On this lesson, I will share some sample subnetting questions para mas ma-practice pa natin ang subnetting skills natin. Mahalaga na makabisado natin at mabilis natin itong magawa dahil malaki ang maitutulong nito para maipasa n’yo ang CCNA exam at syempre sa real networking world.
All of these questions came from subnettingquestions.com(it’s a good site para makapag-practice ka ng subnetting).
Ready?
Let’s do this!
Practice #1

Let’s see kung pano nakuha yung sagot.
Gaya ng formulang pinag-aralan natin nung naunang lessons, gamitin natin ang ating mga daliri. Sa ating question, we need to have 29 subnets that can accomodate 4 hosts each galing sa 192.168.253.0/24.
Given: 192.168.253.0/24
Needed: 29 subnets(or network), 4 hosts each
Find: Bits and NSM
Let’s use our finger subnetting technique. To get the number of bits need to borrow or take from the host portions, need natin ng 5 bits.
Why?
32 ang value ng pang-limang daliri natin kapag nag-bilang tayo ng bits(2,4,8,16,32) from right to left. That can satisfy 29 subnets needed. See the image below that we also used last time.

Ang NSM(New Subnet Mask) naman natin is /29. Why? Eh kasi /24 and old subnet mask natin. Remember our formula for new subnet mask kapag # of networks is NSM = OSM + bits barrowed.
So /24 + 5(bits taken) = 29. Remember, 192 is a Class C ip address na merong 255.255.255.0 or /24 na default subnet mask. Therefore, 192.168.253.0/29 is the subnetted network that can accomodate the question.
/29 = 255.255.255.248. Why? Gaya din ng ginamit natin sa mga naunang lessons, meron na tayong 24 1s galing sa given, so that’s 255.255.255. Then 5 bits or 5 1s ang nadagdag, so count tayo sa ating daliri sa from LEFT TO RIGHT(25, 26, 27, 28, 29).
Ang subnet mask value ng pang-limang daliri from LEFT TO RIGHT is 248(128, 192, 224, 240, 248). See image below(green numbers) na ginamit din natin last time.

Can /29 or 255.255.255.248 accomodate 4 hosts each subnet? Yes! How? Sa ating formula before, para makuha natin ang actual number of hosts ito ang formula.

So in our sample, 2 raise to 3(32 – 29) then minus 2.
2 raise to 3 = 8 minus 2 is 6. So pasok ang hinihingi nating 4 hosts each subnet.
That’s it pansit! So dahil ang tanong is what subnet mask should we use, ang sagot is /29 or 255.255.255.248.
Practice #2
Let see how we got that.
Sa mga ganitong klase ng tanong, kelangan natin makuha ang increment. At gaya ng mga nauna nating lessons, we can get the increment by just getting the increment value of the subnet mask value. Kung ano yung katapat niya. So sa subnet mask na 255.255.255.248, ang value ng increment is 8.
Why? Ang increment value ng 248 is 8. Like what we see on our sample image below.

Then gaya nung mga naunang sample natin sa mga unang lessons, we just need to add the increment to the base network.

As you can see, nag-add lang tayo ng 8(kasi siya yung increment value) sa last octet kung saan nangyari yung change. We keep adding 8 until marating natin yung hinahanap na subnet kung saan pasok yung ip na binigay sa given.
From there we can get the following answers:
Subnet: 192.169.189.128 – 192.168.189.135
Network address: 192.169.189.128
Broadcast address: 192.168.189.135
First valid ip: 192.169.189.129
Last valid ip: 192.169.189.134
So dito natin makikita kung bakit 192.168.189.135 ang sagot. Makse sense?
Practice #3
On this type of question din, kelangan natin makuha ang increment. What’s the increment? 16! Why? 16 ang increment value ng 240 subnet mask. This time nasa 3rd octet lang tayo pero same technique mga idol.

So start tayo from base address(3rd octet) then increment tayo ng 16 kada network. Continue lang hanggang maabot natin ung hinahanap. Ganito dapat mangyayari.

And from there, makikita n’yo kung bakit network 10.173.208.0 ang naging sagot. Why? Kasi yung host na hinahanap natin is pasok sa network na yan. Ang mga valid host IP address niya is 10.173.208.1 – 10.173.223.254.
Make sense idol?
Practice #4
Again, we need to find the increment in this type of question. Ang increment dito is 2! Why? Kasi 2 ang increment value ng 254 subnet mask gaya ng finger subnetting technique na ginagamit natin simula pa nung una.

So to prove na tama yung sagot, let’s compute again. madali lang ito.

