mcAfee Secure Website
23

Cisco 200-301 Bundle

Exam Code: 200-301

Exam Name Cisco Certified Network Associate (CCNA)

Certification Provider: Cisco

Corresponding Certification: CCNA

200-301 Training Materials $44.99
Pass4sure Q&A Sample #1 Pass4sure Q&A Sample #2 Pass4sure Q&A Sample #3 Pass4sure Q&A Sample #4 Pass4sure Q&A Sample #5 Pass4sure Q&A Sample #6 Pass4sure Q&A Sample #7 Pass4sure Q&A Sample #8 Pass4sure Q&A Sample #9 Pass4sure Q&A Sample #10

Reliable & Actual Study Materials for 200-301 Exam Success

The Latest 200-301 Exam Questions as Experienced in the Actual Test!

  • 24
    Questions & Answers

    200-301 Questions & Answers

    760 Questions & Answers

    Includes questions types found on actual exam such as drag and drop, simulation, type in, and fill in the blank.

  • 25
    200-301 Video Course

    200-301 Training Course

    271 Video Lectures

    Based on Real Life Scenarios which you will encounter in exam and learn by working with real equipment.

  • exam =30
    Study Guide

    200-301 Study Guide

    1969 PDF Pages

    Study Guide developed by industry experts who have written exams in the past. They are technology-specific IT certification researchers with at least a decade of experience at Fortune 500 companies.

exam =32

Cisco Certified Network Associate (CCNA) (200-301) Practice Tests

Guarantee

Satisfaction Guaranteed

Pass4sure has a remarkable Cisco Candidate Success record. We're confident of our products and provide no hassle product exchange. That's how confident we are!

99.3% Pass Rate
Total Cost: $194.97
Bundle Price: $149.98

Purchase Individually

  • exam =34
    Questions & Answers

    Questions & Answers

    760 Questions

    $124.99
    exam =35
  • exam =37
    200-301 Video Course

    Training Course

    271 Video Lectures

    $39.99
  • exam =36
    Study Guide

    Study Guide

    1969 PDF Pages

    $29.99

Cisco 200-301 Practice Test 7

Question 60

A branch receives 172.29.104.0/24 and must divide it into four equal networks for separate floors. What prefix length should each resulting subnet use?

  1. /25 for each subnet.
  2. /27 for each subnet.
  3. /26 for each of the four subnets.
  4. /24 for each subnet.

Explanation

Correct Answer

C:

Four equal subnets require two additional network bits because two bits represent four subnet IDs. Extending the original /24 to /26 creates 64-address blocks at .0, .64, .128 and .192. These are nonoverlapping and exhaust the original /24.

Incorrect Answers

A:

Borrowing one bit from /24 yields only two /25 networks rather than the required four. Each /25 consumes half the original block.

B:

Borrowing three bits yields eight /27 networks. That would divide the allocation more finely than four equal groups and is not the specified equal partition.

D:

Repeating the parent /24 four times would overlap the same 256-address allocation. Prefixes must be lengthened to produce independently routed subnet IDs.

Question 61

A router interface uses 172.26.197.62/25. A colleague proposes 172.26.197.126 as a second host on the same on-link subnet. Which statement is accurate?

  1. They lie in different /25 subnets because their last octets differ.
  2. Both addresses are within 172.26.197.0/25 and are ordinary usable hosts.
  3. 172.26.197.126 is the /25 broadcast and cannot be assigned.
  4. 172.26.197.62 is the network identifier of the /25.

Explanation

Correct Answer

B:

A /25 includes final-octet addresses 0 through 127 in its lower half. Values .62 and .126 lie within that range, avoiding reserved .0 network and .127 broadcast boundaries. The second endpoint can be on-link when VLAN, layer-two and host configuration are consistent.

Incorrect Answers

A:

An endpoint subnet includes many different host values; the first 25 network bits must match. Both values are below 128 in the same /25 block.

C:

For lower-half /25, broadcast is .127, not .126. The latter is the highest conventional usable host address in the subnet.

D:

The network ID is .0, since the seven host bits are all zero there. Address .62 is a regular host inside the lower /25.

Question 62

A datacenter allocates 10.182.44.0/23 for two service networks. A proposed subnet 10.182.45.0/24 is paired with 10.182.44.0/24. Do those proposed subnets overlap?

  1. Yes; every pair of /24s in one /23 necessarily overlaps.
  2. No. They are the two separate /24 halves of 10.182.44.0/23.
  3. Yes; 10.182.45.0 is the broadcast of 10.182.44.0/24.
  4. No; because 10.182.44.0/23 has only 256 total addresses.

