Network+: ch 8 Written Labs and Review Questions

Question 1

Practice Example #1C: 255.255.255.128 (/25)
192.168.10.0 = Network address
255.255.255.128 = Subnet mask
Answer the big 5 from network + and the four from subnetting .com

Answer 1

Because 128 is 10000000 in binary, there is only 1 bit for subnetting, and there are 7 bits or hosts. We’re going to subnet the Class C network address 192.168.10.0.

192. 168.10.0 = Network address
193. 255.255.128 = Subnet mask

Now, let’s answer the big five:
 How many subnets? Because 128 is 1 bit on (10000000), the answer is 2^1 = 2. Only looking at the fourth octet in class C, but last 2 in Class b, last 3 a???

How many hosts per subnet? We have 7 host bits off (10000000), so the equation is 2^7 – 2 = 126 hosts.

What are the valid subnets? 256 – 128 = 128. Remember, we’ll start at zero and count in our block size, so our subnets are 0, 128.

What’s the broadcast address for each subnet? The number right before the value of the next subnet is all host bits turned on and equals the broadcast address. For the 0 subnet, the next subnet is 128, so the broadcast of the 0 subnet is 127.

What are the valid hosts? These are the numbers between the subnet and broadcast address. The easiest way to find the hosts is to write out the subnet address and the broadcast address. This way, the valid hosts are obvious. The following table shows the 0 and 128 subnets, the valid host ranges of each, and the broadcast address of both subnets:

Subnet 0 128
First host 1 129
Last host 126 254
Broadcast 127 255

Question 2

Should We Really Use This Mask That Provides Only Two Hosts?

Imagine you are the network administrator for Acme Corporation in San Francisco, with dozens of WAN links connecting to your corporate office. Right now your network is a classful network, which means that the same subnet mask is on each host and router
interface. You’ve read about classless routing where you can have different size masks, but you don’t know what to use on your point-to-point WAN links. Is 255.255.255.252 (/30) a helpful mask in this situation?

Answer 2

Yes, this is a very helpful mask in wide area networks.

If you use the 255.255.255.0 mask, then each network will have 254 hosts, but you only use 2 addresses with a WAN link! That is a waste of 252 hosts per subnet. If you use the 255.255.255.252 mask, then each subnet has only 2 hosts, and you don’t waste precious
addresses.

Question 3

Write the subnet, broadcast address, and valid host range 192.168.100.25/30

Answer 3

192.168.100.25/30. A /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.

Question 4

Write the subnet, broadcast address, and valid host range 192.168.100.37/28

Answer 4

192.168.100.37/28. A /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 are 33–46.

Question 5

Write the subnet, broadcast address, and valid host range 192.168.100.66/27

Answer 5

192.168.100.66/27. A /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. The valid host range is 33–62.

Question 6

Write the subnet, broadcast address, and valid host range 192.168.100.17/29

Answer 6

192.168.100.17/29. A /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 hosts are 17–22.

Question 7

Write the subnet, broadcast address, and valid host range 192.168.100.99/26

Answer 7

192.168.100.99/26. A /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 hosts are 65–126.

Question 8

Write the subnet, broadcast address, and valid host range 192.168.100.99/25

Answer 8

192.168.100.99/25. A /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 hosts are 1–126.

Question 9

You have a Class B network and need 29 subnets. What is your mask?

Answer 9

A default Class B is 255.255.0.0. A Class B 255.255.255.0 mask is 256 subnets, each with 254 hosts. We need fewer subnets. If we use 255.255.240.0, this provides 16 subnets. Let’s add one more subnet bit. 255.255.248.0. This is 5 bits of subnetting, which provides 32 subnets. This is our best answer, a /21.

Question 10

What is the broadcast address of 192.168.192.10/29?

Answer 10

A /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, and broadcast is 15.

Question 11

How many hosts are available with a Class C /29 mask?

Answer 11

A /29 is 255.255.255.248, which is 5 subnet bits and 3 host bits. This is only 6 hosts per subnet.

Question 12

What is the subnet for host ID 172.16.3.65/23?

Answer 12

A /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.

Question 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?
A. 14
B. 15
C. 16
D. 30
E. 31
F. 62

Answer 13

D. 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 number of host bits would never change.

Question 14

You have a class A host of 10.0.0.110/25. It needs to communicate to a host with an
IP address of 10.0.0.210/25. Which of the following devices do you need to use in order
for these hosts to communicate?
A. A layer 2 switch
B. Router
C. DNS server
D. Hub

Answer 14

B. Don’t freak because this is a class A. What is your subnet mask? 255.255.255.128.
Regardless of the class of address, this is a block size of 128 in the fourth octet. The subnets
are 0 and 128. The 0 subnet host range is 1-126, with a broadcast address of 127. The 128
subnet host range is 129-254, with a broadcast address of 255. You need a router for these
two hosts to communicate because they are in differenet subnets.

