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IP Subnetting VLSMs And Troubleshooting IP

Subject :

Instructions:

Answer 50 questions in 30 minutes.
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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. What are the valid hosts in each subnet?
All nodes in the network use the same subnet mask
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
63.255 - 127.255 - 191.255 - 255.255

2. How many subnets?

3. Unable to Ping Local Host
Use different size masks on each router interface
Possible DNS problem
Create many networks using subnet masks of different lengths from one network
Problem with the NIC; Replace the NIC

4. /27
11100000 = 224
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
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 subnet is 172.16.45.12 - with a broadcast of 172.16.45.15

5. To test the IP stack on your local host - which IP address would you ping?
To test the local stack on your host - ping the loopback interface of 127.0.0.1
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination
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.
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

6. 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
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).
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached

7. Given 192.168.10.0/28 - what is the subnet mask?
/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
255.255.255.128
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
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255

8. 192.168.100.37/28
/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 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).
The 10.32 subnet; The broadcast is 10.63
Create many networks using subnet masks of different lengths from one network

9. Subnet Mask
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
Create many networks using subnet masks of different lengths from one network
.0.1 - .63.254 - .64.1 - .127.254 - .128.1 - .191.254 - .192.1 - .255.254
/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

10. 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.
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).
2
.1 - .126 & .129 - .254

11. 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?
The 10.32 subnet; The broadcast is 10.63
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 -
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.
Discard it; by default will discard any broadcast packets

12. 192.168.100.25/30
0.0 - 64.0 - 128.0 - and 192.0
/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
.1 - .126 & .129 - .254
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.

13. What subnet and broadcast address is the IP address 172.16.46.255 255.255.240.0 (/20) a member of?
.1 - .126 & .129 - .254
16 -382
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.
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255

14. /26
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.
.0.1 - .63.254 - .64.1 - .127.254 - .128.1 - .191.254 - .192.1 - .255.254
Each network segment can use a different subnet mask
11000000 = 192

15. You need to configure a server that is on the subnet 192.168.19.24/29. The router has the first available host address. Which of the following should you assign to the server?
Possible DNS problem
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
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

16. 4 Troubleshooting Steps
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.
Possible DNS problem
Ping 127.0.0.1 (loopback) - Ping the local host - Ping Default Gateway (router) - Ping remote destination
/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

17. What is the broadcast address of 192.168.192.10/29?
The number just before the next subnet; the broadcast of the last subnet is always 255
/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
2^x where x is the number of masked bits (or 1's)
/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

18. On a VLSM network - which mask should you use on point-to-point WAN links in order to reduce the waste of IP addresses?
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
10000000 = 128
2
The 10.32 subnet; The broadcast is 10.63

19. 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
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached
11110000 = 240
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.

20. 192.168.100.99/25
/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
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.
You need 5 subnets - each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts
/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

21. /30
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
11111100 = 252
/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
RIPv2 - EIGRP - or OSPF

22. 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?
Create many networks using subnet masks of different lengths from one network
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
Discard it; by default will discard any broadcast packets
Local physical network problem between NIC and router

23. Given 192.168.10.0/28 - What are the valid subnets?
/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
0 & 128
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet
/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

24. What subnet and broadcast address is the IP address 172.16.66.10 255.255.192.0 (/18) a member of?
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
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
Traceroute - Arp -a - Ipconfig /all
.0.1 - .63.254 - .64.1 - .127.254 - .128.1 - .191.254 - .192.1 - .255.254

25. Given 192.168.10.0/28 - What are the valid hosts?
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.
Possible DNS problem
10000000 = 128
.1 - .126 & .129 - .254

26. Additional Cisco Troubleshooting
Tracert - Show ip arp
11111000 = 248
/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
All nodes in the network use the same subnet mask

27. A network administrator is connecting hosts A and B directly thorough 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?

28. How many valid hosts per subnet?
2y - 2 where y is the number of unmasked bits (or 0's)
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).
Reduced network traffic - Optimized network performance - Simplified management - Facilitated spanning of large geographical distances
.0.1 - .63.254 - .64.1 - .127.254 - .128.1 - .191.254 - .192.1 - .255.254

29. Classful Routing
All nodes in the network use the same subnet mask
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
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30
10000000 = 128

30. Unable to Ping Remote Destination
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
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.
.1 - .126 & .129 - .254

31. 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?
Possible DNS problem
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
11000000 = 192
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 -

32. IP Subnet-Zero
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
Allows you to use the first and last subnet in your network design; turned this command on by default
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

33. Which two statements describe the IP address 10.16.3.65/23?
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
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 80.0 and the broadcast address is 95.255
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25

34. What is the subnet and broadcast address of the host 172.16.88.255/20?
/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
All nodes in the network use the same subnet mask
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
The subnet is 80.0 and the broadcast address is 95.255

35. Given 192.168.10.0/28 - How many subnets?
The number just before the next subnet; the broadcast of the last subnet is always 255
2
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
/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

36. Given 172.16.0.0/18 - How many subnets?
Tracert - Show ip arp
63.255 - 127.255 - 191.255 - 255.255
4
127 & 255

37. 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?
2y - 2 where y is the number of unmasked bits (or 0's)
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.
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
/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

38. /25
Traceroute - Arp -a - Ipconfig /all
11111000 = 248
/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
10000000 = 128

39. Class C Subnet Masks
Local physical network problem between NIC and router
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30
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
RIPv1 and IGRP

40. What subnet and broadcast address is the IP address 172.16.50.10 255.255.224.0 (/19) a member of?
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
Use different size masks on each router interface
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 172.16.45.12 - with a broadcast of 172.16.45.15

41. What is the subnet for host ID 10.16.3.65/23?
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21
All interfaces within the classful address space have the same subnet mask
2^x where x is the number of masked bits (or 1's)
/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

42. Given 192.168.10.0/28 - What's the broadcast address for each subnet?
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.
IP stack failure; Reinstall TCP/IP
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.
127 & 255

43. How many hosts are available with a Class C /29 mask?
63.255 - 127.255 - 191.255 - 255.255
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination
11111000 = 248

44. What's the broadcast address of each subnet?
To test the local stack on your host - ping the loopback interface of 127.0.0.1
4
The number just before the next subnet; the broadcast of the last subnet is always 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.

45. Variable Length Subnet Masks (VLSMs)
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
16 -382
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.
Each network segment can use a different subnet mask

46. Classful Routing
/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
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
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.
All interfaces within the classful address space have the same subnet mask

47. Given 172.16.0.0/18 - What's the broadcast address for each subnet?
/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
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
/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
63.255 - 127.255 - 191.255 - 255.255

48. Unable to Ping Default Gateway
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
Local physical network problem between NIC and router
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.
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s

49. If a host on a network has the address 172.16.45.14/30 - what is the subnetwork this host belongs to?
/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.
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.
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.

50. Using the illustration from the previous question - what would be the IP address of S0 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 shoul
The number just before the next subnet; the broadcast of the last subnet is always 255
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.
63.255 - 127.255 - 191.255 - 255.255
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).

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Study | Quiz | Subject: it-skills | Questions: 100+

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