IP address should be made of 4 octets. Each octet should not be greater than 255.
A) 50.50.50.50 : Valid
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The rule for valid subnet mask: From left there should be contious ones. From right, there should be continuous zero. All can be ones. all can be zeros also.
Yes it is valid
Yes. It is valid.
This is NOT valid. Continuity is not there. D) 0.0.0.0 Continuity is there. So it VALID
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Answer for Q4. convert the first octet into binary and check for the first zero. if the first zero comes in the first place, then it is class-a, if the first zero comes in the second place means class-b, third place means class-c... Let us convert 50 into binary......0011 0000. The first zero is in the first place,so, it is claaa-a convert 130 into binary.....1000 0010, the first zero is in the second place. so, it is class-b. convert 200 into binary...11001000, the first zero is in third place, so it is class-c convert 240 into binary....1111 0000, the first zero is in the fifth place, so it is class e |
Answer for Q5: The subnet mask of 192.168.1.39/28 is slash 28. 32 minus 28= 4 bits. 4 bits=16 combinations. So this subnet is having 16 ip addresses in total. If 192.168.1.39 is one among the 16, then the first ip in this group will be 192.168.1.32 and the last will be 192.168.1.47. Only D (192.168.1.45) is coming within this range. So answer is D |
Answer for Q6: A MAC (media access control) address is of 48 bits. The other names for mac address is hardware address, adapter address, physical address, layer 2 address
The mac address is written in human readable form like this: 01-23-45-67-89-ab or 01:23:45:67:89:ab
The first 24 bits represents the company who is manufacturing this network interface card. |
Answer for Q7: The subnet mask of 84.23.5.92 / 0.0.0.0 is slash zero. 32-0=32 bits. 32 bits means all ip addresses from 0.0.0.0 to 255.255.255.255 are in this group. So A,B,C,D are the correct answers
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Answer for Q8:
Other than the following are normally public ip addresses. The private ip addresses are 10.x.x.x 192.168.x.x 172.16.x.x to 172.31.x.x Auto configuration ip addresses: 169.254.x.x Loopback addresses :127.x.x.x
So
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Answer for Q9:
255.255.255.254 has only two ip addresses in its subnet. This cannot be used because we need an ip for network address and broadcast address. so, 255.255.255.252 is the correct answer
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Answer for Q 10:
This question is slightly advanced. So if you are not able to follow, dont worry. A subnet mask is having 29 bits, means, the subnet mask can be rewritten in the following format: 255.255.255. 248. The difference between subnet mask and wild card mask is that: But minus-ing the subnet mask from 255.255.255.255, you will get wild card mask, so 255.255.255.255 minus 255.255.255.248 = 0.0.0.7 the answer is 0.0.0.7
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Answer for Q 11: I am 192.168.1.53/255.255.255.224.
My ip is 192.168.1.53. my subnet mask is "/27". 32-27= 5 bits. 5 bits = 32 combinations. So, totally 32 ip addresses are there in my group and i am one among them. The others can be 192.168.1.32 to 192.168.1.63. So the answer is B and C |
Answer for Q 12: Q12. In the following four IP addresses, Find out which of them are on the same network. A) 192.168.1.1 / 255.255.255.252 B) 192.168.2.1 / 255.255.255.252 C)192.158.1.5 / 255.255.255.252 D)192.168.1.2 / 255.255.255.252 The subnet mask can be rewritten as "/30" 32-30=2 bits. 2 bits=4 combinations. So, only four ip addresses are possible in any of the subnets. Let us take A. Its ip is 192.168.1.1. the range of ip is 192.168.1.0 to 3, The "D" is within the range. so , D
is the anwer.
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