Exam 70-741 All Questions

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Exam 70-741 topic 1 question 98 discussion

Actual exam question from Microsoft's 70-741

Question #: 98
Topic #: 1

HOTSPOT -
You run Get-NetIPAddress and receive the output shown in the following exhibit:

Use the drop-down menus to select the answer choice that completes each statement based on the information presented in the graphic.
NOTE: Each correct selection is worth one point.
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Suggested Answer:

by dan at July 3, 2019, 1:45 p.m.

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coleman

Highly Voted 5 years, 7 months ago

For interface index 20 "Internet": PrefixLength: 24 = 24 bits netmask = 255.255.255.0 For interface index 10 "Ethernet 2" PrefixLength: 8 = 8 bits netmask with 24 bits host ID, (2^24)-2 = 16777214

upvoted 13 times

Ameer

5 years, 7 months ago

how would you know that the host ID bits is 24? Thanks

upvoted 1 times

Blue

5 years, 7 months ago

Ameer: PrefixLength: 24 6th row from the bottom of the exhibit.

upvoted 2 times

Thermal

4 years, 12 months ago

Whilst I've no doubt you seem to know you're stuff you answered Ameer's question incorrectly by taking a PrefixLength from a different ethernet card which just confuses matters. I wanted to point this out for anyone confused when they come across this, refer to the comment by "Arlomini" below for the actual answer as to how they worked out the hosts.

upvoted 2 times

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Blue

5 years, 7 months ago

May i also add Address: 10.10.10.25 00001010 .00001010.00001010.00011001 Netmask: 255.0.0.0 = 8 11111111 .00000000.00000000.00000000 Wildcard: 0.255.255.255 00000000 .11111111.11111111.11111111 => Network: 10.0.0.0/8 00001010 .00000000.00000000.00000000 (Class A) Broadcast: 10.255.255.255 00001010 .11111111.11111111.11111111 HostMin: 10.0.0.1 00001010 .00000000.00000000.00000001 HostMax: 10.255.255.254 00001010 .11111111.11111111.11111110 Hosts/Net: 16777214 (Private Internet)

upvoted 4 times

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Arlomini

5 years, 2 months ago

the prefix lenght is 8, meaning the mask for this network will be 255.0.0.0 and there there are 24 bits left for the hosts. So to count the exact number of host available for that network you do the math 2^24-2= 16777214 hope this explanation helps you pick it up better.

upvoted 6 times

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panda

Most Recent 4 years, 5 months ago

Index of 10 has 8 bit prefix length. Therefor it has 24 bits host address, which is 255*255*255. In test you couldn't calculate strictry. 200*200*200 < 255*255*255 < 300*300*300 8000,000 < 255*255*255 < 27,000,000 So 255*255*255, which is the number of hosts, is 16,777,214.

upvoted 1 times

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lofzee

4 years, 6 months ago

the Interface Identifier is nothing more than a number/name given to the adapter. The prefix length is the subnet mask. prefix length /24 = 255.255.255.0 - pretty standard in most small/medium sized companies and will give you 256-2 = 254 addresses to use. -2 because you cannot use the broadcast or the network/subnet address. As opposed to working out what 2 to the power of 24 is in an exam environment, NO IDEA ! lol. I can't remember if I saw a calculator in my last online proctored exam. Someone else here said there is.

upvoted 1 times

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Vortex_SA

4 years, 6 months ago

Got this Q on my exam - 24 Nov'20

upvoted 3 times

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TA77

4 years, 9 months ago

The index is just a number given to the NIC, so you can refer to a certain NIC using a number instead of a given name. Q1: The NIC with index 20, has a prefix of 24. Which means 24 1's. 11111111.11111111.11111111.00000000 = 255.255.255.0 Just memorize that 255 = 11111111 Q2: The NIC with 10 index have a prefix of 8. You need to calculate the number of addresses for this network. The number of addresses for a given network = 2 ^ the number of 0's in the subnet mask (this is only when there's no subnetting. Because in case of subnetting, there's a different way to calculate the number of addresses. You can tell if there's subnetting if there is other number in the subnet mask other than 255 and 0. Example: 255.255.240.0). In this question: 2 ^ 24 which equals to 16M ---------------------------------------------------------------------

upvoted 2 times

TA77

4 years, 9 months ago

I don't have a specific way to figure out what 2^24 equals to in the exam. but you can just memorize that 2^16 = 65.5k and 2^24 = 16M Because if there's no subnetting, the probable numbers that may appear in a question in the exam are: 2^8 (Based on the prefix 8) 2^16 (Based on the prefix 16) 2^24 (Based on the prefix 24) We all know that 2^8=256, so we left with 2^16 and 2^24. This is just my own opinion.

upvoted 3 times

V1980

4 years, 8 months ago

dang this is overkill! You have 3 octets, each one 255 possible choices, 255 * 255 * 255 = 16,581,375 The exam, at least the online proctored one has a calculator as well as a whiteboard.

upvoted 2 times

V1980

4 years, 8 months ago

Now before some smart a comes in, let me correct my workings out, which doesn't change the answer anyway 256 * 256 * 256 = 16,777,216 16,777,216 minus the IPs reserved for gateway etc.

upvoted 3 times

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Erik93

5 years, 2 months ago

I don't get the first part of the question, an index of 20 (prefix) = 255.255.240.0?

upvoted 1 times

Pox

5 years, 1 month ago

Each network interface has a friendly name you can change e.g. production or management and it has an InterfaceIdentifier which is just a numerical identifier. It has nothing to do with network addresses/subnetting

upvoted 2 times

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Nhan

5 years, 5 months ago

for the cnetwork card its is class C so the subnet mask is 255.255.255.0, for the netwrk card index 10 its class A so there for over 16 million hosts available its all about IP class and subnet mask.

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dan

5 years, 9 months ago

rsmallwo - your link does not support your theory. if the was true then the subnet mask would be 255.255.255.240 and yes the hosts would be 16million. I would like someone else to interpret Microsofts network interface values

upvoted 2 times

minajahan

5 years, 9 months ago

See here (scroll a bit, there is a table): https://www.calculator.net/ip-subnet-calculator.html

upvoted 2 times

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rsmallwo

5 years, 11 months ago

Its the prefix length value that helps you get the answer. Use something like: https://www.calculator.net/ip-subnet-calculator.html to see how many ip's you get dependant on prefix length

upvoted 3 times

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dan

5 years, 11 months ago

can someone provide a link on how this is figured out. know basics in networking but now sure how this interface index plays into subnets and hosts. thks

upvoted 1 times

jhgf

4 years, 5 months ago

I think the 'interface index' bit is just a 'where to look on screen' sort of pointer: "look here, at this bottom one whose interface index says '20' / look here, at this top one whose interface index says '10' " . Then you notice the prefix length (24 = 255.255.255.0 / 8 = 255.0.0.0). Nice of the to cut it clean with these prefixes and not use wild numbers. The bits left past the prefix, are there for hosts. Those covered by prefix are part of the net. IPV4 = 32 bits = 0000 0000 . 0000 0000 . 0000 0000 . 0000 0000 up to 1111 1111 . 1111 1111 . 1111 1111 . 1111 1111 (the 255s and all are decimal readouts of these binary digits).

upvoted 1 times

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