Question

for an mbr - formatted drive that has two partitions, what is the starting sector byte position for the second partition?
a. offset 0x1c4
b. offset 0x1e4
c. offset 0x1d6
d. offset 0x1da

Explanation:

In an MBR (Master Boot Record), each partition entry occupies 16 bytes. The first partition entry starts at offset 0x1BE. The second partition entry begins right after the first, so we calculate its starting offset by adding 16 (0x10 in hexadecimal) to the first entry's offset: $0x1BE + 0x10 = 0x1CE$. Within each 16-byte partition entry, the starting sector (LBA) field is located at the 8th byte (offset +6 to +9 of the entry). Adding 6 to the second entry's starting offset gives $0x1CE + 0x6 = 0x1D4$? No, correction: standard MBR partition entry structure: each entry is 16 bytes. The first entry is 0x1BE-0x1CD, second is 0x1CE-0x1DD. The starting sector (absolute) field is at bytes 8-11 of the entry (offset 6 from the start of the entry). So 0x1CE + 0x6 = 0x1D4? No, wait, standard offset for the second partition's starting sector is 0x1E4? No, correction: Let's list the offsets properly:

1. First partition entry: starts at 0x1BE

• Starting sector (LBA) is at 0x1BE + 8 = 0x1C6? No, no: MBR partition entry fields:

1. Boot indicator (1 byte): 0x00/0x80
2. CHS start (3 bytes)
3. Partition type (1 byte)
4. CHS end (3 bytes)
5. LBA start (4 bytes): this is the starting sector position, located at offset 8 (1+3+1+3=8 bytes into the entry)
6. LBA size (4 bytes)

So first entry's LBA start: 0x1BE + 8 = 0x1C6
Second entry starts at 0x1BE + 16 = 0x1CE, so its LBA start is 0x1CE + 8 = 0x1D6? No, wait 0x1CE + 8 = 0x1D6? 0x1CE + 8 = 0x1D6 (since E+8=16, carry over 1, C+1=D, so 0x1D6). Wait no, 0x1CE is 462 in decimal, 462+8=470, 470 in hex is 0x1D6. Yes. But wait, some sources say the second partition's starting sector offset is 0x1E4? No, no, let's confirm: each partition entry is 16 bytes. First entry: 0x1BE-0x1CD (16 bytes: 0x1BE to 0x1CD inclusive is 16 bytes). Second entry: 0x1CE-0x1DD. The LBA start is the 9th byte of the entry (index 8 if starting at 0). So 0x1CE + 8 = 0x1D6. Yes, that is option c? Wait no, wait 0x1CE + 8: 0x1CE is 1256 + 1216 + 14 = 256 + 192 +14=462. 462+8=470. 470 divided by 16 is 29*16=464, 470-464=6, so 29 is 0x1D, so 0x1D6. Yes.

Wait, no, another way: the first partition's starting sector (LBA) is at 0x1C6, the second is 16 bytes later in the entry, so 0x1C6 + 16 = 0x1D6. Yes, that's correct.

Answer:

c. Offset 0x1D6