Bakit naging 172.18.252.1 ang sagot? Syempre ang tanong is what is the first valid host on the network that node 172.18.252.28, which is pasok sa network na 172.18.252.0 kagaya ng nakikita natin sa image.
Ang range ng valid hosts nito are 172.18.252.1 – 172.18.253.254. Laging tandaan na ang first address ay para sa network address at ang last address ay para sa broadcast, so hindi sila valid host address.
To summarize, here’s what we can get from that answer:
Subnet: 172.18.252.0 – 172.18.253.255
Network address: 172.18.252.0
Broadcast address: 172.18.253.255
First valid ip: 172.18.252.1
Last valid ip: 172.18.253.254
Let’s have a final sample subnetting question.
Practice #5
Sa subnet mask na 255.255.254.0, we used or added 7 bits (FROM LEFT TO RIGHT) from the original subnet mask. To get the actual number of networks, gaya ng pinag-aralan natin last time ang formula is:

So to get the answer, we just need to compute 2 raise to 7.
7 na 1s ang nadagdag natin or pwede rin na /23(.254) – /16 which is 7 din naman ang sagot. 2 raise to 7 = 128. Kaya 128 networks ang sagot.
Paano natin nalaman na 7 ang nadagdag na bits? Eh ang subnet na 172.30.0.0 ay isang class B ip address. Meaning meron siyang default subnet mask na /16 or 255.255.0.0.
From the given, we can see na hindi na default subnet mask ang gamit niya dahil ito ay 255.255.254.0 na. Meaning, ito ay subnetted na. Kaya gaya ng sabi ko, importante na alam mo ang mga ip address classes kasama ng kanilang mga default subnet mask.
Para naman sa actual number of hosts, ang formula natin kung natatandaan n’yo sa mga Part II ng subnetting lessons is ganito:

So to get the answer, 2 raise to 9(0s remaining) minus 2 or pwede rin na 2 raise to 32 – 23(NSM) which is 9 din naman. Ang sagot is 512.
Then syempre minus 2 kasi hind kasali ang network at broadcast address kaya 510 valid hosts addresses.
Gotcha? Let me show it to you below.

