid subnet is 192.168.100.24 - broadcast is 192.168.100.27 - and valid hosts are 192.168.100.25 and 26
11111000 = 248

19. What is the broadcast address of 192.168.192.10/29?
/29 is 255.255.255.248. This is a block size of 8 in the fourth octet. 0 - 8 - 16. The host is in the 8 subnet - broadcast is 15
126
63.255 - 127.255 - 191.255 - 255.255
10000000 = 128

20. 192.168.100.37/28
2y - 2 where y is the number of unmasked bits (or 0's)
/28 is 255.255.255.240. The fourth octet is a block size of 16. Just count by 16s until you pass 37. 0 - 16 - 32 - 48. The host is in the 32 subnet - with a broadcast address of 47. Valid hosts 33-46
11000000 = 192
Create many networks using subnet masks of different lengths from one network

21. What is the subnet and broadcast address of the host 172.16.88.255/20?
The 10.32 subnet; The broadcast is 10.63
A /29 (255.255.255.248) has a block size of 8 in the fourth octet. This means the subnets are 0 - 8 - 16 - 24 - etc. 10 is in the 8 subnet. The next subnet is 16 - so 15 is the broadcast address.
A /28 is a 255.255.255.240 mask. We need to count to the eighth subnet - not starting at subnet-zero. 16 - 32 - 48 - 64 - 80 - 96 - 112 - 128. The ninth subnet is 144 (we need this to help us find the 128 subnet broadcast address - which is 143).The
The subnet is 80.0 and the broadcast address is 95.255

22. Given 192.168.10.0/28 - How many subnets?
/27 is 255.255.255.224. The fourth octet is a block size of 32. Count by 32s until you pass the host address of 66. 0 - 32 - 64. The host is in the 32 subnet - broadcast address of 63. Valid host range of 33-62.
32-bit value that allows the recipient of IP packets to distinguish the network ID portion of the IP address from the host ID portion of the IP address
A /28 is a 255.255.255.240 mask. We need to count to the eighth subnet - not starting at subnet-zero. 16 - 32 - 48 - 64 - 80 - 96 - 112 - 128. The ninth subnet is 144 (we need this to help us find the 128 subnet broadcast address - which is 143).The
2

23. Additional Windows Troubleshooting
Traceroute - Arp -a - Ipconfig /all
Discard it; by default will discard any broadcast packets
11100000 = 224
A Class C subnet mask of 255.255.255.224 is 3 bits on and 5 bits off (11100000) and provides 8 subnets - each with 30 hosts. However - if the command ip subnet-zero is not used - then only 6 subnets would be available for use.

24. Which two statements describe the IP address 10.16.3.65/23?
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21
This is a pretty simple question. A /28 is 255.255.255.240 - which means that our block size is 16 in the fourth octet. 0 - 16 - 32 - 48 - 64 - 80 - etc. The host is in the 64 subnet.
11100000 = 224
The mask 255.255.254.0 (/23) used with a Class A means that there are 15 subnet bits and 9 hosts bits. The block size in the third octet is 2 (256 - 254). So this makes the subnets in the interesting octet 0 - 2 - 4 - 6 - etc. - all the way to 254. T

25. Given 192.168.10.0/28 - How many hosts per subnet?
/23 is 255.255.254.0. The third octet is a block size of 2. 0 - 2 - 4. The subnet is in the 16.2.0 subnet - the broadcast address is 16.3.255
Problem with the NIC; Replace the NIC
The number just before the next subnet; the broadcast of the last subnet is always 255
126

26. 192.168.100.25/30
Each network segment can use a different subnet mask
A /21 is 255.255.248.0 - which means we have a block size of 8 in the third octet - so we just count by 8 until we reach 66. The subnet in this question is 64.0. The next subnet is 72.0 - so the broadcast address of the 64 subnet is 71.255.
/30 is 255.255.255.252. The valid subnet is 192.168.100.24 - broadcast is 192.168.100.27 - and valid hosts are 192.168.100.25 and 26
IP stack failure; Reinstall TCP/IP

27. Your router has the following IP address on Ethernet0: 172.16.2.1/23. Which of the following can be valid host ID's on the LAN interface attached to the router?
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts
The routers IP address on the E0 interface is 172.16.2.1/23 - which is a 255.255.254.0. This makes the third octet a block size of 2. The routers interface is in the 2.0 subnet - the broadcast address is 3.255 because the next subnet is 4.0. The vali
255.255.255.128
A /28 is a 255.255.255.240 mask. The first subnet is 16 (remember that the question stated not to use subnet zero) and the next subnet is 32 - so our broadcast address is 31. This makes our host range 17-30. 30 is the last valid host.