Explanation

Correct Answer

B:

The parent /23 spans third-octet values 44 and 45. Each child /24 fixes all 24 network bits, so one covers 10.182.44.0–.255 and the other covers 10.182.45.0–.255. Neither contains an address owned by the other.

Incorrect Answers

A:

Subnets derived by adding a prefix bit can be disjoint portions of the parent. Their shared parent address space does not mean the two children overlap.

C:

A /24 broadcast ends with last octet .255, so 10.182.44.255 is the relevant broadcast. The 45.0 address starts a different /24.

D:

A /23 has 9 host bits and therefore 512 addresses. Two /24 children partition those 512 addresses without overlap.

Question 63

A security engineer needs the bitwise inverse of the /28 subnet mask 255.255.255.240 for ACL matching. Which IPv4 wildcard mask has the host bits set?

  1. 255.255.255.240.
  2. 0.0.0.240.
  3. 0.0.0.31.
  4. 0.0.0.15.

Explanation

Correct Answer

D:

The /28 subnet mask leaves four host bits, and an ACL wildcard marks those variable bits with ones. Subtracting each subnet-mask octet from 255 gives 0.0.0.15. This wildcard matches an aligned 16-address IPv4 block when paired with the correct network address.

Incorrect Answers

A:

That is the subnet mask itself, not its inverse wildcard. Reusing the subnet mask in an IOS wildcard field changes which bits are compared and which are ignored.

B:

The value 240 marks the four high bits of the last octet as variable, opposite the /28 host-bit position. The correct inverse must mark the low four bits.

C:

A wildcard of 31 corresponds to five variable host bits and thus a /27-sized block. A /28 has only four variable bits and needs 15.

Question 64

A network team has existing VLAN subnet 172.23.129.0/27. A new proposed VLAN is 172.23.129.16/28. Can both operate as nonoverlapping routed VLAN subnets?

  1. Yes; a longer prefix is always disjoint from a shorter parent prefix.
  2. Yes; because the first 16 addresses belong to the /27 and the last 16 to the /28.
  3. No; the proposed /28 sits entirely inside the existing /27.
  4. No; because 172.23.129.16 is outside all IPv4 private-use space.

Explanation

Correct Answer

C:

The /27 allocation covers last-octet addresses 0 through 31. The /28 beginning at 16 covers addresses 16 through 31, a subset of that /27. Assigning both as separate routed networks creates overlapping connected prefixes and ambiguous addressing for hosts.

Incorrect Answers

A:

A longer prefix can describe a contained portion of the shorter block. Here the two explicitly share every address in the new /28.

B:

The original /27 includes all 32 addresses, not just the first half. A child subnet cannot automatically remove its addresses from the parent.

D:

The address is in RFC 1918 172.16.0.0/12. The actual conflict is overlapping subnets, not public-versus-private classification.

Question 65

A /24 allocation begins at 10.183.133.0. When it is divided into /27s, what is the network ID of its third /27 block in ascending order?

  1. 10.183.133.32/27.
  2. 10.183.133.63/27.
  3. 10.183.133.64/27.
  4. 10.183.133.96/27.

Explanation

Correct Answer

C:

Each /27 block contains 32 addresses. The /27 subnet IDs of this /24 start at .0, .32, .64 and so on, so the third begins at last octet 64. Counting block boundaries, rather than device host addresses, avoids placing a network ID at an invalid offset.

Incorrect Answers

A:

That is the second 32-address block starting at .32. Counting the first block at .0 makes the third boundary .64.

B:

The .63 address is the directed broadcast of the second /27, not an aligned subnet ID. A /27 network starts at multiples of 32.

D:

This is the fourth /27 network after .0, .32 and .64. The question asks for the third, so it is one block too far.

Question 66

An interface is 172.16.222.109/27. A monitoring agent proposes using 172.16.222.127 for a second endpoint in the same VLAN. Why is that address unsuitable?

  1. It belongs to the next /27 subnet starting at .128.
  2. It is the /27 network identifier and all host bits are zero.
  3. It is prohibited because all 172.16.x.x addresses are globally routed public space.
  4. It is the directed broadcast of 172.16.222.96/27, not an ordinary host address.

Explanation

Correct Answer

D:

The /27 block containing .109 begins at .96 and ends at .127. The final address has all five host bits set, which is the conventional directed broadcast. A second endpoint should use an unused host address between .97 and .126 instead of the reserved boundary.

Incorrect Answers

A:

The next network begins only after .127. The proposed address is still inside the current block but is its broadcast, not its neighbor network ID.

B:

The network identifier is .96. At .127 the host bits are all ones, so the issue is broadcast rather than a zero-host-bit network address.

