NIMCET Computer Architecture Previous Year Questions (PYQs) – Page 2 of 2
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Showing 16 questions
FFFF will be the last memory location in a memory of size
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Solution
✅ Finding the Memory Size:
The last memory location "FFFF" is a hexadecimal number.
Each hexadecimal digit represents 4 bits, so:
- "FFFF" is a 4-digit hexadecimal number, meaning it represents 16 bits or 2 bytes.
- "FFFF" in decimal is equivalent to 65535.
- Therefore, the total memory size is 65536 bytes or 64 KB.
✅ Final Answer:
The memory size is: 64 KB
Which is a stored-program machine?
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Solution
Stored-program concept = microprocessor
Access time = $45\text{ ns}$, gap = $5\text{ ns}$
Bandwidth = ?
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Solution
Cycle time = $45 + 5 = 50\text{ ns}$
Bandwidth = $\frac{1}{50\times10^{-9}} = 20\text{ MHz}$
CPU has 12-bit address bus.
Total memory = 16 KB.
Word length = ?
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Solution
$2^{12}=4096$ locations
Word size = $16384/4096 = 4$ bytes
Let $A_k$ be the arithmetic mean of squares of $k$ natural numbers. If $\sum_{k=1}^{n}(6A_k-3k)=31$, find the value of $n$.
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Solution
$A_k=\frac{1^2+2^2+3^2+\cdots+k^2}{k}$
$A_k=\frac{\frac{k(k+1)(2k+1)}{6}}{k}$
$A_k=\frac{(k+1)(2k+1)}{6}$
Now,
$6A_k-3k=(k+1)(2k+1)-3k$
$=2k^2+3k+1-3k$
$=2k^2+1$
Given,
$\sum_{k=1}^{n}(6A_k-3k)=31$
$\sum_{k=1}^{n}(2k^2+1)=31$
For $n=3$,
$(2\cdot1^2+1)+(2\cdot2^2+1)+(2\cdot3^2+1)$
$=3+9+19=31$
Therefore,
$n=3$
Which one of the following is correct order of memory types in increasing access speed, from slowest to fastest?
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Solution
Access speed increases as we move closer to the CPU.
Hard Drive is the slowest among the given options.
RAM is faster than Hard Drive.
Cache memory is faster than RAM.
CPU Registers are the fastest because they are inside the CPU.
So, the correct order from slowest to fastest is:
Hard Drive $\to$ RAM $\to$ Cache $\to$ CPU Registers
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