It is simple hexa-decimal addition.
8+F = 17
Explanation:
8+15 = 23 (which is 17 in hexadecimal)
Further explanation:
8+15 = 23
23-16 = 7 (here 16 is hexadecimal base) 1 is carried
so it becomes 17

I hope now you can understand why 8+F = 7 written in the answer above.

Sorry for the little mistake, I mistakenly wrote 01FF as 011F. Please check now.
This calculation can be expressed as: Physical address = SS (hex) x 16 + SP (hex).
SS (7698)h x 16 results as 76980
same in the second question.

Answer 1.
The value of the stack segment register (SS) = 7698h
The value of the stack pointer (SP) which is Offset = 01FFh
Thus, Physical address of the top of the stack is:

7

6

9

8

0

+

0

1

F

F

7

6

B

7

F

This calculation can be expressed as: Physical address = SS (hex) x 16 + SP (hex). Answer 2.
The value of the code segment register (CS) = 5526h
The value of the Instruction Pointer, holding address of the instruction = 8874h
Physical address of the instruction
Physical Address = CS (Hex) x 16 + IP

## aksit

i am also finding the ans of this qsn how you wrote 8+F=7 plz tell

## admin

It is simple hexa-decimal addition.

8+F = 17

Explanation:

8+15 = 23

(which is 17 in hexadecimal)Further explanation:

8+15 = 23

23-16 = 7

(here 16 is hexadecimal base)1 is carriedso it becomes 17

I hope now you can understand why 8+F = 7 written in the answer above.

## Vandana

SS=76980

SP=01FF

01FF kaise 011F ho gaya

9+1=10=A hota hai but apne B kyu likha

Similarly 8+1=9 =9 he hona chahey

Plz admin explain ?

## admin

Sorry for the little mistake, I mistakenly wrote 01FF as 011F. Please check now.

This calculation can be expressed as: Physical address = SS (hex) x 16 + SP (hex).

SS (7698)h x 16 results as 76980

same in the second question.

## Vandana

Thanks

## admin

This answer was edited.Answer 1.The value of the stack segment register (SS) = 7698h

The value of the stack pointer (SP) which is Offset = 01FFh

Thus, Physical address of the top of the stack is:

This calculation can be expressed as: Physical address = SS (hex) x 16 + SP (hex).

Answer 2.The value of the code segment register (CS) = 5526h

The value of the Instruction Pointer, holding address of the instruction = 8874h

Physical address of the instruction

Physical Address = CS (Hex) x 16 + IP