GATE 2026 EC Q.42: solution and answer

GATE 2026 · ECQ.42 · 2 marks · MCQ

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.

(A)
$(2 F F F)_{H}$
(B)
$(124 F F)_{H}$
(C)
$(424 F F)_{H}$
(D)
$(324 F F)_{H}$

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Answer and solution

ANSWER

The last location of the 256 KB memory is at $(424 F F)_{H}$ . The answer is option (C).

Official answer: CSource: Official answer key, GATE 2026 EC (IIT Guwahati)

Take the memory to be byte-addressable, so each address holds one byte. The number of locations is $256 KB = 2^{8} \times 2^{10} = 2^{18}$ .
Write the size in hexadecimal. Since $2^{16} = (10000)_{H}$ and $2^{18} = 4 \times 2^{16}$ , we get $2^{18} = (40000)_{H}$ .
A memory of $N$ locations that starts at address $A$ ends at $A + N - 1$ . The offset of the last location from the first is $N - 1 = (40000)_{H} - 1 = (3 F F F F)_{H}$ .
Add this offset to the start address in hexadecimal. Write $(3 F F F F)_{H} = (40000)_{H} - 1$ : $(2500)_{H} + (3 F F F F)_{H} = (2500)_{H} + (40000)_{H} - 1 = (42500)_{H} - 1 = (424 F F)_{H}$
Check: $(424 F F)_{H} - (2500)_{H} + 1 = (3 F F F F)_{H} + 1 = (40000)_{H} = 2^{18}$ bytes $= 256 KB$ . This is option (C).

KEY IDEA

A byte-addressable block of $N$ locations that starts at address $A$ occupies $A$ to $A + N - 1$ . Here $256 KB = 2^{18} = (40000)_{H}$ , so the last address is $(2500)_{H} + (3 F F F F)_{H} = (424 F F)_{H}$ .

WATCH OUT FOR

One mistake is to add the full size, $(2500)_{H} + (40000)_{H} = (42500)_{H}$ . That is the first address after the memory, so you still have to subtract 1. The other is to write $256 KB$ as $(10000)_{H}$ or $(30000)_{H}$ instead of $(40000)_{H}$ ( $2^{16}$ is only 64 KB). That mistake leads to distractors such as (B) or (D).

From the GATE 2026 EC paper. Every question of this paper · Download the full paper · All 9 Processors & memory questions, with notes

GATE 2026 EC Q.42: solution and answer

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