Subnetting and CIDR Quiz

Note: A good set of questions can also be found at CCNP Exam Cram: Extending IP Addresses

Fere to the folloing Netmask translation table

Q1. A subnet’s mask is 255.255.255.224. How many nodes can this subnet have?

A: 254-224=30

Q2. A subnet’s mask is 255.255.255.240. How many nodes can this subnet have?

A: 254-40=14

Q3. A subnet’s mask is 255.255.255.192. How many nodes can this subnet have?

A: 254-192=62

Q4. You have a class C subnet. You would like divide it up into 30 more subnets by "subnetting" ? What would be the subnet mask?

A: 255.255.255.248

     The length of subnet section to accommodate 30 subnet should be 5 bits.  That means that host portion of the address contains 3 bits

     Mask = 256-8=248

Q5. You have a class C subnet. You would like to divide it up into 10 more Subnets by "subnetting" . How many hosts can you have per subnet?

A: 14

     Host part will be 4 bits.    Max hosts per subnet = 2^4 -2 = 14

Q6. you have address 10.1.1.0/23. How many hosts can be in this network

A: /24 is class C that has 256 hosts. So it can have 256*2 -2 = 510 hosts.

Q7. You have a class C subnet and need 48 nodes per subnet. How many subnets can you have ?

A: 2

      # max hosts/subnet = 2^N -2 , that means host part should be 6 bits

     That means M can be at most 2 bits

     # max subnets = 2^M  -2 = 2

Q8. Your class C network has been subnet-ed. Mask for the new subnets is 255.255.255.224. How many subnets and how many nodes per subnet can you have ?

A: 254-224=30

Q9. You have divided your class C network into 2 subnets. Your subnet mask is  255.255.255.192. What are the ranges of IPs for your subnets ?   Your Network address before subnetting was 200.252.144.0

A:  200.252.144.65-200.252.144.126 and 200.252.144.129-200.252.144.190

explanation:  M=2 N=6

                The First range =  2^6+1  to 2^7-2 = 65 to 126

                The Second range =  65+2^6 to 126+2^6 = 129 to 190

Q10.  Your previous Network Administrator had subneted your class C network      into many subnets. Only info you have is that one of the IP range was  200.252.144.33 through 128.252.144.46. How many subnets did he make ? what was the mask ?

A: 14, 255.255.255.240

explanation:  We see that the subnet has 14 nodes. That means N=4 (2^4-2 = 16).

                M= 8-4 =4 . Number of Subnets = 2^M -2 = 14.

                Mask = 128+64+32+16 (add 4 bits)  255.255.255.240

Q11. You have a class B subnet. You would like divide it up into 30 more subnets by "subnetting" ? What would be the subnet mask ?

 A: 255.255.248.0

     Refering to the formulas in the "SUBNETTING FORMULAS" section :

     M = 5 (since max number of subnets would be 30 if M = 5)

     Mask = 2^7+2^6+2^5+2^4+2^3 = 248

Q12. You have a class B subnet. You would like to divide it up into 10 more subnets by "subnetting" . How many hosts can you have per subnet?

A: 4094

     Refering: to the formulas in the "SUBNETTING FORMULAS" section.

     M = 4 (since max number of subnets would be 6 if M = 3)

     N = 16-4 = 12

     Max hosts per subnet = 2^12 -2 = 4094

Q13. You have a class B subnet. You would like to divide it up into 12 more subnets by "subnetting". However you also have to have at least 4000 hosts per subnet. Can you do it?

A: Yes.

     Refering: to the formulas in the "SUBNETTING FORMULAS" section.

     M = 4 (since max number of subnets would be 6 if M = 3)

     N = 16-M = 12

     Max hosts per subnet = 2^12 -2 = 4094

Q14. You have a class B subnet (128.252.0.0). You want to divide it into 2 subnets.      What will the ranges of IPs?

A: 128.252.65.1 to 128.252.126.254 AND 128.252.129.1 to 128.252.190.254

explanation: Referring to the 'SUBNETTINg FORMULA" section:

               M=2 N=14.  2^(14-8)+1=65  2^(14-8+1)-2=126

Q15. What IP network addresses are set-aside for class B (for private use)?

a. 128.0.0.0 (mask 255.255.0.0)

b. 128.252.0.0 (mask 255.255.0.0)

c. 172.16.0.0 (mask 255.240.0.0)

d. 172.16.0.0 (mask 255.255.0.0)

e. 172.240.0.0 (mask 255.255.0.0)

A: C

Q16. IP classes and their ranges (A-C):

 a. 1-127, 128-191, 192-223

b. 1-127, 128-191, 192-254

c. 1-126, 128-192, 193-223

d. 1-126, 128-191, 192-223

A: D

Q17. What IP network addresses are set aside for class C (for private use) ?

a. 172.16.0.0 (mask 255.240.0.0)

b. 172.16.0.0 (mask 255.255.0.0)

c. 196.168.0.0 (mask 255.255.0.0)

d. 192.168.0.0 (mask 255.255.0.0)

e. 198.168.0.0 (mask 255.255.255.0)

A: D

Q18. What leading bit values in the IP address indicate a class A

     and Class B address ?