We’re good?
With that 5 practice questions and samples, I hope by this far you now know understand the process and the reasons why we need to know subnetting. Kapag nakabisado mo ito, it will make your life easier. Hindi lang sa CCNA kundi pati na rin sa ‘real world.
If you’ve helped by this article, share it to others or post it on social media. That way, we can help and inspire more aspiring network engineers like you.
This is the end of our subnetting topic. In the CCNA Fundamentals Ebook, we cover advance subnetting katulad ng VLSM(variable subnet mask) and Route summarization.
The ebook covers other CCNA 200-301 topics na wala at hindi kasama dito sa blog. It’s also in taglish like what you’re reading right now. In fact, all these free lessons are taken from the ebook. Hindi nga lang lahat.
If you’d like to know more about the ebook, see the details here.
We’re now moving into a new lesson.
Hi, Ask ko lang po,
bakit 172.26.12.0 and answer dito
“Which subnet does host 172.26.12.2/28 belong to?”
since 12 bits ang ndagdag from OSM /16
na confuse lang po
Tama lang naman idol norman. 16 yung increment n’ya at ito yung mga subnets n’ya.
172.26.12.0 – 172.26.12.15 (usable address 172.26.12.1 – 172.26.12.14)
172.26.12.16 – 172.26.12.31 (usable address 172.26.12.17 – 172.26.12.30)
172.26.12.32 – 172.26.12.63 etc.
Yung 172.26.12.2 pasok sa unang subnet. So tama ang sagot.
172.26.12.0 – network address
172.26.12.15 – broadcast address
Let me know if you need help. Thank you!
Thanks po, dami q nakukuha tip sa blog mo idol,
Nakatulong lalo pra masagutan q ng mabilis yung mga subnetting sample questions, ?
172.26.12.0 – 172.26.12.15 (usable address 172.26.12.1 – 172.26.12.14)
mali ka ng type sir, pero thanks ( usable host)
Thanks Rameces. Typo lang, hehe. It’s been updated. 🙂
Idol pasagot naman
Question: What valid host range is the IP address 172.20.160.8/20 a part of?
Here’s the answer: Valid host range = (172.20.160.1 – 172.20.175.254). Why?
Subnet mask: 255.255.240.0
Increment: 16
Network range:
172.20.160.0 – 172.20.175.255 (172.20.160.1 – 172.20.175.254)
172.20.176.0 –
172.20.192.0 –
Kung susundan mo yung tutorials dito idol, makukuha mo yan. I hope it helps. Godbless!
thankyou so much Sir!
Welcome idol Jack! 🙂
Doon sa question na what is the broadcast address of the network 192.168.189.128 and answer is 192.168.189.135. nag gawa ako ng table like 192.168.189.0- 192.168.189.7 and so on. So bali dagdg lang ako ng 8 kada network and so on. Kilan at paano ko malalaman kung kilan ako titigil? Kc yung 135 pde pa dagdagan ng another 8 and so on. What is the clue kung kilan ako titigil sa pag add ng 8. Thanks
When you got the answer. That’s it! Sa mga ganyang type ng tanong hahaba talaga if wala kang hinahanap idol, once na nakuha mo na yung sagot yun na yun. 🙂
Hi Sir,,
Share ko lang sa ans. ko..(trying my best lang para makalearn)
Network address : 192.168.189.128
Increment: 8
192.168.189.128
192.168.189.129 – 1st valid host
192.168.189.130 – 2nd valid host
+ 8 192.168.189.131 – 3
192.168.189.132 – 4
192.168.189.133 – 5
192.168.189.134 – 6th valid host or LAST VALID HOST
192.168.189.135 – ito yung broadcast address
192.168.189.136 – Ito naman yung address for the next network
FLex ko lang…hehe
Idol billy i have question.got this from subnettingquestions.com. If you’re desingning a network 172.22.0.0 and you want 200 subnets and 190 hosts. What subnet mask should you use. The answer is 255.255.255.0. Tanung ko idol paano naging 255.255.255.0 ang sagot. Thanks idol billy
Hello idol billy, alam kuna pala ang sagot. I borrowed 8 bits which is equivalent to 256 which is pasok ang 200 na subnets. So 16 + 8 = 24 and 24 bits is equal to 255.255.255.0 hehehe ?
Tama idol. Good job. Ngayon ko lang nakita ‘to, cnxa na. 😀
San po galing yung 16?
What valid host range is the IP address 192.168.82.99/27 a part of?
Answer: 192.168.82.97 through to 192.168.82.126
Tama ba ito idol? Hindi ba dapat 192.168.82.96 through 192.168.82.127?
Tama lang idol Carlo. Ang tanong kasi is “valid host range” so dapat hindi kasama ang network address(192.168.82.96) tska broadcast address(192.168.82.127). Dapat yung in between lang nilang dalawa which is 192.168.82.97 – 192.168.82.126. Hope it helps. Thanks!
idol ask lang dito
Question: What valid host range is the IP address 192.168.237.185/29 a part of?
Answer: 192.168.237.185 through to 192.168.237.190
bkit through 192.168.237.190? 8 ung increment bali 185+8 = 193 dba pdeng 192.168.237.191?
Tama lang ang sagot idol Paul. Ang tanong kasi is valid host range kung nasaan si 192.168.237.185. Bale ganito siya:
192.168.237.176 – 192.168.237.183
192.168.237.184 – 192.168.237.191 (si .184 kasi is network address at si .191 naman is brodcast address. Hindi sila kasama sa valid host range)
I hope this helps. Thanks!
Sir baket inaadd ung .185 sa increment ng 8? D po ba host din sya? at ang hinahanap is kung saan syang subnet belong? if ever ano ang pinaka first network ng subnet na /29. sa given IP address na yan. 🙂 Thanks po!
Hindi siya ini-add Ruth, pinakita ko lang as sample, tumapat lang. Lol
