09.your final binary exam

8/13/2019 09.Your Final Binary Exam

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The Ultimate CCNA Study Package - ICND 1 And 2

Chris Bryant, CCIE #12933 http://www.thebryantadvantage.com

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Your Final Binary ExamNow that youve reviewed the approach to take for the various types ofbinary math and subnetting questions you might see on the CCNAexams, its time to get even more practice in. Remember, you pass the

exambefore you ever enter the exam room and you pass it by theamount of practice and study you put in.

The following questions are presented to you in the order in which thequestion type was discussed in this book. When youre solid with the

fundamentals, youre halfway home. If you haveany hesitancy with anyof the question types, go back to that section and practice them untilyou can answer the question flawlessly and without hesitation.For the first few sections, you are not presented with choices. Your skill

level will be more than enough to answer the questions without multiplechoice.

Lets go!Converting Binary To Dotted DecimalWhat is the dotted decimal equivalent of the binary number 01010101

11100010 01101010 01001010?

Answer: 85.226.106.74

What is the dotted decimal equivalent of the binary number 1111000000001111 01111111 10000000?

Answer: 240.15.127.128.

128 64 32 16 8 4 2 1 Total

First Octet 0 1 0 1 0 1 0 1 85

Second Octet 1 1 1 0 0 0 1 0 226

Third Octet 0 1 1 0 1 0 1 0 106

Fourth Octet 0 1 0 0 1 0 1 0 74

128 64 32 16 8 4 2 1 Total

First Octet 1 1 1 1 0 0 0 0 240

Second Octet 0 0 0 0 1 1 1 1 15

Third Octet 0 1 1 1 1 1 1 1 127

Fourth Octet 1 0 0 0 0 0 0 0 128


2/20

What is the dotted decimal equivalent of the binary number 1100110100000011 11110010 00100101?

Answer: 205.3.242.37.

What is the dotted decimal equivalent of the binary string 00110010

00100011 11110011 00100111?

Answer: 50.35.243.39.

What is the dotted decimal equivalent of the binary string 1000011100111111 01011111 00110010?

Answer: 135.63.95.50

At this point in your studies, you should have had no trouble answeringthe previous five questions. If you did, or were unsure of the approach,review Section Two.

If not, lets proceed to the next section.

Exam TipAs you can see in the previous question, the more you practice binarymath, the more of it youll be able to do in your head. You may havesolved the last question without even writing it down. Thats great,

but be careful with it in the exam room. It only takes about thirtyseconds to write it out, and since youll be a binary math masterbefore you walk into the testing room, time will not be a factor for you.

128 64 32 16 8 4 2 1 Total

First Octet 1 1 0 0 1 1 0 1 205

Second Octet 0 0 0 0 0 0 1 1 3

Third Octet 1 1 1 1 0 0 1 0 242Fourth Octet 0 0 1 0 0 1 0 1 37

128 64 32 16 8 4 2 1 Total

First Octet 0 0 1 1 0 0 1 0 50

Second Octet 0 0 1 0 0 0 1 1 35

Third Octet 1 1 1 1 0 0 1 1 243

Fourth Octet 0 0 1 0 0 1 1 1 39

128 64 32 16 8 4 2 1 Total

First Octet 1 0 0 0 0 1 1 1 135Second Octet 0 0 1 1 1 1 1 1 63

Third Octet 0 1 0 1 1 1 1 1 95

Fourth Octet 0 0 1 1 0 0 1 0 50


3/20

Converting Decimal To Binary

What is the binary equivalent of the dotted decimal IP address195.29.37.126?

Answer: 11000011 00011101 00100101 01111110.


Answer: 11001111 01011101 00110001 10111101.


Answer: 00010101 11001000 01010000 11110101.


