Enjoy this latest quiz and let me know how you scored in the comments section here!

CCENT (100-101 ICND1) Sample Quiz 1

Enjoy this sample quiz where subnetting is the focus!

Start

Congratulations - you have completed CCENT (100-101 ICND1) Sample Quiz 1.
You scored %%SCORE%% out of %%TOTAL%%.
Your performance has been rated as %%RATING%%

Your answers are highlighted below.

Question 1

The decimal number 205 converted into a binary number is _____.

A

11011101

B

11001101

C

110001019

D

11000101

Question 2

What is the decimal equivalent of the binary number 11000111?

A

218

B

199

C

179

D

208

Question 3

Using six subnet bits, how many usable subnets are created?

A

58

B

62

C

64

D

128

Question 4

What is the last usable host address on the third subnet for the following address and mask combination - 192.168.1.0 255.255.255.248?

A

192.168.1.24

B

192.168.1.23

C

192.168.1.25

D

192.168.1.22

Once you are finished, click the button below. Any items you have not completed will be marked incorrect.
Get Results

There are 4 questions to complete.

←

List

→

Return

Shaded items are complete.

1

2

3

4

End

Return

You have completed

questions

question

Your score is

Correct

Wrong

Partial-Credit

You have not finished your quiz. If you leave this page, your progress will be lost.

I use something similar to this. https://www.pearsonitcertification.com/articles/article.aspx?p=433757&seqNum=6 Notice how easy this question becomes. There are five subnet bits in use here (255.255.255.248). If you go 5 bits deep into the binary conversion chart – you see the increment in use here is 8. So the first subnet is 192.168.1.0, the second is 192.168.1.8, the third is 192.168.1.16, and the forth is 192.168.1.24. They are asking about the last usable in the third subnet. The broadcast for that subnet is 192.168.1.23 and the last usable is 192.168.1.22.

I loved the questions scored 100% please post more this kind of questions are fun and boost our confidence to do the exam

OK – you got it. Just the other day I realized that I need to do more of these – thanks for confirming that!

the last question is a bit….weird.

Since subnet-zero was NOT mentioned. Why on earth would it be, 192.168.1.22?

The subnet-zero doesn’t exist.

Subnet-1) 192.168.1.0–>1st host=192.168.1.1 & last=192.168.1.6

Subnet-2) 192.168.1.8–>1st host=192.168.1.9 & last=192.168.1.14

*subnet-3) 192.168.1.16 is the network address so the first host is 192.168.0.17 and the last usable address is 192.168.1.22.

The question is referring to the *Third subnet

1st subnet 192.168.1.0 /29 = increment of 8

2nd subnet .8/29

3rd subnet .16/29

usable .17 – .22(Last Usable)

broadcast .23

4th subnet .24/29

I’m a bit confused on the last question despite getting it right. What method do you guys use for subnetting?

I use something similar to this. https://www.pearsonitcertification.com/articles/article.aspx?p=433757&seqNum=6 Notice how easy this question becomes. There are five subnet bits in use here (255.255.255.248). If you go 5 bits deep into the binary conversion chart – you see the increment in use here is 8. So the first subnet is 192.168.1.0, the second is 192.168.1.8, the third is 192.168.1.16, and the forth is 192.168.1.24. They are asking about the last usable in the third subnet. The broadcast for that subnet is 192.168.1.23 and the last usable is 192.168.1.22.

Thanks for explaining the solution. This helped me understand subnetting some more.