READ Free Dumps For CISCO- 100-105
| Question ID 14469 | An administrator is working with the 192.168.4.0 network, which has been subnetted with a
/26 mask. Which two addresses can be assigned to hosts within the same subnet?
(Choose two.)
|
| Option A | 192.168.4.61
|
| Option B | 192.168.4.63
|
| Option C | 192.168.4.67
|
| Option D | 192.168.4.125
|
| Option E | 192.168.4.128
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| Option F | 192.168.4.132
|
| Correct Answer | C,D |
Explanation Explanation: Increment: 64 (/26 = 11111111.11111111.11111111.11000000) The IP 192.168.4.0 belongs to class C. The default subnet mask of class C is /24 and it has been subnetted with a /26 mask so we have 2(26-24) = 22 = 4 sub-networks: 1st subnet: 192.168.4.0 (to 192.168.4.63) 2nd subnet: 192.168.4.64 (to 192.168.4.127) 3rd subnet: 192.168.4.128 (to 192.168.4.191) 4th subnet: 192.168.4.192 (to 192.168.4.225) In all the answers above, only answer C and D are in the same subnet. Therefore only IPs in this range can be assigned to hosts.
| Question ID 14470 | Given an IP address of 192.168.1.42 255.255.255.248, what is the subnet address?
|
| Option A | 192.168.1.8/29
|
| Option B | 192.168.1.32/27
|
| Option C | 192.168.1.40/29
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| Option D | 192.168.1.16/28
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| Option E | 192.168.1.48/29
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| Correct Answer | C |
Explanation Explanation: 248 mask uses 5 bits (1111 1000) 42 IP in binary is (0010 1010) The base subnet therefore is the lowest binary value that can be written without changing the output of an AND operation of the subnet mask and IP... 1111 1000 AND 0010 1010 equals 0010 1000 - which is .40 /24 is standard class C mask. Adding the 5 bits from the .248 mask gives /29