128 64 32 16 8 4 2 1

1stOctet: 195 1 1 0 0 0 0 1 1

2ndOctet: 29 0 0 0 1 1 1 0 1

3rdOctet: 37 0 0 1 0 0 1 0 1

4thOctet: 126 0 1 1 1 1 1 1 0

128 64 32 16 8 4 2 1

1stOctet: 207 1 1 0 0 1 1 1 1

2ndOctet: 93 0 1 0 1 1 1 0 1

3rdOctet: 49 0 0 1 1 0 0 0 1

4th

Octet: 1891 0 1 1 1 1 0 1

128 64 32 16 8 4 2 1

1stOctet: 21 0 0 0 1 0 1 0 1

2ndOctet: 200 1 1 0 0 1 0 0 0

3rdOctet: 80 0 1 0 1 0 0 0 0

4thOctet: 245 1 1 1 1 0 1 0 1

128 64 32 16 8 4 2 1

1stOctet: 105 0 1 1 0 1 0 0 1

2ndOctet: 83 0 1 0 1 0 0 1 1

3rdOctet: 219 1 1 0 1 1 0 1 1


4/20


5/20

How many valid subnets exist on the 111.0.0.0 /14 network?This is a Class A network, with a network mask of 255.0.0.0, or /8. The

subnet mask is /14, indicating that there are six subnet bits. 2 to the

6thpower is 64.

How many valid subnets exist on the 172.12.0.0 /19 network?

This is a Class B network, with a network mask of 255.255.0.0, or /16.The subnet mask is /19, indicating there are three subnet bits. 2 to the

3rdpower is 8.

How many valid subnets exist on the 182.100.0.0 /27 network?

This is a Class B network, with a network mask of 255.255.0.0, or /16.The subnet mask is /27, indicating there are three subnet bits. 2 to the

11thpower is 2048.

How many valid subnets exist on the 231.23.19.0 /30 network?This is a Class C network, with a network mask of 255.255.255.0,

or /24. The subnet mask is /30, indicating there are six subnet bits. 2

to the 6th

power is 64.

1stOctet 2ndOctet 3rdOctet 4thOctet

Default Network Mask

255.0.0.0

11111111 00000000 00000000 00000000

Subnet Mask255.252.0.0

11111111 11111100 00000000 00000000


Default Network Mask255.255.0.0

11111111 11111111 00000000 00000000

Subnet Mask

255.255.224.0

11111111 11111111 11100000 00000000



11111111 11111111 00000000 00000000

Subnet Mask

255.255.255.224

11111111 11111111 11111111 11100000


6/20

How many valid subnets exist on the 17.0.0.0 255.240.0.0 network?

This is a Class A network, with a network mask of 255.0.0.0. The subnetmask here is 255.240.0.0, indicating there are four subnet bits. 2 to the

4thpower is 16.

How many valid subnets exist on the 234.12.200.0 255.255.255.248

network?This is a Class C network, with a network mask of 255.255.255.0. The

subnet mask here is 255.255.255.248, indicating there are five subnet

bits. 2 to the 5thpower is 32.

How many valid subnets exist on the 155.200.0.0 255.255.255.128 network?

This is a Class B network, with a network mask of 255.255.0.0. Thesubnet mask here is 255.255.255.128, indicating there are nine subnet

bits. 2 to the 9thpower is 512.



11111111 11111111 11111111 00000000

Subnet Mask255.255.255.252

11111111 11111111 11111111 11111100



11111111 00000000 00000000 00000000

Subnet Mask

255.240.0.0

11111111 11110000 00000000 00000000



255.255.255.0

11111111 11111111 11111111 00000000

Subnet Mask255.255.255.248

11111111 11111111 11111111 11111000



255.255.255.0

11111111 11111111 00000000 00000000

Subnet Mask255.255.255.128

11111111 11111111 11111111 10000000


7/20

At this point in your studies, you should have gotten each questioncorrect. If you had difficulty with the last nine questions, review SectionFour. If not, lets press on!

