Networking - MCQ Practice Questions

Understanding:

We need to find the minimum subnet mask (fewest borrowed bits) that provides at least 6 subnets from a /24 block.

Formula:

The number of subnets created by borrowing $n$ bits is:

Step 1: Determine bits to borrow

We need $2^n\ge 6$.

So we must borrow $n=3$ bits.

Step 2: Calculate the new prefix length

Starting from a /24 network and borrowing 3 bits:

The new subnet mask is $/27$.

Step 3: Verify host capacity

With a /27 mask, each subnet has $32-27=5$ host bits, giving $2^5-2=30$ usable hosts per subnet — reasonable for most scenarios.

Answer:

The minimum subnet mask required to create at least 6 subnets from a /24 block is /27.

Quick Tip:

Always use $2^n\ge \text{required subnets}$ and pick the smallest $n$ that satisfies the inequality. Borrowing fewer bits than needed is the most common mistake.

Understanding:

We must find the number of usable host addresses for the subnet mask $\mathrm{255.255.255.192}$.

Formula:

The number of usable hosts in a subnet is:

where $h$ is the number of host bits (the $-2$ accounts for the network address and broadcast address).

Step 1: Convert the mask to prefix notation

$\mathrm{255.255.255.192}$ in binary:

So the last octet has $2$ network bits. Total prefix length:

Step 2: Find the number of host bits

Step 3: Calculate usable hosts

Answer:

The number of usable host addresses in a /26 subnet is 62.

Quick Tip:

The octet value $192=128+64$, which corresponds to $11000000_2$ — 2 bits set. This is a quick way to read subnet masks without full binary conversion.

Understanding:

We need to find the broadcast address of the subnet $172.16.5.64/26$.

Formula:

The broadcast address is obtained by setting all host bits to $1$:

Step 1: Determine the block size

With a /26 mask, the host portion has $32-26=6$ bits.

Step 2: Identify the subnet range

The network address is $\mathrm{172.16.5.64}$. The next subnet starts at:

So this subnet spans $\mathrm{172.16.5.64}$ to $\mathrm{172.16.5.127}$.

Step 3: Identify the broadcast address

The broadcast address is the last address in the range:

Verification:

Setting all 6 host bits of $\mathrm{172.16.5.64}$ to 1:

Confirmed: $\mathrm{172.16.5.127}$.

Answer:

The broadcast address for the subnet $172.16.5.64/26$ is $\mathrm{172.16.5.127}$.

Quick Tip:

Broadcast = Network address + Block size $-1$. Here: $64+64-1=127$.

Understanding:

We must find the prefix length that gives at least $500$ usable hosts per subnet.

Formula:

where $h=32-\text{prefix length}$.

Step 1: Solve for minimum host bits

So the minimum value of $h=9$.

Step 2: Calculate the prefix length

Step 3: Verify

A /23 subnet has:

A /24 subnet has:

Answer:

The prefix length required to support at least 500 hosts per subnet is /23.

Quick Tip:

For host requirements, always find the smallest subnet (largest prefix) that still accommodates the needed count. /24 is the most common wrong answer here because it is the most familiar mask.

Understanding:

We must determine which of the given addresses falls within the subnet $192.168.1.128/25$.

Formula:

A /25 subnet has a block size of $2^{32-25}=2^7=128$.

The range of a subnet starting at network address $N$ is:

Step 1: Determine the subnet range

The network address is $\mathrm{192.168.1.128}$.

So the subnet range is $\mathrm{192.168.1.128}$ to $\mathrm{192.168.1.255}$.

Step 2: Test each option against the range $[128,255]$

Step 3: Confirm the valid host

$\mathrm{192.168.1.200}$ is a usable host address within $192.168.1.128/25$.

Answer:

The address that belongs to the subnet $192.168.1.128/25$ is $\mathrm{192.168.1.200}$.

Quick Tip:

For a /25, the two subnets within any /24 are $.0$ to $.127$ and $.128$ to $.255$. Memorising this split saves time in exams.

Understanding:

We need a subnet mask that simultaneously satisfies:

Formula:

where $n$ = subnet bits borrowed, $h$ = host bits remaining, and $n+h=8$ (since we start with a /24).

Step 1: Find required subnet bits

Step 2: Find required host bits

Step 3: Check feasibility with $n+h=8$

With $n=2$ and $h=6$: $2+6=8$ ✓

Both conditions are satisfied.

Step 4: Calculate the subnet mask

Prefix length $=24+2=26$.

The last octet of the mask has $2$ bits set:

Subnet mask $=\mathrm{255.255.255.192}$.

Verification:

With /27 ($n=3,h=5$): usable hosts $=2^5-2=30<50$ — insufficient.

Answer:

The most efficient subnet mask is $\mathrm{255.255.255.192}$, providing 4 subnets with 62 usable hosts each.

Quick Tip:

Always verify BOTH constraints — subnets and hosts. A /27 gives more subnets but fails the host requirement, making it a common trap.

Understanding:

We start with a /16 network and apply a /20 mask.

Formula:

Step 1: Calculate the number of subnets

Step 2: Calculate usable hosts per subnet

Verification:

Total addresses: $2^{16}=65536$. Each /20 subnet holds $2^{12}=4096$ addresses. $16\times 4096=65536$ ✓

Answer:

Subnetting $10.10.0.0/16$ with a /20 mask yields 16 subnets with 4094 usable hosts each.

Quick Tip:

The number of borrowed bits is always (new prefix $-$ original prefix). Borrowing 4 bits from a /16 gives $2^4=16$ subnets, each with $2^{12}-2=4094$ hosts.

Understanding:

We must identify the class and the classful default subnet mask of the IP address $\mathrm{192.168.1.1}$.

Formula:

Classful IP address ranges:

Step 1: Identify the class

The first octet of $\mathrm{192.168.1.1}$ is $192$.

Step 2: Apply the default mask

Class C networks have a 24-bit network portion:

Step 3: Confirm it is a private address

$192.168.0.0/16$ is the private address space for Class C, confirming the classification.

Answer:

The IP address $\mathrm{192.168.1.1}$ is a Class C address with default subnet mask $\mathrm{255.255.255.0}$.

Quick Tip:

The leading bits determine the class: Class A starts with $0$, Class B with $10$, Class C with $110$. Since $192=11000000_2$, the leading bits are $110$ — confirming Class C.

Understanding:

We need to find the starting network address of the subnet immediately after $192.168.20.0/26$.

Formula:

In VLSM, the next subnet begins immediately after the broadcast address of the current subnet:

where:

Step 1: Calculate the block size of the /26 subnet

Step 2: Find the broadcast address of the first subnet

Step 3: Find the next network address

Verification:

The /26 subnet $192.168.20.0/26$ spans addresses $.0$ to $.63$. The address $.64$ is the first address of the next available block, making it a valid network address for the next subnet.

Answer:

The next available subnet immediately after $192.168.20.0/26$ starts at $\mathrm{192.168.20.64}$.

Quick Tip:

In VLSM, always add the block size to the current network address to get the next subnet's network address. The block size for /26 is 64, so subnets start at $.0$, $.64$, $.128$, $.192$.