Ok so first check out professor messers video on 7 second subnetting. Its difficult but once you get it. You got it. The answer is C The subnet mask is incorrectly configured. A /30 would give you a subnet mask of 255.255.255.252 First available would be 10.10.255.253 If the (ISP) is conNECTED to the 253 itS WRONG It should be 252 which is the network ID stop giving bogus replies because you got it wrong. Your wrong simple. Study subnetting and stop posting bs replies because you got the answer 100,000% WRONG.

Both IP addresses are private IP, the subnetting will change the Subnet make value. hence the given answer is correct.

I think the issue is that they have assigned a private IP address to the serial interface which is connected to the ISP, and really should be a public IP address. I don't believe there is such a thing as "ip overlap issue" the way it has been described here. remember, you need an ip address AND a subnet mask to work. and Two machines with similar ip's but different subnet masks won't see each other. try it out in a lab. :) - if that's hard to grasp, consider how many different subnets could have the network ID of 192.168.0.0.

C is the right answer. Although A could be correct too. Changing the internal subnet mask into a /25 would fix the issue since that would divide the 10.10.0.x network into two, avoiding a collision of address space with .252 - .255

Given answer is correct, because the IPs are overlapping due to the subnet. 10.10.0.1/24 gives host IDs of 10.0.0.1 - 10.0.0.254 10.10.0.253/30 gives host IDs of 10.0.0.253 - 10.0.0.254 If two subnets overlap, when a router needs to send a packet to an IP address inside that range of overlapped addresses, the router may forward the packet to the wrong subnet.

Oh god.... this is so simple... stop overthinking it. On one it’s /24 bits.... there are 8 host bits reserved On the other same subnet there /30 there’s two host bits left...??? It’s simple how can there be /30 on the same subnet it would over lap with the /24??? All others subnet mask must be /24

Basic subnet chart for anyone who needs it: magic # 128 64 32 16 8 4 2 1 CIDR /25 /26 /27 /28 /29 /30 /31 /32 Mask 128 192 224 240 248 252 254 255

I do not see why the mask is wrong when they provided an IP address for Gigabit and Serial this is taken from a subnet calculator: Clients: IP Address: 10.10.0.1 Network Address: 10.10.0.0 Usable Host IP Range: 10.10.0.1 - 10.10.0.254 Broadcast Address: 10.10.0.255 Total Number of Hosts: 256 Number of Usable Hosts: 254 Subnet Mask: 255.255.255.0 Wildcard Mask: 0.0.0.255 Binary Subnet Mask: 11111111.11111111.11111111.00000000 IP Class: C CIDR Notation: /24 IP Type: Private Short: 10.10.0.1 /24 ------------------------------------------------------------------------------------------------------------- ISP: IP Address: 10.10.0.253 Network Address: 10.10.0.252 Usable Host IP Range: 10.10.0.253 - 10.10.0.254 Broadcast Address: 10.10.0.255 Total Number of Hosts: 4 Number of Usable Hosts: 2 Subnet Mask: 255.255.255.252 Wildcard Mask: 0.0.0.3 Binary Subnet Mask: 11111111.11111111.11111111.11111100 IP Class: C CIDR Notation: /30 IP Type: Private Short: 10.10.0.253 /30

They didn't give the router interface address so that rule A and B out (i think A or B will have been correct if there was a wrong interface address for the router in the question) D is off in a million years. C is correct without not even doing the calculation.

simply put. C. incorrect subnet mask? the subnet masks /24 and /30 are okay. but a private IP address(10.10.0.253) was used for Serial port to ISP -> yes, so the answer is A. It should be a public IP address.

The router could have an incorrect IP address configured, so A could be right. The router may also have the incorrect subnet mask, so C could be right. Everyone is correct as we know the two interfaces have overlapping addressing. Since both A and C are entirely possible and even likely, the Q is obviously not "recorded" correctly because enough information is not given to choose the "BEST" answer. Keep in mind, exam dumps are mostly created by test takers' "memory". So the questions are most likely not exact, nor are the answers always correct. Know the "whys" and you'll do fine on the actual exam.

This is a tricky question, and the explanation is that the ip address configured on the serial connection is wrong because it's using the broadcast which isn't configured to any host/port. While it's true the subnet mask was wrong, in the real world you will use a point-to-point connection with the ISP, and the /30 will be used to avoid ip waste, and to only make room for 2 addresses. This also adds security to that connection. But for this question, the wrong mask is the 'appropiate' answer.

Private IPs are non-routable. Therefore it's Switching and not Routing anything. Bones and Mimi are correct. Answer is A

Answer is A. The Router has the wrong IP address on either the serial interface or the Ethernet interface. As configured, both the serial interface and the interface are on the same network (10.10.0.0/24) but they should be on different networks. As an example, assume that the serial interface is correctly configured. This network has two IPS: one that is assigned to the serial interface (10.10.0.253/30) and other is assigned to the ISPs interface (10.10.0.254/30). The Ethernet interface would need to be a different network, such as with an IP address of 192.168.1.1/24

I agree with Eye_Tea. Different IP assigned to the LAN interface could be .2 .3 etc. and it would be still causing the issue. Moreover, its totally normal when WAN IP is a private one

I believe the answer is correct, just as "fluff's" answer is most closely correct... The router cannot have overlapping network assignment's on different interfaces as it wouldn't know which to route to. By using a /24, which is simply a default class c mask, it's usable IP range includes the assignment associated to the ISP. With a /25 or more (255.255.255.128), the IP range would have been out of the scope of the ISP, and thus on separate networks. Router should be 10.10.0.1 /25 subnet mask = 255.255.255.128

I believe the answer is correct, just as "fluff's" answer is partly correct... The router cannot have overlapping network assignment's on different interfaces as it wouldn't know which to route to. By assigning that IP a /24, which is only a default class c mask, it's usable IP range includes the ISP.