C:

The entire 172.16/12 reservation is designated private under RFC 1918. The host assignment problem arises from the local subnet boundary, not global routing.

Question 67

A host needs to use 10.6.2.255/23 on a routed LAN. A technician rejects every IPv4 address ending in .255 as a broadcast. Is this particular interface address necessarily unusable?

  1. No; the entire 10.6.2.0/23 has only two usable hosts.
  2. Yes; every .255 address is universally a directed broadcast regardless of mask.
  3. Yes; a /23 permits only addresses with odd final octets.
  4. No. 10.6.2.255 is a usable host in 10.6.2.0/23; the directed broadcast is 10.6.3.255.

Explanation

Correct Answer

D:

A /23 mask fixes the first 23 bits, spanning third-octet values 2 and 3 in this block. The subnet begins at 10.6.2.0 and ends at 10.6.3.255, so 10.6.2.255 is an interior address, not its directed broadcast. Whether an address ends with .255 is not sufficient without the actual subnet mask.

Incorrect Answers

A:

A /23 has nine host bits and contains 512 addresses, normally 510 usable. The host-count assumption confuses it with a /30.

B:

Broadcast status depends on all host bits of the configured network, not the final octet alone. This /23 broadcast is 10.6.3.255.

C:

Host eligibility is based on the prefix boundaries and reservation of network/broadcast addresses. Odd/even last-octet parity is not a rule.

Question 68

A router has 10.47.167.0/24 for a lab. The address team needs one /25 for a large floor and two /26 networks for smaller floors. Which layout uses the /24 without overlap?

  1. 10.47.167.0/25, 10.47.167.64/26 and 10.47.167.128/26.
  2. 10.47.167.0/25, 10.47.168.0/26 and 10.47.168.64/26.
  3. 10.47.167.0/25, 10.47.167.160/26 and 10.47.167.224/26.
  4. 10.47.167.0/25, 10.47.167.128/26 and 10.47.167.192/26.

Explanation

Correct Answer

D:

The /25 consumes the lower half, .0–.127. The upper half can be divided into two /26 blocks, .128–.191 and .192–.255. The three networks meet their intended sizes and cover the /24 exactly once rather than overlapping existing ranges.

Incorrect Answers

A:

The .64/26 subnet lies inside the lower .0/25, so the first two overlap and the upper .192/26 remains unallocated. The allocation is not disjoint.

B:

The two /26s start in the next /24, outside the supplied address block. A VLSM solution must stay within 10.47.167.0–.255.

C:

The latter two values are not aligned /26 network boundaries. Valid /26 IDs are multiples of 64 in the final octet.

Question 69

A point-to-point link has subnet 172.19.184.128/29 but a technician wants to configure an IP with last octet 135. What is the most accurate rejection reason?

  1. It is the network address of 172.19.184.128/29.
  2. It belongs to 172.19.184.136/29, the next subnet.
  3. 172.19.184.135 is the directed broadcast of that /29 subnet.
  4. It is an IPv4 multicast address by virtue of its last octet.

Explanation

Correct Answer

C:

The /29 block holds eight addresses from .128 through .135. The last address sets all three host bits and is the conventional directed broadcast, while usable hosts run .129 through .134. The technician should select an unassigned usable host instead of using the boundary.

Incorrect Answers

A:

The network identifier has all host bits cleared and is .128. Address .135 has all host bits set, so it is the opposite reserved boundary.

B:

The next /29 starts at .136, not .135. The candidate is still part of the preceding /29 but is unsuitable as a conventional host.

D:

IPv4 multicast classification depends on the leading bits of the first octet, 224/4, not last-octet 135. The address is in a private IPv4 range.

Question 70

A host uses 172.17.58.242/26 as its IP address and needs to ARP for its gateway. Which proposed gateway is in the same /26 and is an ordinary host?

  1. 172.17.58.193.
  2. 172.17.58.191.
  3. 172.17.58.192.
  4. 172.17.59.1.

Explanation

Correct Answer

A:

With /26, addresses .192 through .255 form one subnet, with .192 network and .255 broadcast. Both .242 and .193 are host values in that shared block, so the endpoint can resolve the gateway directly over the local VLAN. A gateway in the preceding /26 would need routing and is not directly on-link.

Incorrect Answers

B:

The .191 address ends the previous /26 block .128–.191 and is its broadcast. It is not an eligible gateway on the host’s .192/26 subnet.

C:

The .192 address begins the required subnet but is the network identifier, not a conventional usable interface address. The gateway must use an eligible host ID.

D:

That address is outside the 172.17.58.192/26 block and cannot be directly ARPed on the stated subnet. It would require a separate network path.