Your story is very inspiring bro. Salamat sa mga taong katulad mo na nagshe-share ng knowledge para sa aspiring Cisco enthusiast. Na-inspire ako ng sobra sa story mo dahil matagal ko nang plano ang mag-study at mag-take ng Cisco Certification,actually I’m a Computer Eng’g.grad. but I did not practice my studies since naiba ang line of work ko d2 sa abroad. Malayong malayo ang work ko sa tinapos ko pero sabi ko nga sa sarili ko “it’s never too late”. Still im in the early 30’s. Nag-start na ko mag-study coz im planning to take the bootcamp and exam too on my vacation next year by May siguro..Sabi mo nga bro tiyaga lang at hardwork. Gusto kong magshift ng career na related talaga sa tinapos ko although ayos naman ang salary ko d2 sa work ko. ‘Nung nabasa ko ang mga basic sa networking bro, biglang nag flash back sken mga inaral ko dati sa networking nung college times ko.Sabi ko medyo malapit na ko sa katotohanan. Anyway, maraming salamat bro.Malaking tulong ang ginagawa mong ito hindi lang sa mga katulad ko maging sa mga estudyanteng bumubuo pa lng ng kanilang mga pangarap. God bless u bro!
Hi idol! Thanks sa mga tips mo. Mas lalo ako nainspire i-pursue cisco career ko.
Ask ko lang din sa no. 4, dun sa range of valid hosts, should it be 172.18.252.1 – 172.18.253.254?
Yes Japoy. Thanks!
Sir may tanong lang po ako:
What is the first valid host on the subnetwork that the node 172.19.47.195 255.255.255.224
Answer: 172.19.47.193
Bakit po naging 172.19.47.193?
Bigla lang ako naguluhan kasi diba po +32 ung increment nya? Tas mag minus 1 ka na lang para mabuo ung isang network?
Last valid kasi ung tanong idol, hindi naman valid ip yung 194 kasi broadcast address yun. Kaya siya 193. Ang first at last ip ay not valid host ip kasi sila ay network address at broadcast address respectively. 😉
Hi Sir Billy,
Sir nasa industry na din po ako at real world na akong nagde-design ng buong subnets ng isang company, both switching and routing subnetting, even applying it on different flavor of Firewall IOS. I am just worried on what I am currently know about subnetting.
I’ve found an online subnetting and here is the scenario – How many network/s and host/s are there in a network of 192.168.1.0/24, the answer was obvious, NETWORK=1 and the HOST=254. Am I right sir Billy? The online said that the NETWORK=256 and the HOST=254. Hence, the same application on classful CIDR notation network like /8, and /16 respectively.
Hmmm. Tricky yung question idol pero tama ka kung literally susundin yung mismong context na tanong. Iniisip ko ang gusto niyang palabasin is yung buong networks na kasama si 192.168.1.0/24 so lalabas yung sagot niya. Kaso mo yun nga ndi naman sinabi tska pwedeng mali din iniisip ko ksi walang nakalagay. Hehe. Good catch anyways. 😉
Sa number 3, di ba ang 10.173.212.244 ay class A (less than 126 ung 1st Octet: 10)?
Bakit sa 3rd octet nag-increment at hindi sa 2nd Octet? Thanks.
HI Umar, the given is already subnetted. You just need to find the subnet(network) where those hosts belong. THanks!
Sir,
Sa number 4 naman ung first valid host sa 172.18.252.28 with a subnet mask of 255.255.255.254.0.
since ang given lang is yung node. San niyo nakuha ung 172.18.252.0 na unang subnet? kasi dito po ako magbabase kung ano ung mga increments na susunod.
For example po eto yung isosolve.
172.26.127.19 255.255.248.0
Answer: 172.26.120.1 (ayon sa internet)
di ko alam sir kung pano makuha ung subnet niyan. naguluhan ako sa naging 3rd Octet nung answer which is san nakuha ung .120 (3rd octet)
Thank you sir 🙂
Idol Jay, kindly check the tutorial on how to get the increment. It’s all there. Salamats!
Sir, san nakukuha ang OSM sa question 5? kase ung ibang lesson nakaindicate na ang OSM
Sa default na class niya kapag ndi given. 🙂
Question: How many subnets and hosts per subnet can you get from the network 172.24.0.0 255.255.255.240?
Answer: 4096 subnets and 14 hosts
Sir paanong naging 12 borrowed bits? Ano basehan dun?
Originally, ang formula ng NSM ay:
NSM = OSM + bits
As per the problem, NSM is given, OSM is also given, pano natin nasabi?
OSM = /16, why? kasi yug yung IP address sa problem ay nagsisimula sa 172, which is Class B, and Class has 16 bits by default correct?
NSM is also given na from the subnet mask 255.255.255.240
Subnet Mask Long Format: 255.255.255.240
8 + 8 + 8 + 4 = 28 ==> NSM
Summary:
Formula is NSM = OSM + bits
Derive natin para makuha yung Bits
Bits = NSM – OSM
= 28 – 16
= 12 ==> # of Bits
Actual Networks: 2^(nsm-osm)=2^(28-16)=2^12=4096 networks
AH: 2^(32-nsm) -2= 2^(32-28) -2= 2^(4) -2= 16-2=14 hosts
Billy, medyo nalito ako sa example number 2.
dun sa illustration mo na
192.168.189.0- 192.168.189.7
192.168.189.8- 192.168.189.15
……
…….
192.169.189.120- 192.168.189.127
192.169.189.128- 192.168.189.135
192.169.189.136
tanong ko bakit naging 169 instead of 168?
Typo error lang yan sir.
What is the last valid host on the subnetwork 192.168.25.128, 255.255.255.192?
Answer: 192.168.25.190
sir di ba dapat ang sagot ay 191? kasi yung 192 ay broadcast na.
192.168.25.128
LF: 255.255.255.192
NSM: /26
Bits: /32-/26=6
inc: 64
Range:
192.168.25.0
192.168.25.64
192.168.25.128 – 192.168.25.191 >>> Valid Range for subnetwork 192.168.25.128
192.168.25.192
Valid Host Range:
192.168.25.129 – 192.168.25.190
pano sir makuha ang cidr ng IP?
pasagot naman idol what is the default gateway on the subnetwork that the node 172.18.252.28 255.255.44.0 belongs to