Determining The Number Of Valid HostsHow many valid host addresses exist on the 211.24.12.0 /27 subnet?

This is a Class C network, with a network mask of /24. The subnet maskis /27, indicating that there are three subnet bits and five host bits. 2 to

the 5thpower is 32; subtract two, and 30 valid hosts remain.

How many valid host addresses exist on the 12.0.0.0 /19 subnet?

This is a Class A network, with a network mask of /8. The subnet maskis /19, indicating that there are eleven subnet bits and 13 host bits:

2 to the 13thpower is 8192; subtract two, and 8190 valid hosts remain.

How many valid host addresses exist on the 178.34.0.0 /28 subnet?This is a Class B network, with a network mask of /16. The subnet maskis /28, indicating there are twelve subnet bits and four host bits:



11111111 11111111 11111111 00000000

Subnet Mask255.255.255.224

11111111 11111111 11111111 11100000


Default Network

Mask255.0.0.0

11111111 00000000 00000000 00000000


11111111 11111111 11100000 00000000



255.255.0.0

11111111 11111111 00000000 00000000

Subnet Mask

255.255.255.240

11111111 11111111 11111111 11110000


8/20

2 to the 4th power is 16; subtract two, and 14 valid host addressesremain.

How many valid host addresses exist on the 211.12.45.0 /30 subnet?This is a Class C network, with a network mask of /24. The subnet maskis /30, indicating there are six subnet bits and two host bits:

2 to the 2nd power is 4; subtract two, and two valid host addressesremain.

How many valid host addresses exist on the 129.12.0.0 /20 subnet?This is a Class B network, with a network mask of /16. The subnet mask

is /20, indicating there are four subnet bits and twelve host bits:


How many valid host addresses exist on the 210.34.24.0

255.255.255.248 subnet?This is a Class C network, with a network mask of 255.255.255.0. The

subnet mask is five bits longer, indicating there are five subnet bits andthree host bits:



11111111 11111111 11111111 00000000

Subnet Mask

255.255.255.252

11111111 11111111 11111111 11111100



11111111 11111111 00000000 00000000


11111111 11111111 11110000 00000000



11111111 11111111 11111111 00000000

Subnet Mask

255.255.255.248

11111111 11111111 11111111 11111000


9/20

2 to the 3rdpower is 8; subtract two, and six valid hosts remain.

How many valid host addresses exist on the 145.100.0.0 255.255.254.0subnet?

This is a Class B network, with a network mask of 255.255.0.0. Thesubnet mask is seven bits longer, indicating there are seven subnet bitsand nine host bits:

2 to the 9thpower is 512; subtract two, and 510 valid host addressesremain.

How many valid host addresses exist on the 23.0.0.0 255.255.240.0subnet?This is a Class A network, with a network mask of 255.0.0.0. The subnetmask is 12 bits longer, indicating there are 12 subnet bits and 12 hostbits:


Your skill set should have allowed you to answer each of the previouseight questions perfectly. If you had difficulty in either the approach orthe execution, review Section Five.

Determining The Subnet Number Of A Given IP Address

On what subnet can the IP address 142.12.38.189 /25 be found?



11111111 11111111 00000000 00000000


11111111 11111111 11111110 00000000



11111111 00000000 00000000 00000000


11111111 11111111 11110000 00000000

Octet 1 Octet 2 Octet 3 Octet 4

IP Address

142.12.38.189

10001110 00001100 00100110 10111101

Subnet Mask /25 11111111 11111111 11111111 10000000


10/20

Convert the Boolean AND result to dotted decimal, and the resultingsubnet is 142.12.38.128 /25.




Convert the Boolean AND result to dotted decimal, and the resulting

subnet is 10.178.39.200 /29.



On what subnet can the IP address 203.23.189.205 255.255.255.240 be

found?

