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Cisco Certified Network Associate (CCNA) 200-301 Practice Test 7

Question 60

Two routers have a dedicated IPv4 point-to-point link and need exactly two addresses. Both support RFC 3021 and the interface has no traditional Ethernet broadcast hosts. Which address allocation can conserve the most IPv4 space without breaking the router-to-router link?

  1. Use a /32 on each end and expect both routers to discover the other endpoint without further routing.
  2. Use a /30 solely because every IPv4 prefix must reserve network and broadcast addresses.
  3. Use a /29 because RFC 3021 requires six usable addresses before the link comes up.
  4. Use a /31 on the point-to-point link, assigning the two consecutive IPv4 values to the two endpoints.

Explanation

Correct Answer

D:

RFC 3021 allows both address values in a 31-bit IPv4 prefix to identify the two endpoints of a supported point-to-point link. Unlike a conventional multiaccess subnet, the two patterns are not reserved as network and directed-broadcast host values for this use. A /31 therefore consumes two addresses instead of four while giving the endpoints unique identifiers. Verify support on both sides, interface addressing, and basic adjacency after installation. Do not apply the two-endpoint exception indiscriminately to shared Ethernet networks with ordinary hosts.

Incorrect Answers

A:

A /32 denotes one specific IPv4 address, typically a host route or loopback address. Configuring unrelated isolated /32 endpoint addresses does not automatically create an on-link two-address subnet or neighbor relationship. The requested standard space-saving link is a mutually supported /31 with both endpoints assigned from the same two-address block.

B:

The network/broadcast reservation convention applies to many ordinary IPv4 subnets, but RFC 3021 deliberately defines special behavior for point-to-point links. A /30 would work in many environments yet spends four address values when only two endpoints exist. With RFC 3021 support, /31 is the more efficient valid answer.

C:

A /29 spans eight addresses with six conventional host values and is much larger than a dedicated two-router link needs. Nothing in RFC 3021 imposes that minimum. The distinguishing requirement is a supported point-to-point circuit with only two endpoint addresses, making a /31 more efficient.

Question 61

You must allocate nonoverlapping subnets inside `10.10.8.0/24` for departments needing 100, 45 and 14 usable IPv4 hosts. Which initial allocation is valid and leaves contiguous space for smaller groups?

  1. Assign /26 to the 100-host department, /27 to the 45-host department and /28 to the 14-host group.
  2. Assign 10.10.8.0/25 and 10.10.8.64/26 to the first two departments.
  3. Assign 10.10.8.0/25 to the 100-host department, 10.10.8.128/26 to the 45-host department, and 10.10.8.192/28 to the 14-host group.
  4. Assign /24 to the 100-host department, then reuse the same /24 for the other two departments.

Explanation

Correct Answer

C:

A /25 offers 126 conventional usable hosts, a /26 offers 62 and a /28 offers 14. Aligning largest-to-smallest produces nonoverlapping blocks at .0–.127, .128–.191 and .192–.207 within the available /24.

Incorrect Answers

A:

A /26 has only 62 usable hosts and a /27 has only 30, so the first two departments cannot fit.

B:

The /26 at .64 lies inside the .0/25 block, creating overlapping addresses rather than separate departments.

D:

Allocating the entire /24 to one independent subnet leaves no separate disjoint address space for the others.

Question 62

A switch SVI is configured for IPv4 subnet 192.0.2.64/27. Which dotted-decimal mask corresponds to the /27 prefix length?

  1. 255.255.255.224.
  2. 255.255.255.240.
  3. 255.255.255.192.
  4. 255.255.255.248.

Explanation

Correct Answer

A:

The /27 mask has 27 network bits, including three high-order bits in the last octet (11100000 binary = 224).

Incorrect Answers

B:

255.255.255.240 has 28 network bits and changes the SVI subnet boundaries.

C:

255.255.255.192 corresponds to /26, a 64-address block rather than the requested 32-address block.

D:

255.255.255.248 is /29, providing only eight addresses per subnet.

Question 63

An engineer must advertise 10.20.8.0/24, 10.20.9.0/24, 10.20.10.0/24, and 10.20.11.0/24 as one aligned CIDR route without including 10.20.12.0/24. Which summary prefix is exact?

  1. 10.20.8.0/21, which also includes 10.20.12.0/24 through 10.20.15.0/24.
  2. 10.20.9.0/22, which is not aligned on the four-/24 boundary.
  3. 10.20.8.0/23, which includes only the first two /24 networks.
  4. 10.20.8.0/22

Explanation

Correct Answer

D:

Four contiguous /24 networks can be represented by a /22 when their initial third-octet value is a multiple of four. The sequence 8, 9, 10, and 11 is aligned to that boundary, so 10.20.8.0/22 covers exactly those addresses, from 10.20.8.0 through 10.20.11.255. It does not include 10.20.12.0/24. Confirm the aggregate is actually reachable before advertising it; CIDR arithmetic does not prove the underlying component networks have working next-hop paths. A Null0 summary, if used, must be considered separately.

