Latest TCS Aptitude Question SOLUTION: A Two-dimensional array X(7,9) is stored linearly column-wise in a computer's memory. Each element requires 8 bytes for storage of the value. If the first byte address of X(1,1) is

• Placement Papers
• /
• TCS
• /
• Programming
• /
• Arrays

TCS Question Papers: TCS Interview Experience(90) TCS Company Profile

(#M40165507) TCS Programming question TCS Logical Keep an EYE

A Two-dimensional array X(7,9) is stored linearly column-wise in a computer's memory. Each element requires 8 bytes for storage of the value. If the first byte address of X(1,1) is 3000, what would be the last byte address of X(2,3)?
(a) 3247 (b) 3567 (c) 3088 (d) 3225 (e) 3765

Asked In TCS Mehul KOTHARI (5 years ago)

Unsolved Read Solution (9)

Is this Puzzle helpful?   (52)   (12) Submit Your Solution Arrays

Read Solution (9) :

venkateswarlu reddy khambam  (5 years ago)

formula :
Baseaddress+byte required {N(i-1)+(j-1)}
3000+8{9(2-1)+(3-1)}
3000+88
3088

3088 is the answer

i represents the row
j represents the column
N means column size total 9 here

Is this solution Helpfull? Yes (22) | No   

Mohammad Sarfaraz Ali  (4 years ago)

3088
if 1,1=3000
1,3=3016
1,9=3064
2,1=3072
2,3=3088.
Simple

Is this solution Helpfull? Yes (9) | No   

Sureddy veera babu  (5 years ago)

C by the address of each arrays it is 3088

Is this solution Helpfull? Yes (1) | No (3)  

Ankush batra  (5 years ago)

ans: c
1 -> 1-9 -> 3000+(8*7)= 3056
2 -> 0-3 -> 3056+(8*4)= 3088

Is this solution Helpfull? Yes (1) | No (2)  

Ravi Ranjan  (4 years ago)

each element takes 8 bytes , and x(1,1) is 3000 according to this we add 8 bytes after each elements. in first row total 9 elements so x(1,9) becomes 3064 now we enter in 2nd row ,we have to find x(2,3) so in 2nd row we add 3 elements & ac to this 24 also added in address .so x(2,3) becomes 3088

Is this solution Helpfull? Yes (1) | No (1)  

D.V.R.S.Akhil  (5 years ago)

C by the address of each arrays it is 3088

Is this solution Helpfull? Yes | No (1)  

Akanksha Belsare  (4 years ago)

Formula:
Base address+byte required{N(i-1)+(j-1)}
Here Base addres is 3000.
Byte required is 4.
N=Number of columns in an array.
i=row of the array element of the address to be found.
j=column of the array element of the address to be found.
Solution;
3000+4{7(7)+4}=3212.

Is this solution Helpfull? Yes | No (2)  

Manisha Nandi  (4 years ago)

formula
baseaddress+{n(i-1)+(j-1)}
n=9, i=2, j=3
ans=3088

Is this solution Helpfull? Yes | No   

Shubham Tariyal  (4 years ago)

3088 x(I, j) =b+w(m(i-l1)+(j-l2))

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.