Networking - MCQ Practice Questions

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 $.0$, $.64$, $.128$, $.192$.