Question 15

What is the subnetwork address for a host with the IP address 200.10.5.68/28?
A. 200.10.5.56
B. 200.10.5.32
C. 200.10.5.64
D. 200.10.5.0

Answer 15

C. 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, and so on. The host is in the 64 subnet.

Question 16

The network address of 172.16.0.0/19 provides how many subnets and hosts?
A. 7 subnets, 30 hosts each
B. 7 subnets, 2,046 hosts each
C. 7 subnets, 8,190 hosts each
D. 8 subnets, 30 hosts each
E. 8 subnets, 2,046 hosts each
F. 8 subnets, 8,190 hosts each

Answer 16

F. A CIDR address of /19 is 255.255.224.0. This is a Class B address, so that is only 3 subnet
bits, but it provides 13 host bits, or 8 subnets, each with 8,190 hosts.

Question 17

You receive a call from a user that is complaining that they cannot get on the Internet. You
have them verify their IP address, mask, and default gateway. The IP address is 10.0.37.144,
with a subnet mask of 255.255.254.0. The default gateway is 10.0.38.1. What is the problem?
A. Invalid IP address
B. Invalid subnet mask
C. gateway IP is incorrect
D. IP address and mask are not compatible.

Answer 17

C. The host ID of 10.0.37.144 with a 255.255.254.0 mask is in the 10.0.36.0 subnet (yes,
you need to be able to subnet in this exam!). The third octet has a block size of two, so the
next subnet is 10.0.28.0, which makes the broadcast address 10.0.37.255. The default gateway
address of 10.0.38.1 is not in the same subnet as the host. Even though this is a Class A
address, you still should easily be able to subnet this because you look more at the subnet
mask and find your interesting octet, which is the third octet in this question. 256-254 = 2.
Your block size is 2. Class A subnetting is covered in Appendix A.

Question 18

If a host on a network has the address 172.16.45.14/30, what is the subnetwork this host
belongs to?
A. 172.16.45.0
B. 172.16.45.4
C. 172.16.45.8
D. 172.16.45.12
E. 172.16.45.16

Answer 18

D. 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, and so on. Address 14 is obviously
in the 12 subnet.

Question 19

On a network, which mask should you use on point-to-point WAN links in order to reduce
the waste of IP addresses?
A. /27
B. /28
C. /29
D. /30
E. /31

Answer 19

D. A point-to-point link uses only two hosts. A /30, or 255.255.255.252, mask provides
two hosts per subnet.

Question 20

On which of the following devices are you most likely to be able to implement NAT?
A. Hub
B. Ethernet switch
C. Router
D. Bridge

Answer 20

C. Devices with Layer 3 awareness, such as routers and firewalls, are the only ones that can
manipulate the IP header in support of NAT.

Question 21

You have an interface on a router with the IP address of 192.168.192.10/29. Including the
router interface, how many hosts can have IP addresses on the LAN attached to router
interface?
A. 6
B. 8
C. 30
D. 62
E. 126

Answer 21

A. A /29 (255.255.255.248), regardless of the class of address, has only 3 host bits. Six hosts
is the maximum number of hosts on this LAN, including the router interface.

Question 22

When configuring the IP settings on a computer on one subnet to ensure it can communicate
with a computer on another subnet, which of the following is desirable?
A. Configure the computer with the same default gateway as the other computer.
B. Configure the computer with the same subnet mask as the other computer.
C. Configure the computer with a default gateway that matches the IP address of the
router’s interface that is attached to the same subnet as the computer.
D. Configure the computer with a subnet mask that matches the IP

Answer 22

C. A computer should be configured with an IP address that is unique throughout the
reachable internetwork. It should be configured with a subnet mask that matches that of
all other devices on its local subnet, but not necessarily one that matches the mask used on
any other subnet. It should also be configured with a default gateway that matches its local
router’s interface IP address.

Question 23

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. 192.168.192.15
B. 192.168.192.31
C. 192.168.192.63
D. 192.168.192.127
E. 192.168.192.255

Answer 23

A. 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, and so on. 10 is in the 8 subnet. The next subnet is 16, so 15 is
the broadcast address.

Question 24

What is the highest usable address on the 172.16.1.0/24 network?
A. 172.16.1.255
B. 172.16.1.254
C. 172.16.1.253
D. 172.16.1.23

Answer 24

B. A 24-bit mask, or prefix length, indicates the entire fourth octet is used for host
identification. In a special case, such as this, it is simpler to visualize the all-zeros value
(172.16.1.0) and the all-ones value (172.16.1.255). The highest usable address, the last one
before the all-ones value, is 172.16.1.254.

