ANS(1) array is given as a=$\begin{bmatrix} 1& 2 & 3 & 4 &5 & 6 & 7 &8 &9 \\ 10& 11& 12 & ..& ..& &.. & &18 \\ 19 &20 &21 & .. &. & ..& &. . &27 \\ .. &.. & & .. & & ...&... &.. & ..\\ & & & & & & & & \\ & & & & & & & & \\ & & & & & & & & \\ & & & & & & & & \\ ..& ... ..&. .. ..& &.. & .. .& ..& .. & \\ 82 & 83 & 84 &85 &86 &87 &88 &89 &90 \end{bmatrix}$
now the address 67 first find 67-29=38
word size =2
now assume you have to find out address of element 10 , it will be 29 +2=31 as array is stored in column major order
similarly , for address 67 we can find number of elements between these addresses i.e. 67-29=38
#elements=38/2=19 it will be a[10][2] since a[1][2]=2 a[2][2]=11 similarly (like A.P.) a[10][2] =2+(10-1)9=83
ANS(2) I think option A is correct
using column major order 29+[(7-1)*10+6]*2=161
