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Comptia Network+ MP N10-007 > Final Exam - Practice Examples > Flashcards

Final Exam - Practice Examples Flashcards

1
Q

How many host bits needed for 99 hosts?

A

7 bits

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2
Q

how many host bits needed for 120 hosts?

A

7 bits

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3
Q

How many host bits needed for 45 hosts?

A

6 bits

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4
Q

How many host bits needed for 8hosts?

A

4 bits

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5
Q

How many host bits needed for 5 hosts?

A

3 bits

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6
Q

How many host bits needed for 32 hosts?

A

6 bits

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7
Q 
For Class C 1 to 6 bits may be borrowed leaving 7 down to 2 bits(Minimum 2
bits—Calculate the # of bits, then subtract 2 bits for the network ID and
Broadcast. This value is # valid hosts.
2
2
- 2 = 2 valid hosts
2
3
- 2 = 6 valid hosts (8-2)
2
4
- 2 = 14 valid hosts (16-2)
2
5
- 2 = 30 valid hosts (32-2)
2
6
- 2 = 62 valid hosts (64-2)
2
7
- 2 = 126 valid hosts (128-2)
A 
For Class C 1 to 6 bits may be borrowed leaving 7 down to 2 bits(Minimum 2
bits—Calculate the # of bits, then subtract 2 bits for the network ID and
Broadcast. This value is # valid hosts.
2
2
- 2 = 2 valid hosts
2
3
- 2 = 6 valid hosts (8-2)
2
4
- 2 = 14 valid hosts (16-2)
2
5
- 2 = 30 valid hosts (32-2)
2
6
- 2 = 62 valid hosts (64-2)
2
7
- 2 = 126 valid hosts (128-2)
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8
Q

How many bits borrowed for 3 subnets?

A

2 bits

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9
Q

How many bits borrowed for 15 subnets

A

4

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10
Q

How many bits borrowed for 33 subnets

A

6

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11
Q

How many bits borrowed for 60 subnets

A

6

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12
Q

How many bits borrowed for 70 subnets

A

n/a

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13
Q

How many bits borrowed for 6 subnets

A

3 bits

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14
Q 
(Class C can borrow 1 to 6 bits—must leave a minimum of 2 bits)
2
1 = 2 subnets
2
2 = 4 subnets
2
3 = 8 subnets
2
4 = 16 subnets
2
5 = 32 subnets
2
6 = 64 subnets
A 
(Class C can borrow 1 to 6 bits—must leave a minimum of 2 bits)
2
1 = 2 subnets
2
2 = 4 subnets
2
3 = 8 subnets
2
4 = 16 subnets
2
5 = 32 subnets
2
6 = 64 subnets
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15
Q

For Class A and Class B, many more bits are available to borrow or to be remaining. The chart can
be extended to accommodate larger values.

A 
If a company needs 1000 hosts on a network, how many bits are required?
Continuing the chart above…
2
7
- 2 = 126 valid hosts (128-2)
2
8 – 2 = 254 valid hosts (256-2)
2
9 – 2 = 510 valid hosts (512-2(
2
10 – 2 = 1022 valid hosts (1024-2)
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16
Q

Class C– Organize your calculations for each step since a calculation may help answer more than one
question.

A

Using Binary
The first 3 octets will remain the same since this is a class C network to start with and the last octet
was used for creating subnets and hosts.
The /27 indicates 3 bits borrowed: 205.25.25. _ _ _| _ _ _ _ _
Showing the last octet may help: 205.25.25. 0 1 0 0 1 0 1 1
64 | 8 2 1 = 75 Network is 64—bits before the line
Using increment values
/27 means 3 bits borrowed and last bit to be borrowed was 32, so subnets are
205.25.25.0
205.25.25.32
205.25.25.64 broadcast 205.25.25.95 host range 205.25.25.65—205.25.25.94 (75 between 64 & 96)
205.25.25.96
205.25.25.128
205.25.25.160
205.25.25.192
205.25.25.224

17
Q

What is the default subnet mask for 205.25.25.75/27?

A

First octet value is 205 which means a class C so default mask is 255.255.255.0 (default prefix is /24)

18
Q

hat is the custom subnet mask for 205.25.25.75/27?

A

255.255.255.224

19
Q

For the IP address of 205.25.25.75/27, how many bits were borrowed to create the subnet?

A

Default prefix is /24 and custom prefix is /27 so 3 bits were borrowed.

20
Q

What is the broadcast address for the subnet (network) that 205.25.25.75/27 is a part of?

A

Using increment value: (Previously calculated)

  1. 25.25.0
  2. 25.25.32
  3. 25.25.64 broadcast 205.25.25.95 (one less than next subnet)
  4. 25.25.96
  5. 25.25.128
  6. 25.25.160
  7. 25.25.192
  8. 25.25.224
21
Q

What is the total # of valid hosts for the subnet (network) that 205.25.25.75/27 is a part of?

A

3 bits borrowed, leaves remaining 5 bits, 25

-2 = 32 – 2 = 30 valid hosts

22
Q

What is the subnet # for the subnet that 205.25.25.75/27 resides on?