Question 25

A network administrator is connecting hosts A and B directly through their Ethernet interfaces,
as shown in the illustration. Ping attempts between the hosts are unsuccessful. What
can be done to provide connectivity between the hosts? (Choose two.)
IP Address: 192.168.1.20
Mask 255.255.255.240

IP Address: 192.168.1.201
Mask 255.255.255.240
Straight-through Cable

A. A crossover cable should be used in place of the straight-through cable.
B. A rollover cable should be used in place of the straight-though cable.
C. The subnet masks should be set to 255.255.255.192.
D. A default gateway needs to be set on each host.
E. The subnet masks should be set to 255.255.255.0.

Answer 25

A, E. 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 easy solution is just to set both masks
to 255.255.255.0 (/24).

Question 26

If an Ethernet port on a router were assigned an IP address of 172.16.112.1/25, what would
be the subnet address of this host?
A. 172.16.112.0
B. 172.16.0.0
C. 172.16.96.0
D. 172.16.255.0
E. 172.16.128.0

Answer 26

A. 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. Because there is only 1 bit in the fourth octet, the bit is either off or on—which
is a value of 0 or 128. The host in the question is in the 0 subnet, which has a broadcast
address of 127 because 128 is the next subnet.

Question 27

Using the illustration in question 16, 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 0 subnet should not be considered valid for this question.
A. 192.168.10.142
B. 192.168.10.66
C. 192.168.100.254
D. 192.168.10.143
E. 192.168.10.126

Answer 27

A. A /28 is a 255.255.255.240 mask. Let’s count to the ninth subnet (we need to find the
broadcast address of the eighth subnet, so we need to count to the ninth subnet). Starting
at 16 (remember, the question stated that we will not use subnet 0, so we start at 16, not 0),
16, 32, 48, 64, 80, 96, 112, 128, 144. The eighth subnet is 128, and the next subnet is 144,
so our broadcast address of the 128 subnet is 143. This makes the host range 129–142. 142
is the last valid host.

Question 28

what would be the IP address of E0 if you were using the first
subnet? The network ID is 192.168.10.0/28, and you need to use the last available IP address
in the range. Again, the zero subnet should not be considered valid for this question.

A. 192.168.10.24
B. 192.168.10.62
C. 192.168.10.30
D. 192.168.10.127

Answer 28

C. A /28 is a 255.255.255.240 mask. The first subnet is 16 (remember that the question
stated not to use subnet 0), 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.

Question 29

If you are forced to replace a router that has failed to the point that you are unable to access
its current configuration to aid in setting up interface addresses on the new router, which of
the following can you reference for assistance?
A. The default-gateway settings on computers from each subnet that the old router interconnected.
B. The router’s configuration that was periodically cached on the DHCP server.
C. The router’s configuration that was periodically cached on the DNS server.
D. The new router will auto-configure itself with the correct settings.

Answer 29

A. The best method here is to check the configuration of devices that were using the old
router as a gateway to the rest of the internetwork. Routers do not periodically cache their
configurations to servers of any sort. You might have copied the old router’s configuration
to a TFTP server or the like; but failing that, you will have to rebuild the configuration
from scratch, which might well be much more than interface addresses. Therefore, keeping
a copy of the router’s current configuration somewhere other than on the router is a wise
choice. Routers don’t auto-configure themselves; we wouldn’t want them to.

Question 30

You have a network with a subnet of 172.16.17.0/22. Which is the valid host addresses?
A. 172.16.17.1 255.255.255.252
B. 172.16.0.1 255.255.240.0
C. 172.16.20.1 255.255.254.0
D. 172.16.16.1 255.255.255.240
E. 172.16.18.255 255.255.252.0
F. 172.16.0.1 255.255.255.0

Answer 30

E. 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 listed, and 172.16.18.255
is a valid host.

Question 31

Your router has the following IP address on Ethernet0: 172.16.2.1/23. Which of the following
can be valid host IDs on the LAN interface attached to the router? (Choose two.)
A. 172.16.0.5
B. 172.16.1.100
C. 172.16.1.198
D. 172.16.2.255
E. 172.16.3.0
F. 172.16.3.255

Answer 31

D, E. The router’s 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 router’s interface is in the 2.0 subnet, and the
broadcast address is 3.255 because the next subnet is 4.0. The valid host range is 2.1 through
3.254. The router is using the first valid host address in the range.

Question 32

You have one IP address provided from your ISP with a /30 mask. However, you have 300
users that need to access the Internet. What technology will you use to implement a solution?
A. PAT
B. VPN
C. DNS
D. LANs

Answer 32

A. Network Address Translation can allow up to 65,000 hosts to get onto the Internet with
one IP address by using Port Address Translation (PAT).