Exam 200-301 topic 1 question 109 discussion

You can do this fairly easily by process of elimination. Starting with R7, a /26 is .192, so that leaves us with A or D. The first difference between A and D is the last octet, 127 or 126 (respectively). Do whatever process you prefer for subnetting and we figure that .126 is the last usable. .127 is the broadcast. Answer is D.

D is correct

Just by eliminating options.. the correct for R7 is option D

As mentioned in the question we need the last usable IP address so remember the rule Network ID Even First IP Odd Last IP Even Broadcast Address Odd So starting with Router 7 we know that the subnet is 192 so now we have 2 options A and D and we know last IP should be even so answer is D.

D is the correct answer.

it's enough to look only R7 and answer is D

Just look at the subnets mask.for r7 “/26” = .192 because 128 plus 64. Then for r8 “/28 “ = .240 because 128+64+32+16 that only leaves option d

10.88.31.64 - 10.88.31.127/26 (FUH 10.88.31.65-10.88.31.126 LUH) inc = 64 Last usable host for R7(fa1/0) 10.88.31.126 255.255.255.192 10.19.63.80 - 10.19.63.95/28 (FUH 10.19.63.81-10.19.63.94 LUH) inc = 16 Last usable host for R8(fa0/0) 10.19.63.94 255.255.255.240 10.23.98.128 - 10.23.98.159/27 (FUH 10.23.98.129 - 10.23.98.158 LUH) inc = 32 Last usable host for R9(fa1/1) 10.23.98.158 255.255.255.224

it's D

Subnet Mask of R7 is /26 thus 255.255.255.192

I agree that D is the right answer

it's D the subnet of first address in B is Wrong

Its D the subnet masks in B are incorrect

The right answer is B. As for R9 the last usable address should be .158.