A 
(Previously calculated)
Subnet #0--205.25.25.0
Subnet #1--205.25.25.32
Subnet #2--205.25.25.64 (205.25.25.75—75 between 64 and 96)
Subnet #3--205.25.25.96
Subnet #4--205.25.25.128
Subnet #5--205.25.25.160
Subnet #6--205.25.25.192
Subnet #7--205.25.25.224
23
Q

What is the network ID of 199.198.197.51/28? 199.198.197.48/28

A

*Using binary:
The first 3 octets will remain the same since this is a class C network to start with and the last octet
was used for creating subnets and hosts.
The /28 indicates 3 bits borrowed: 199.198.197. _ _ _ _| _ _ _ _
Showing the last octet may help: 199.198.197. 0 0 1 1| 0 0 1 1
32+16 +2+1This tells us the network is 48
*Using Increments:
/28 means borrowing 4 bits and last bit borrowed is 16 so 16 is the Increment.
Subnet #0 199.198.197.0 /28
Subnet #1 199.198.197.16/28
Subnet #2 199.198.197.32/28
Subnet #3 199.198.197.48/28 Broadcast 199.198.197.63 Host Range 199.198.197.49-199.198.197.62
Subnet #4 199.198.197.64/28
…
Subnet #15 199.198.197.240 /28

24
Q

What is the default subnet mask for 199.198.197.51/28?

A

First octet value is 199 which means a class C so default mask is 255.255.255.0 (default prefix is /24)

25
Q

What is the custom subnet mask for 199.198.197.51/28

A

255.255.255.240

26
Q

For the IP address of 199.198.197.51/28, how many bits were borrowed to create the subnet?

A

Default prefix is /24 and custom prefix is /28 so 4 bits were borrowed.

27
Q

What is the broadcast address for the subnet (network) that 199.198.197.51/28 is a part of?

A 
Using increment value:
/28 means borrowing 4 bits and last bit borrowed is 16 so 16 is the Increment.
Subnet #0 199.198.197.0 /28
Subnet #1 199.198.197.16/28
Subnet #2 199.198.197.32/28
Subnet #3 199.198.197.48/28 ---If next network is .64 then broadcast is 199.198.197.63/28
Subnet #4 199.198.197.64/28
…
Subnet #15 199.198.197.240 /28
28
Q

What is the total # of valid hosts for the subnet (network) that 199.198.197.51/28 is a part of?

A

4 bits borrowed, leaves remaining 4 bits, 24

-2 = 16 – 2 = 14 valid hosts

29
Q

What is the subnet # for the subnet that 205.25.25.75/27 resides on? Subnet#3

A 
Previous work shows last octet with the subnet values:
Subnet#0 199.198.197.0 /28
Subnet#1 199.198.197.16/28
Subnet#2 199.198.197.32/28
Subnet#3 199.198.197.48/28
Subnet#4 199.198.197.64/28
…
Subnet#15 199.198.197.240 /28
30
Q

What is the network ID of 166.166.5.61/24? 166.166.5.0/24

A

The address 166.166.5.61 with /24 is on the 5th subnet (3rd octet) all hosts are in 4th octet.
*Using binary:
The first 2 octets will remain the same since this is originally a Class B address. 8 bits were borrowed,
so the value of the 3rd octet identifies the subnet#. The entire 4th octet is zeroes for the Network ID.
The /24 indicates a total of 8 bits were borrowed: 166.166.0 0 0 0 0 1 0 1 . 0 0 0 0 0 0 0 0
Showing the last octet may help: 166.166. 4 + 1. 0 0 0 0 0 0 0 0
(3rd octet value 4 +1 tells us the subnet is 5
*Using Increments:
/24 means borrowing 8 bits and last bit borrowed is 1 so 1 is the Increment.
Subnet #0 166.166.0.0/24
Subnet #1 166.166.1.0/24
Subnet #2 166.166.2.0/24
Subnet #3 166.166.3.0/24
Subnet #4 166.166.4.0/24
Subnet #5 166.166.5.0/24 Broadcast is 166.166.5.255 Host Range 166.166.5.1-166.166.5.254
Subnet #6 166.166.6.0/24
Subnet #7 166.166.7.0/24
…
Subnet#255 166.166.255.0/24

31
Q

What is the default subnet mask for 166.166.5.0/24

A

255.255.0.0

32
Q

What is the custom subnet mask for 166.166.5.0/24

A

255.255.255.0

33
Q

For the IP address of 166.166.5.61/24, how many bits were borrowed to create the subnet?

A

Default prefix is /16 and custom prefix is /24 so 8 bits were borrowed.

34
Q

What is the broadcast address for the subnet (network) that 166.166.5.61/24 is a part of?

A 
*Using Increments (Previously calculated)
Subnet #0 166.166.0.0/24
Subnet #1 166.166.1.0/24
Subnet #2 166.166.2.0/24
Subnet #3 166.166.3.0/24
Subnet #4 166.166.4.0/24
Subnet #5 166.166.5.0/24 Broadcast is 166.166.5.255
Subnet #6 166.166.6.0/24
…
Subnet #255 166.166.255.0/24
35
Q

What is the total # of valid hosts for the subnet (network) that 166.166.5.61/24 is a part of?

A

8 bits borrowed, leaves remaining 8 bits, 28

-2 = 256 – 2 = 254 valid hosts

36
Q

What is the subnet # for the subnet that 166.166.5.61/24 resides on? Subnet#5

A 
Using Increments (Previously calculated)
Subnet #0 166.166.0.0/24
Subnet #1 166.166.1.0/24
Subnet #2 166.166.2.0/24
Subnet #3 166.166.3.0/24
Subnet #4 166.166.4.0/24
*Subnet #5 166.166.5.0/24
Subnet #6 166.166.6.0/24
Subnet #7 166.166.7.0/24

Comptia Network+ MP N10-007 (9 decks)

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