Previous year gate question modification(disks)

Question

Consider a hard disk with 16 recording surfaces (1-16) having 16384 cylinders (1-16384) and each cylinder contains 64 sectors (1-64). Data storage capacity in each sector is 512 bytes. Data are organized cylinder-wise and the addressing format is <cylinder no., surface no., sector no.> . A file of size 42797 KB is stored in the disk and the starting disk location of the file is <1200, 9, 40>. What is the cylinder number of the last sector of the file, if it is stored in a contiguous manner?

(A) 1281 (B) 1282 (C) 1283 (D) 1284

This is same as previous year gate question except for change in indexing. It starts from 1-16384 and so on..

Please the tell me the <cylinder no,surfaceno, sector no> of last sector in this case.

Thanks.

Comments

Please someone answer.

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Someone please give a proper answer.

Answer 1

if it starts from Cylinder $1$ ,then Address of the last sector will be $\frac{42797 \times 1024}{512}=85594$

$\text{I can write addressing equation as}$

$x \times 16 \times 64+16 \times 16+64=85594$

$\Rightarrow x \times 16 \times 64=85594-64-256$

$\Rightarrow x=\frac{85594-64-256}{16 \times 64}=83.27\approx 83$

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Final ADDRESS will be it will $< 83,9,26>$

Comments

please clearify your question .you are confusing me 

You have wriiten complete question of https://gateoverflow.in/1540/gate2013_29  and then at the last 

This is same as previous year gate question except for change in indexing. It starts from 1-16384 and so on..

I only got to know from your question that what will be What is the cylinder number of the last sector of the file, if it is stored in a contiguous manner and starts from cylinder $1$.

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But the starting location is still <1200,9,40>  and not <0,0,0> . You are getting confused in the question. I changed only the way cylinders are numbered, that's it.

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i think it is 1284....

1283.12  means here we will take some surfaces of 1284th cylinder...