Which of the following IP addresses can be assigned to host devices?(Choose two)
A.
205.7.8.32/27
B.
191.168.10.2/23
C.
127.0.0.1
D.
224.0.0.10
E.
203.123.45.47/28
F.
10.10.0.1/13
Explanation:
We can’t assign the Broadcast address, Network ID, Loopback, Multicast address on host.A is incorrect because it is a Network ID of 121 subnet B is correct and can be assigned to host
C is incorrect because it is loopback address
D is incorrect because it is multicast address
E is incorrect because it is broadcast address of 202.123.45.32/28 subnetB is correct because IP address from 191.168.10.0/23 can be assigned to host
F is correct because IP address from 10.10.0/13 can be assigned to host
191.168.10.2/23 is not a network address?
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Absolutely!
With 191.168.10.2/23,
the network address is 191.168.10.0/23
the host address range is from 191.168.10.1 to 192.168.11.254
So B is correct.
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I Forget that, Thanks!!
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Actually you are wrong, it is impossible to change from a 191 to a 192 network address with a /23 subnet mask.
The range will be from 191.168.10.1 – 191.168.11.254 which is not a valid private IP address according to RFC1918
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A – 205.7.8.32/27 (incorrect) : Network Address
C – 127.0.0.1 (incorrect) :
is the loopback Internet protocol (IP) address also referred to as the “localhost.” The address is used to establish an IP connection to the same machine or computer being used by the end-user. The same convention is defined for computer’s that support IPv6 addressing using the connotation of ::1. Establishing a connection using the address 127.0.0.1 is the most common practice
D – 224.0.0.10 (incorrect) :
The EIGRP adjacency process begins with R1 sending an EIGRP Hello packet to the multicast address 224.0.0.10 in an attempt to find potential neighbors.
B & F Are Correct
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