Boolean AND Result 10001110 00001100 00100110 10000000


IP Address170.17.209.36

10101010 00010001 11010001 00100100

Subnet Mask /19 11111111 11111111 11100000 00000000



IP Address10.178.39.205

00001010 10110010 00100111 11001101

Subnet Mask /29 11111111 11111111 11111111 11111000



IP Address190.34.9.173

10111110 00100010 00001001 10101101

Subnet Mask /22 11111111 11111111 11111100 00000000



11/20

Convert the Boolean AND result to dotted decimal, and the resultingsubnet is 203.23.189.192 255.255.255.240.

On what subnet can the IP address 109.89.45.204 255.255.224.0 befound?


On what subnet can the IP address 49.210.83.201 255.255.255.248 befound?




IP Address

203.23.189.205

11001011 00010111 10111101 11001101

Subnet Mask

255.255.255.240 (/28)

11111111 11111111 11111111 11110000



IP Address109.89.45.204

01101101 01011001 00101101 11001100

Subnet Mask255.255.224.0 (/19)

11111111 11111111 11100000 00000000



IP Address49.210.83.201

00110001 11010010 01010011 11001001

Subnet Mask255.255.255.248 (/29)

11111111 11111111 11111111 11111000



IP Address31.189.234.245

00011111 10111101 11101010 11110101


12/20


subnet is 31.189.128.0 /17.


Convert the Boolean AND result into dotted decimal, and the resulting

subnet is 100.10.8.0 /21.



subnet is 190.98.35.0 /27.

You should have gotten each of these questions correct. If you had

trouble, review Section Six. If you answered each one correctly, letsmove on!

Determining Broadcast Addresses and Valid IP Address Ranges

What is the valid range of IP addresses for the subnet100.100.45.32 /28?

Subnet Mask /17 11111111 11111111 10000000 00000000



IP Address100.10.13.176

01100100 00001010 00001101 10110000

Subnet Mask /21 11111111 11111111 11111000 00000000



IP Address190.98.35.17

10111110 01100010 00100011 00010001

Subnet Mask /27 11111111 11111111 11111111 11100000



Subnet Address100.100.45.32

01100100 01100100 00101101 00100000

Subnet Mask /28 11111111 11111111 11111111 11110000


13/20

We know that the last four bits are the host bits. If these are all zeroes,this is the subnet address itself, and is not valid. If there are all ones,this is the broadcast address for this subnet, and is not valid. Alladdresses between the two are valid.

All-Zeroes Subnet Address: 100.100.45.32All-Ones Broadcast Address: 100.100.45.47

Valid IP Addresses: 100.100.45.33 46

What is the valid range of IP addresses for the subnet 208.72.109.8 /29?


Valid IP Addresses: 208.72.109.9 208.72.109.14

What is the valid range of IP addresses for the subnet 190.89.192.0

255.255.240.0?

All-Zeroes Subnet Address: 190.89.192.0

All-Ones Broadcast Address: 190.89.207.255Valid IP Addresses: 190.89.192.1 190.89.207.254

What is the broadcast address of the subnet 101.45.210.52 /30?


Subnet Address

208.72.109.8

11010000 01001000 01101101 00001000

Subnet Mask /29 11111111 11111111 11111111 11111000



10111110 01011001 11000000 00000000


11111111 11111111 11110000 00000000


Subnet Address

101.45.210.52

01100101 00101101 11010010 00110100

Subnet Mask /30 11111111 11111111 11111111 11111100


14/20

All-Zeroes Subnet Address: 101.45.210.52All-Ones Broadcast Address: 101.45.210.55Valid IP Addresses: 101.45.210.53, 101.45.210.54

What is the broadcast address of the subnet 90.34.128.0 /18?

All-Zeroes Subnet Address: 90.34.128.0All-Ones Broadcast Address: 90.34.191.255Valid IP Addresses: 90.34.128.1 90.34.191.254

What is the valid range of IP addresses for the subnet205.186.34.64 /27?