Incorrect Answers

A:

A /21 aggregates eight consecutive /24 networks. Beginning at third octet 8, it would cover 10.20.8.0 through 10.20.15.255 and therefore include third-octet values 12 to 15, which the requirement explicitly excludes. Broadening the prefix is not an exact four-network summary.

B:

A /22 must start on a third-octet boundary divisible by four: 0, 4, 8, 12, and so on. The address 10.20.9.0 lies within the 10.20.8.0/22 network but is not that /22's canonical network address. Advertise the aligned prefix instead of an address with host bits set. Otherwise, downstream routers may reject the unaligned network statement.

C:

A /23 summarizes two aligned adjacent /24 networks rather than four. The prefix 10.20.8.0/23 covers only 10.20.8.0 and 10.20.9.0; it does not include 10.20.10.0 or 10.20.11.0. The requested aggregate needs two fewer prefix bits than /24, which gives /22. A smaller aggregate necessarily omits two requested subnets.

Question 64

A host should use address `192.0.2.78/27` on VLAN 20, whose subnet is `192.0.2.64/27`. It incorrectly sends ARP for `192.0.2.97` as if that destination were local. Which fault best explains this behavior?

  1. The host is using a correct /27 and ARP for all destinations is expected.
  2. The far-end switch has an STP alternate port, which changes the IPv4 host subnet mask.
  3. The host is using a mask broader than /27, such as /26, causing it to treat .97 as on-link.
  4. The destination is the directed broadcast address of 192.0.2.64/27.

Explanation

Correct Answer

C:

The /27 spans .64–.95; .97 belongs to a different /27. With an incorrect /26 covering .64–.127, a host ARPs for .97 rather than forwarding to its default gateway, producing the symptom.

Incorrect Answers

A:

With a correct /27, .97 is off-subnet and the host normally ARPs for its default gateway, not directly for .97.

B:

STP path state can affect frame forwarding but cannot expand an endpoint IP prefix from /27 to /26.

D:

That broadcast address is .95, not .97; interpreting .97 as local is evidence of incorrect prefix logic.

Question 65

A VLAN 70 DHCP pool is 192.168.70.0/28. The network address and broadcast are reserved by subnet rules. Its gateway uses .1 and two printers use reserved host addresses .2 and .3. Twelve additional clients now need distinct addresses. What is the capacity problem?

  1. A /28 always provides sixteen client addresses, so twelve clients plus the gateway and printers will fit.
  2. Only eleven assignable addresses remain after the gateway and two printer reservations, so twelve further clients exceed capacity by one.
  3. The subnet has just two usable host addresses, and one is the broadcast address.
  4. The DHCP server must permit addresses 192.168.70.16 through .19 inside the same /28.

Explanation

Correct Answer

B:

The /28 192.168.70.0/28 covers .0 through .15, with usable .1–.14. The network (.0) and broadcast (.15) are not host pool members. A gateway at .1 plus printer reservations at .2 and .3 consume three of the fourteen usable addresses, leaving eleven for other clients. The request for twelve additional leases therefore exceeds capacity by exactly one. Verify the actual lease scope, reservations and active leases; neither increasing DHCP retry time nor changing a default route adds an address to the subnet. The engineer must enlarge or subdivide the plan correctly or reduce reservations/demand according to policy. This is a capacity calculation, not simply checking whether the /28 mask is syntactically valid.

Incorrect Answers

A:

A /28 contains sixteen total addresses, not sixteen assignable conventional host addresses. The network ID and broadcast cannot serve as ordinary clients, leaving fourteen usable .1 through .14. Three of those are consumed by the gateway and printer reservations. The claim that all sixteen can go to clients ignores both protocol reservations and already assigned addresses.

C:

A conventional /30 offers two usable hosts; a /28 offers fourteen. The broadcast on this /28 is .15, not one of the two usable /30 addresses. Confusing the two prefix lengths would substantially understate DHCP capacity and could prompt unnecessary network changes. Calculate 2^(32−28) total addresses and subtract the normal network/broadcast reservations.

D:

The addresses .16 through .19 belong to the next /28 boundary, 192.168.70.16/28, and are not within the original .0/28 scope. Simply entering them as additional leases under the old mask would create a subnet mismatch and could break gateway reachability. The solution requires a deliberate correct network renumbering or additional scoped subnet, not a wider numerical pool outside the prefix.

Question 66

Two routed handoffs use adjacent disjoint blocks `192.0.2.32/29` and `192.0.2.40/29`. The second router has host address `192.0.2.42` but its mask was mistakenly set to `/28`. Which unintended connected-prefix overlap results?

  1. The router treats `192.0.2.32–192.0.2.47` as one /28 connected subnet, which includes the first handoff's `.32/29` range.
  2. The /28 mask on 192.0.2.42 creates 192.0.2.48/28 and reaches only higher .48–.63 addresses.
  3. The /28 is functionally identical to /29, so the connected route cannot change.
  4. The 192.0.2.42 host address is the broadcast address of 192.0.2.40/29.

