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Ltat42a wrote: » That looks like one of those matching questions. In questions like these, look at the class of address you are subnetting from, then look at the mask you are given. The 172.16.22.0 = Class B. The default mask for Class B is 255.255.0.0. You are given 255.255.255.192. 10 bits were taken from the host portion for subnets, which leaves 6 bits for hosts. Using your subnets formula, 2(n) n= 10, you have 1024 subnets. You have 6 bits remaining for hosts, using your hosts formula 2(n)-2, n = 6, 64-2 = 62 - you have 62 hosts per subnet. Your answer is the 2nd one down from the top on the right. hth
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