Valid IP Addresses: 205.186.34.65 94

What is the valid range of IP addresses for the subnet 175.24.36.0

255.255.252.0?





01011010 00100010 10000000 00000000

Subnet Mask /18 11111111 11111111 11000000 00000000


Subnet Address

205.186.34.64

11001101 10111010 00100010 01000000

Subnet Mask /27 11111111 11111111 11111111 11100000



10101111 00011000 00100100 00000000

Subnet Mask

255.255.252.0 (/22)

11111111 11111111 11111100 00000000


15/20

What is the broadcast address for the subnet 10.10.44.0 /25?



What is the broadcast address for the subnet 120.20.240.0 /21?

All-Zeroes Subnet Address: 120.20.240.0

All-Ones Broadcast Address: 120.20.247.255


What is the range of valid IP addresses for the subnet200.18.198.192 /26?





00001010 00001010 00101100 00000000

Subnet Mask /25 11111111 11111111 11111111 10000000



01111000 00010100 11110000 00000000

Subnet Mask /21 11111111 11111111 11111000 00000000


Subnet Address

200.18.198.192

11001000 00010010 11000110 11000000

Subnet Mask /26 11111111 11111111 11111111 11000000

Exam TipA single operation the one youve used to answer the last ten

questions, and that you first learned in Section Seven gives us thesubnet address, the broadcast address, and the valid IP addresses of a

given subnet. Make sure you give the correct answer when yourepresented with multiple choice. Dont give a subnet address for ananswer when a broadcast address has been asked for.


16/20

You should have the methodology down cold for determining subnetaddresses, broadcast addresses, and valid IP address ranges. If you hadany trouble with this section, review Section Seven. If not, lets head to

the last part of your exam!

Meeting The Stated Design Requirements

Youre working with the network address 135.13.0.0. Your networkmanager requests that you create a subnetting scheme that allows for atleast 500 valid subnets, while each subnet should contain as many hosts

as possible without exceeding 100. What is the best subnet mask touse?

A. 255.255.0.0

B. 255.255.192.0C. 255.255.254.0D. 255.255.255.128E. 255.255.255.192

F. 255.255.255.224

This is a Class B network, with 16 network bits and 16 host bits.

The first requirement is that we have at least 500 subnets. Using the

same formula weve used throughout this book, we see that nine subnetbits would give us 512 valid subnets:

2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 512

This would leave seven host bits, though. That would result in 126 valid

host addresses, which violates the second requirement. (2 to the 7th

power is 128; subtract two, and 126 valid host addresses remain.)

What about six host bits? That would yield 62 valid host addresses,which meets the second requirement. A combination of ten subnet bitsand six host bits gives us 1024 valid subnets and 62 valid hostaddresses, meeting both requirements.


Network Bits 11111111 11111111

Subnet Bits Host Bits


Network Bits 11111111 11111111

Subnet Bits 11111111 11

Host Bits 000000


17/20

The resulting mask is 255.255.255.192. (Note that a mask of255.255.255.224 would work, but it does not yield the maximumnumber of host addresses without going over 100.)

You are working with the network 160.10.0.0. Your network managerhas requested you develop a subnetting scheme that allows for at least230 subnets, but no fewer than 200. At present, each subnet willcontain no more than 20 hosts, but thats expected to grow by 20 hostsper year for the next five years. What is the best subnet mask to use?

A. 255.255.192.0B. 255.255.224.0

C. 255.255.240.0D. 255.255.248.0E. 255.255.255.0F. None of the above.

This is a Class B network, with a mask of 16 network bits, and 16 hostbits that can be borrowed for subnetting.

The only number of subnet bits that results in between 200 and 230subnets is eight, resulting in 256 valid subnets. That leaves eight host

bits. Is this enough for the second requirement? Yes. If a subnet has20 hosts now, and will grow by 20 hosts per year for the next five years,

thats a total of 120 host addresses. Eight host bits is plenty.