Explanation

Correct Answer

A:

A /28 prefix aligns on 16-address boundaries. Host .42 with mask /28 belongs to 192.0.2.32/28, combining the .32/29 and .40/29 blocks that were intended as separate routed subnets; changing the mask back to /29 restores the .40–.47 scope.

Incorrect Answers

B:

The /28 containing .42 is .32/28, not .48/28; align the subnet base by 16-address increments.

C:

A /28 contains 16 addresses while a /29 contains eight; the broader connected route can absorb an adjacent /29.

D:

For .40/29, .40 is network, .47 is broadcast, and .41–.46 are usable, so the defect is the mask, not the host identifier.

Question 67

Two routed interfaces are intended to use `192.0.2.8/30`. Instead, one was assigned `192.0.2.8/30` and the other `192.0.2.11/30`. What is the addressing error?

  1. The interfaces were given the two valid /30 host addresses.
  2. One interface should use /31 while the other stays /30 for compatibility.
  3. The failure is caused only by an absent route to an off-site /24.
  4. Both routers were assigned the network and broadcast values rather than usable host IPs .9 and .10.

Explanation

Correct Answer

D:

In a conventional /30 subnet, .8 denotes the network and .11 is the directed broadcast. The two usable addresses are .9 and .10; configuring the reserved endpoints causes an addressing error before investigating routes.

Incorrect Answers

A:

The valid host values are .9 and .10; .8 and .11 are the subnet identifiers reserved by conventional IPv4 broadcast semantics.

B:

Peers must agree about which addresses are on-link; mixing /31 and /30 is not a remedy for invalid /30 host assignments.

C:

The question already supplies invalid local interface IPs. Static routing does not make the network and broadcast addresses ordinary /30 hosts.

Question 68

A point-to-point routed link uses RFC 3021 `192.0.2.20/31` on one end and `192.0.2.21/31` on the other. A legacy script rejects .21 as a broadcast address. Why is its check wrong?

  1. 192.0.2.21 is the broadcast address of every IPv4 /31 and must never be assigned.
  2. The only valid endpoint IPs for that link are .19 and .22 with mask /31.
  3. For a /31 on a point-to-point interface, both addresses in the two-address prefix are usable endpoints; a /30 network/broadcast rule does not apply.
  4. A router must translate both private /31 addresses into public addresses to form local adjacency.

Explanation

Correct Answer

C:

RFC 3021 permits a /31 prefix on suitable point-to-point links, treating both addresses as host endpoints without the conventional all-zero/all-one host reservations. The legacy script assumed a /30-style network and broadcast.

Incorrect Answers

A:

The special point-to-point /31 design uses both prefix addresses; ordinary /30 broadcast calculations are not applicable.

B:

Those addresses belong to different aligned /31 blocks, not the .20/31 link.

D:

The two stated addresses are documentation-range IPv4 values on one point-to-point link; NAT is unrelated to whether /31 endpoints are usable.

Question 69

An address plan reserves `10.20.0.0/27` for a small LAN, then tries to give `10.20.0.0/25` to a 100-host department. Why does the later allocation conflict?

  1. The /25 covers .0–.127 and contains the previously reserved .0–.31 block, so the subnets overlap.
  2. The /25 has only 30 usable IPv4 addresses and cannot support 100 hosts.
  3. The /27 always starts at .128, so both entries are necessarily separate.
  4. IPv4 routers reject all subnets whose first octet begins with 10.

Explanation

Correct Answer

A:

VLSM prefixes must be aligned and disjoint when assigned to different layer-3 networks. A /25 beginning at .0 encompasses the full first /27, creating duplicate routing/on-link ranges; reserve large aligned blocks first or relocate one subnet.

Incorrect Answers

B:

A conventional /25 provides 126 usable hosts; capacity is not the reason this particular plan fails.

C:

The explicitly reserved subnet is 10.20.0.0/27, inside the .0/25. Valid /27s can start at multiples of 32, including .0.

D:

10.0.0.0/8 is RFC1918 private space; it is usable internally and does not explain the overlapping assignment.

Question 70

An IPv4 ACL should match only `192.0.2.64/27`. It instead uses `192.0.2.64 0.0.0.63`, allowing packets from .96 onward. Which change corrects the match?

  1. Keep wildcard 0.0.0.63 because /27 has exactly 64 addresses.
  2. Use wildcard `0.0.0.31` with network `192.0.2.64` for the intended /27.
  3. Set wildcard 0.0.0.255, which restricts the ACL to 32 consecutive addresses.
  4. Use subnet mask 255.255.255.224 as the wildcard on the ACL line.

Explanation

Correct Answer

B:

An ACL wildcard is the bitwise complement of the subnet mask. /27 is 255.255.255.224, so its final wildcard octet is 31. The current wildcard 63 matches a /26-sized range (.64–.127).

Incorrect Answers

A:

A /27 has 32 addresses, while a 63 wildcard permits 64 values in the low-order six bits.

C:

A 255 wildcard would ignore the full final octet, allowing 256 address values.

D:

ACL network matching uses an inverse/wildcard mask in this context; 255.255.255.224 is the regular subnet mask, not its wildcard complement.

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