Data structure subjective question for competitive exam(3)

Q1.Given a base address of an array B[1400...1800] as 1020 and size of each element is 2 bytes in the memory. Find the address of B[1700].

base address=1020
LB=1400
UB=1800
W=2
I=1700
address of A[I]=B+W*(I-LB)
address of A[1700]=1020+2*(1700-1400)
A[1700]=1020+600=1620

Q2.Calculate the address of x[4,3] in two dimensional array x[1....5,1....4] stored in row major order in the main memory.Base address to be 1000 and that each requires 4 words of storage.

x[4][3]=B+W(C(I-1)+(J-1))
x[4][3]=1000+4(4(4-1)+(3-1))
x[4][3]=1000+4(12+2))
x[4][3]=1000+56=1056                

Q3.An array x[-15....10,15.....40]requires one byte of storage.If begining location is 1500 determine the location of x[15][20] using row major order.

number of column=40-15+1=26
x[15][20]=1500+1((15-(-15))*26+(20-15))
x[15][20]=2285

Q4.An array x[-15....10,15.....40]requires one byte of storage.If begining location is 1500 determine the location of x[15][20] using column major order.

number of rows=10-(-15)+1=26
x[15][20]=1500+1((15-(-15))+*26(20-15))
x[15][20]=1500+1(30+130)
x[15][20]=1500+160
x[15][20]=1660 

Q5.Each element of an array arr[15][20] requires W bytes of storage.If the address of arr[6][8] is 4440 and the base address at arr[1][1] is 4000,find the width W of each cell in the array arr[][] when the array is stored as column major wise.

4440=[4000+w((6-1)+15(8-1))
4440=4000+w[5+105]
4440-4000=110w
w=4400/110
w=4