Question:
The address of the first location of a 256 kilo byte (KB) memory is \((2500)_H\).
Choose the correct address of the last location of the memory.
The address of the first location of a 256 kilo byte (KB) memory is \((2500)_H\).
Choose the correct address of the last location of the memory.
Choose the correct address of the last location of the memory.
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Last address = first address + total bytes - 1, all worked out in hexadecimal.
Updated On: Jul 20, 2026
- \((2FFF)_H\)
- \((124FF)_H\)
- \((424FF)_H\)
- \((324FF)_H\)
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Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Convert the memory size into bytes.
A memory of $256$ KB has \[ 256\times1024=262144 \] bytes, since $1$ KB $=1024$ bytes.
Step 2: Write this size in hexadecimal.
\(262144=4\times65536=4\times16^4\), so in hex the size is \((40000)_H\).
Step 3: Recall how the last address is found.
If the first address is $A_1$ and the memory holds $S$ bytes, the last address is \[ A_2=A_1+S-1 \] because the first byte itself already occupies one of the $S$ locations.
Step 4: Add the starting address and the size in hex.
\[ (2500)_H+(40000)_H=(42500)_H \]
Step 5: Subtract 1 to get the last address.
\[ (42500)_H-1=(424FF)_H \]
Step 6: Check the other options.
$(2FFF)_H$ treats the memory as far too small, close to $4$KB only. $(124FF)_H$ and $(324FF)_H$ come from an error in converting $256$KB to hex, for example using $(20000)_H$ or $(30000)_H$ in place of the correct $(40000)_H$.
Step 7: Final conclusion.
The last address of the memory is \[ \boxed{(424FF)_H} \]
A memory of $256$ KB has \[ 256\times1024=262144 \] bytes, since $1$ KB $=1024$ bytes.
Step 2: Write this size in hexadecimal.
\(262144=4\times65536=4\times16^4\), so in hex the size is \((40000)_H\).
Step 3: Recall how the last address is found.
If the first address is $A_1$ and the memory holds $S$ bytes, the last address is \[ A_2=A_1+S-1 \] because the first byte itself already occupies one of the $S$ locations.
Step 4: Add the starting address and the size in hex.
\[ (2500)_H+(40000)_H=(42500)_H \]
Step 5: Subtract 1 to get the last address.
\[ (42500)_H-1=(424FF)_H \]
Step 6: Check the other options.
$(2FFF)_H$ treats the memory as far too small, close to $4$KB only. $(124FF)_H$ and $(324FF)_H$ come from an error in converting $256$KB to hex, for example using $(20000)_H$ or $(30000)_H$ in place of the correct $(40000)_H$.
Step 7: Final conclusion.
The last address of the memory is \[ \boxed{(424FF)_H} \]
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