28. If an Ethernet port on a router were assigned an IP address of 172.16.112.1/25 - what would be the valid subnet address of this host?
A /25 mask is 255.255.255.128. Used with a Class B network - the third and fourth octets are used for subnetting with a total of 9 subnet bits - 8 bits in the third octet and 1 bit in the fourth octet. Since there is only 1 bit in the fourth octet -
2y - 2 where y is the number of unmasked bits (or 0's)
Local physical network problem between NIC and router
2^x where x is the number of masked bits (or 1's)

29. What subnet and broadcast address is the IP address 172.16.46.255 255.255.240.0 (/20) a member of?
First - if you have two hosts directly connected - as shown in the graphic - then you need a crossover cable. A straight-through cable won't work. Second - the hosts have different masks - which puts them in different subnets. The easily solution is
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
IP stack failure; Reinstall TCP/IP
/25 is 255.255.255.128. The fourth octet is a block size of 128. 0 - 128. The host is in the 0 subnet - broadcast of 127. Valid host 1-126

30. What is the subnetwork number of a host with an IP address of 172.16.66.0/21?
A /28 is a 255.255.255.240 mask. The first subnet is 16 (remember that the question stated not to use subnet zero) and the next subnet is 32 - so our broadcast address is 31. This makes our host range 17-30. 30 is the last valid host.
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination
/28 is 255.255.255.240. The fourth octet is a block size of 16. Just count by 16s until you pass 37. 0 - 16 - 32 - 48. The host is in the 32 subnet - with a broadcast address of 47. Valid hosts 33-46
A /21 is 255.255.248.0 - which means we have a block size of 8 in the third octet - so we just count by 8 until we reach 66. The subnet in this question is 64.0. The next subnet is 72.0 - so the broadcast address of the 64 subnet is 71.255.

31. /28
127 & 255
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15
A CIDR address of /19 is 255.255.224.0. This is a Class B address - so that is only 3 subnet bits but provides 13 host bits - or 8 subnets - each with 8 -190 hosts.
11110000 = 240

32. Able to ping but still unable to communicate
Possible DNS problem
/23 is 255.255.254.0. The third octet is a block size of 2. 0 - 2 - 4. The subnet is in the 16.2.0 subnet - the broadcast address is 16.3.255
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
First - if you have two hosts directly connected - as shown in the graphic - then you need a crossover cable. A straight-through cable won't work. Second - the hosts have different masks - which puts them in different subnets. The easily solution is

33. Given 172.16.0.0/18 - How many hosts per subnet?
A /29 is 255.255.255.248 - which is a block size of 8 in the fourth octet. The subnets are 0 - 8 - 16 - 24 - 32 - 40 - etc. 192.168.19.24 is the 24 subnet - and since 32 is the next subnet - the broadcast address for the 24 subnet is 31. 192.168.19.2
16 -382
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
IP stack failure; Reinstall TCP/IP

34. /26
A /28 is a 255.255.255.240 mask. The first subnet is 16 (remember that the question stated not to use subnet zero) and the next subnet is 32 - so our broadcast address is 31. This makes our host range 17-30. 30 is the last valid host.
The subnet is 80.0 and the broadcast address is 95.255
11000000 = 192
16 -382

35. 192.168.100.66/27
/27 is 255.255.255.224. The fourth octet is a block size of 32. Count by 32s until you pass the host address of 66. 0 - 32 - 64. The host is in the 32 subnet - broadcast address of 63. Valid host range of 33-62.
The routers IP address on the E0 interface is 172.16.2.1/23 - which is a 255.255.254.0. This makes the third octet a block size of 2. The routers interface is in the 2.0 subnet - the broadcast address is 3.255 because the next subnet is 4.0. The vali
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
126

36. Subnet Mask
/30 is 255.255.255.252. The valid subnet is 192.168.100.24 - broadcast is 192.168.100.27 - and valid hosts are 192.168.100.25 and 26
The mask 255.255.254.0 (/23) used with a Class A means that there are 15 subnet bits and 9 hosts bits. The block size in the third octet is 2 (256 - 254). So this makes the subnets in the interesting octet 0 - 2 - 4 - 6 - etc. - all the way to 254. T
A /28 is a 255.255.255.240 mask. The first subnet is 16 (remember that the question stated not to use subnet zero) and the next subnet is 32 - so our broadcast address is 31. This makes our host range 17-30. 30 is the last valid host.
32-bit value that allows the recipient of IP packets to distinguish the network ID portion of the IP address from the host ID portion of the IP address

37. Unable to Ping Default Gateway
Local physical network problem between NIC and router
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination
Problem with the NIC; Replace the NIC
The routers IP address on the E0 interface is 172.16.2.1/23 - which is a 255.255.254.0. This makes the third octet a block size of 2. The routers interface is in the 2.0 subnet - the broadcast address is 3.255 because the next subnet is 4.0. The vali

38. 192.168.100.17/29
/29 is 255.255.255.248. The fourth octet is a block size of 8. 0 - 8 - 16 - 24. The host is in the 16 subnet - broadcast of 23. Valid host 17-22
A /30 - regardless of the class of address - has a 252 in the fourth octet. This means we have a block size of 4 and our subnets are 0 - 4 - 8 - 12 - 16 - etc. Address 14 is obviously in the 12 subnet.
/30 is 255.255.255.252. The valid subnet is 192.168.100.24 - broadcast is 192.168.100.27 - and valid hosts are 192.168.100.25 and 26
Tracert - Show ip arp

