Q1. Paper 3, j15. For the 8 bit word 00111001.the check bits stored would be 0111. When word is read from memory, the check bits are calculated to be 1101.what is the data word that was read from memory?

       1.  10011001.

2.    00011001. 
3.   00111000.
4. 11000110.

Answer 2.

Explanation .
Given 8 bit word 00111001. encoded in d8=0 d7=0 d6=1 d5=1 d4=1 d3=0 d2=0 d1=1.
parity is calculated as 01111  which are written as p8p4p2p1.
p1=1
p2=1
p4=1
p8=0.

encode the string as 
d8d7d6d5p8d4d3d2p4d1p2p1.
001101001111.
check bit at other end is 1101.
c1=1
c2=0
c4=1
c8=1. 
xor(1101,0111)=1010 . its decimal conversion is 10. so 10th bit is changed.
 10th bit is d6.  so we flip it. and we get
00011001