Passed CompTIA Security+ SY0-301

Passed Security + with 894.

1. Want information and pure information read (helps little in EXAM)-

CompTIA Security+™ EXAM GUIDE (Exam SY0 - 301) third editionWm. Arthur Conklin, Gregory White, Dwayne Williams, Roger Davis, Chuck Cothren.

2. Concise information -

CompTIA® Security+™ SY0-301 Exam Cram, Third Edition
Diane Barrett, Kalani K. Hausman, and Martin Weiss

3. Practice tests do plenty.

4. Google all topics and read as much.

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Comments

HLRS

congrats, I am lookin to knock out sec + also.

Nytrocide

Congrats! I hope to take and pass it soon too!

FrankTh

Congratulations.
Almost a perfect score you had.
Are you planning to get more certification in the future?

FrankTh

quinnyfly

WOW, that's not far from perfect, perhaps the best score I have ever seen, well done and congrats.

sameerjan77

@ FrankTH
CISSP - all the way now...have to have it at ANY cost.

Dakinggamer87

Congrats on pass!!

whatthehell

Congratz on the pass bhayaa.

sameerjan77

@FrankTh

thanks. Now going for Baap of all - CISSP

FrankTh wrote: »

Congratulations.
Almost a perfect score you had.
Are you planning to get more certification in the future?

FrankTh

_Marauder

congrats

qwertyu1234567

Hi everyone,
There is a question on Security+ SY0-301:

Which of the following IP address would be hosts on the same subnet given the subnet mask 255.255.255.224? (Select Two)
A. 10.4.4.125
B. 10.4.4.158
C. 10.4.4.165
D 10.4.4.189
E. 10.4.4.199
How do I find the right answer?
Thank you.

abramsgunner

Let's see if I can do this.....

For that subnet mask you convert the last octet to binary.. 11100000 (using Windows built-in calculator)

Note the bits that aren't zero (111)00000 those show you which bits are the 'network mask' bits

Convert the last octet for each IP address

.125 = 1111101 <- calculator doesn't display leading zero, so add enough to show 8 bits = 01111101
.158 = 10011110
.165 = 10100101
.189 = 10111101
.199 = 11000111

Mask of (111)00000 tells us we need to look at first three bits......

.165 and .189 have identicle first three bits, so they should be on the same subnet.

At least that's the way I think it works... LOL

PS.. Best to post a question in it's own thread, makes it easier to find later.

movingzachb

abramsgunner wrote: »

Let's see if I can do this.....

For that subnet mask you convert the last octet to binary.. 11100000 (using Windows built-in calculator)

Note the bits that aren't zero (111)00000 those show you which bits are the 'network mask' bits

Convert the last octet for each IP address

.125 = 1111101 <- calculator doesn't display leading zero, so add enough to show 8 bits = 01111101
.158 = 10011110
.165 = 10100101
.189 = 10111101
.199 = 11000111

Mask of (111)00000 tells us we need to look at first three bits......

.165 and .189 have identicle first three bits, so they should be on the same subnet.

At least that's the way I think it works... LOL

PS.. Best to post a question in it's own thread, makes it easier to find later.

Thanks abram

Can anyone second to confirm this is correct?

Darril

Yes, that's a good explanation abramsgunner.

danny069

Congrats on the pass! and abramsgunner you should teach subnetting 101 lol

abramsgunner

Nah.. not my work man; I read how to do it last week some time, so it was fresh on my mind....

SharkDiver

Congratulations on the pass!
Very nice score.

Nice method for finding the answer, abramsgunner.

Just to offer another way,
I do it like this:

I know the following:
.0 is one group of 255
.128 is two groups of 128
.192 is four groups of 64
and
.224 is eight groups of 32

32 is the key

So, I can figure out:
0 - 31 is a subnet
32 - 63 is a subnet
64 - 95 is a subnet
96 - 127 is a subnet
128 - 159 is a subnet
160 - 191 is a subnet
192 - 223 is a subnet
and
224 - 255 is a subnet

From there I look to see where all of the IP Addresses they gave fall.
.125 - falls between 96 and 127
.158 - falls between 128 and 159
.165 - falls between 160 and 191
.189 - falls between 160 and 191
.199 - falls between 192 and 223

So, .165 and .189 are in the same subnet.