39. Given 172.16.0.0/18 - What are the valid hosts?
.1 - .126 & .129 - .254
This is a pretty simple question. A /28 is 255.255.255.240 - which means that our block size is 16 in the fourth octet. 0 - 16 - 32 - 48 - 64 - 80 - etc. The host is in the 64 subnet.
Each network segment can use a different subnet mask
.0.1 - .63.254 - .64.1 - .127.254 - .128.1 - .191.254 - .192.1 - .255.254

40. Additional Cisco Troubleshooting
Tracert - Show ip arp
2y - 2 where y is the number of unmasked bits (or 0's)
/29 is 255.255.255.248. The fourth octet is a block size of 8. 0 - 8 - 16 - 24. The host is in the 16 subnet - broadcast of 23. Valid host 17-22
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination

41. /27
A Class C subnet mask of 255.255.255.224 is 3 bits on and 5 bits off (11100000) and provides 8 subnets - each with 30 hosts. However - if the command ip subnet-zero is not used - then only 6 subnets would be available for use.
11100000 = 224
Create many networks using subnet masks of different lengths from one network
The number just before the next subnet; the broadcast of the last subnet is always 255

42. You have a network with a subnet of 172.16.17.0/22. Which are valid host addresses?
A Class B network ID with a /22 mask is 255.255.252.0 - with a block size of 4 in the third octet. The network address in the question is in subnet 172.16.16.0 with a broadcast address of 172.16.19.255. Only option E even has the correct subnet mask
A /28 is a 255.255.255.240 mask. We need to count to the eighth subnet - not starting at subnet-zero. 16 - 32 - 48 - 64 - 80 - 96 - 112 - 128. The ninth subnet is 144 (we need this to help us find the 128 subnet broadcast address - which is 143).The
2
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30

43. The network address of 172.16.0.0/19 provides how many subnets and hosts?
A CIDR address of /19 is 255.255.224.0. This is a Class B address - so that is only 3 subnet bits but provides 13 host bits - or 8 subnets - each with 8 -190 hosts.
2y - 2 where y is the number of unmasked bits (or 0's)
Use different size masks on each router interface
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached

44. Unable to Ping Remote Destination
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination
A /28 is a 255.255.255.240 mask. We need to count to the eighth subnet - not starting at subnet-zero. 16 - 32 - 48 - 64 - 80 - 96 - 112 - 128. The ninth subnet is 144 (we need this to help us find the 128 subnet broadcast address - which is 143).The
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
RIPv2 - EIGRP - or OSPF

45. Classful Routing
To test the local stack on your host - ping the loopback interface of 127.0.0.1
32-bit value that allows the recipient of IP packets to distinguish the network ID portion of the IP address from the host ID portion of the IP address
All nodes in the network use the same subnet mask
11100000 = 224

46. A router receives a packet on an interface with a destination address of 172.16.46.191/26. What will the router do with this packet?
255.255.255.128
Discard it; by default will discard any broadcast packets
11111100 = 252
.1 - .126 & .129 - .254

47. You have an interface on a router with the IP address of 192.168.192.10/29. What is the broadcast address the hosts will use on this LAN?
A /29 is 255.255.255.248 - which is a block size of 8 in the fourth octet. The subnets are 0 - 8 - 16 - 24 - 32 - 40 - etc. 192.168.19.24 is the 24 subnet - and since 32 is the next subnet - the broadcast address for the 24 subnet is 31. 192.168.19.2
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
A /29 (255.255.255.248) has a block size of 8 in the fourth octet. This means the subnets are 0 - 8 - 16 - 24 - etc. 10 is in the 8 subnet. The next subnet is 16 - so 15 is the broadcast address.
A Class B network ID with a /22 mask is 255.255.252.0 - with a block size of 4 in the third octet. The network address in the question is in subnet 172.16.16.0 with a broadcast address of 172.16.19.255. Only option E even has the correct subnet mask

48. Given 192.168.10.0/28 - What are the valid subnets?
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances
To test the local stack on your host - ping the loopback interface of 127.0.0.1
A CIDR address of /19 is 255.255.224.0. This is a Class B address - so that is only 3 subnet bits but provides 13 host bits - or 8 subnets - each with 8 -190 hosts.
0 & 128

49. Unable to Ping Local Host
To test the local stack on your host - ping the loopback interface of 127.0.0.1
Problem with the NIC; Replace the NIC
Local physical network problem between NIC and router
/29 is 255.255.255.248. The fourth octet is a block size of 8. 0 - 8 - 16 - 24. The host is in the 16 subnet - broadcast of 23. Valid host 17-22

50. Variable Length Subnet Masks (VLSMs)
A 240 mask is 4 subnet bits and provides 16 subnets - each with 14 hosts. We need more subnets - so let's add subnet bits. One more subnet bit would be a 248 mask. This provides 5 subnet bits (32 subnets) with 3 hosts bits (6 host per subnet).
11111000 = 248
0 & 128
Use different size masks on each router interface