The resulting mask is 255.255.255.0.

Youre working with the network number 223.12.23.0. Your network

manager has asked you to develop a subnetting scheme that allows atleast 30 valid hosts per subnet, but yields no more than five validsubnets. Which of the following is the appropriate network mask to use?

A. /17

B. /24


Network Bits 11111111 11111111

Subnet Bits

Host Bits


Network Bits 11111111 11111111

Subnet Bits 11111111

Host Bits

00000000


18/20

C. /25

D. /26E. /27F. /28

G. /29H. /30

This Class C networks network mask is /24, leaving eight host bits:

Using the formulas weve used throughout the book, we know that we

need five host bits for at least 30 hosts per subnet. (2 to the 5th

power,minus two, equals exactly 30.) Does this meet the second requirement?

No. That would leave three bits for subnetting, which yields eight valid

subnets. To meet the second requirement, you can have only twosubnet bits, which yields only two valid subnets.

This yields a mask of 255.255.255.192, which in prefix notation isexpressed as /26.


manager has directed you to develop a subnetting scheme that will allow

for at least 5000 subnets while allowing the maximum number of hostsper subnet. What is the appropriate subnet mask?

A. 255.255.0.0B. 255.255.128.0C. 255.255.192.0D. 255.255.224.0E. 255.255.240.0F. 255.255.248.0

G. 255.255.252.0This Class A network has eight network bits, and 24 host bits that can

be borrowed for subnetting:


Network Bits 11111111 11111111 11111111

Subnet Bits

Host Bits


Network Bits 11111111 11111111 11111111

Subnet Bits 11

Host Bits 000000


19/20

For this question, we simply need to decide the minimum number ofsubnet bits that will yield at least 5000 subnets. Using a calculator orpencil and paper, you find that it takes 13 subnet bits to come up with atleast that many subnets; in this case, 8192 usable subnets.

The resulting mask is 255.255.248.0.


manager has requested that you develop a subnetting scheme thatallows at least fifty subnets. No subnet should contain more than 1000hosts. What is the best subnet mask to use?

A. 255.255.255.0B. 255.255.254.0C. 255.255.224.0

D. 255.255.192.0E. 255.255.128.0F. 255.255.64.0

G. None of the above.This Class B network has 16 network bits, and 16 host bits that can beborrowed for subnetting:

You quickly determine that for fifty subnets, you only need six subnetbits. That gives you 64 valid subnets. Does this meet the secondrequirement?No. That would leave 10 host bits, which yields 1022 valid host

addresses. (2 to the 10thpower equals 1024; subtract two, and 1022

remain.) But by borrowing one more bit for subnetting, giving us seven


Network Bits 11111111

Subnet Bits

Host Bits


Network Bits 11111111

Subnet Bits 11111111 11111

Host Bits 000 00000000


Network Bits 11111111 11111111

Subnet Bits

Host Bits


20/20

subnet bits and nine host bits, both requirements are met. Seven

subnet bits yield 128 valid subnets; nine host bits yield 510 valid hostaddresses. The appropriate mask is 255.255.254.0.

Congratulations! Youve completed this 65-question final exam. If youhad any difficulty with the final section, please review Section Eight. If

you nailed all five of the final questions great work!

Thanks !

Thank you for choosing The Bryant Advantage CCNA Study Packageto help you on your way to your next Cisco certification.

I know from experience that passing the CCNA exams is tough,especially now that binary math and subnetting are so prevalent.

Practice the questions in this book until you can approach each questiontype with confidence, and youll enter the CCNA testing room withsupreme confidence!

And drop me a line [email protected] you pass

your CCNA exams!

To your success,

Chris BryantCCIE #12933


Network Bits 11111111 11111111

Subnet Bits 1111111

Host Bits 0 00000000

Copyright @ 2011 The Bryant Advantage. All Rights Reserved.

09.your final binary exam

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