a. 0 and 10

b. 0 and 11

c. 1 and 11

d. 01 and 10

e. 10 and 110

A: A

Q19. What leading bit values in the IP address indicate a class C address ?

a. 0

b. 01

c. 10

d. 11

e. 110

A: C

Q20. Consider the IP address 128.252.144.84. What is the network

     id and what is the node id ? assume classful networking.

a. 0.0.0.0 and 128.252.144.84

b. 128.0.0.0 and 252.144.84

c. 128.252.0.0 and 144.84

d. 128.252.144.0 and 84

A: C  (class B network)

Q21. You have a class C subnet. You would like divide it up into 8 more subnets by "subnetting" ? What would be the subnet mask ?

a. 255.255.255.192

b. 255.255.255.224

c. 255.255.255.240

d. 255.255.255.248

d. 255.255.255.252

A: C

Explanation:

     Refer to the formulas in the "SUBNETTING FORMULAS" section.

     M = 4 (since max number of subnets would be 6 if M = 3)

     Mask = 2^7+2^6+2^5+2^4 = 240

Q22. You have a class C subnet. You would like divide it up into 5 more subnets by "subnetting" . How many hosts can you have per subnet ?

a. 30

b. 31

c. 30

d. 62

e. 126

A: C

Explanation:

Refering: to the formulas in the "SUBNETTING FORMULAS" section.

M = 3 (since max number of subnets would be 2 if M = 2)

N = 8-M = 5

Max hosts per subnet = 2^5 -2 = 30

Q23. You have a class C subnet. You would like divide it up into 5 more subnets by "subnetting". However you also have to have at least 31 hosts per subnet. Can you do it ?

a. Yes

b. No

ans: B

Explanation:

     Answer is NO (Even though you are asking for 5x31=155 nodes)

     Referring: to the formulas in the "SUBNETTING FORMULAS" section.

     M = 3 (since max number of subnets would be 2 if M = 2)

     N = 8-M = 5

     Max hosts per subnet = 2^5 -2 = 30

Q24. You have a class C subnet. You would like divide it up into subnets      by "subnetting". You have to have at least 24 nodes per subnet.    How many subnets can you have ?

a. 2

b. 6

c. 7

d. 8

e. 14

 

A: B

 Explanation:

     Refering: to the formulas in the "SUBNETTING FORMULAS" section:

     # max hosts/subnet = 2^N -2 , that means N has to be at least 5

     That means M can be at most 3

     # max subnets = 2^M  -2 = 6

Q25. Your class C network has been subnet-ed. Mask for the new subnets is 255.255.255.248. How many subnets and how many nodes per subnet can you have ?

a. 30, 6

b. 30, 7

c. 62, 2

d. 62, 6

e. 62, 7

A: A

Q26. You have divided your class C network into 6 subnets. Your subnet mask is  255.255.255.224. What are the ranges of IPs for your subnets ?  Your Network address before subnetting was 200.252.144.0

a. 322

 Q26. Your previous Network Administrator had subnneted your Class C network into many subnets. Only info you have is that the mask was 255.255.255.224  How many subnets did he make ? How many nodes were there per subnet ?

a. 2, 62

b. 7, 30

c. 6, 30

d. 6, 62

e. 7, 62

A: C

Explanation:  224=128+64+32 . That means number borrowed 1 bits = 3. M=3.

                Subnets = 2^M -2 = 6 . Nodes per subnet = 2^N - 2 = 30 .

Q27. You have a class B subnet. You would like divide it up into 5 more subnets by "subnetting" . How many hosts can you have per subnet ?

a. 30

b. 1022

c. 2046

d. 4094

e. 8190

 

A: E

 

Explanation:

 

     Refering: to the formulas in the "SUBNETTING FORMULAS" section.

     M = 3 (since max number of subnets would be 2 if M = 2)

     N = 16-M = 13

     Max hosts per subnet = 2^13 -2 = 8190

Q30. You have a class B subnet. You would like divide it up into 5 more subnets by "subnetting". However you also have to have at least 6000 hosts per subnet. Can you do it ?

a. Yes

b. No.

A: B

Explanation:

     Refering to the formulas in the "SUBNETTING FORMULAS" section.

     M = 3 (since max number of subnets would be 2 if M = 2)

     N = 16-M = 13

     Max hosts per subnet = 2^13 -2 = 3070

Q31. Class D IP address range an its use?

a. 192-223, broadcast

b. 224-239, multicast

c. 224-255, broadcast

d. 192-255, multicast

 

A: B

Q32. Class E IP address range and its use ?

a. 240-255, experimental

b. 224-239, multicast

c. 224-255, multicast

d. 192-255, experimental

A: A

Q33. CIDR stands for ____________________?

a. Carrier International Domain Routing

b. Carrier Independent Domain Routing

c. Classless Inter-Domain Routing

d. Cross-platform Inter-Domain Routing

A: C

Q34. Which IP-class provides the fewest numbers of Hosts?

a. Class A

b. Class B

c. Class C

d. Class D

Answer: D

Q35. You see the following subnet addresses ; what is the subnet mask ?

128.252.4.0

128.252.8.0

128.252.12.0

128.252.16.0

a. 255.255.255.0

b. 255.255.192.0

c. 255.255.240.0

d. 255.255.252.0

Answer: D

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