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Each cell needs 4 words

- A[0][0], A[0] [1],.....A[0][9] needs 4x10 =40 words.
- Since array starts at 100, A[0][9] will be at 100+40=140 th location.
- Similarly, from A[0][0] to A[10][9] there are 11x10=110 elements.
- From A[11][0] to A[11][5]there are 5 elements.
- Total 115 elements are present before A[11][5].
- Hence A[11][5] will be at 100+ 4x115 = 560

Answer is **A**

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A[20][10]

to reach A[11][5]

we have to cross 11 rows and each row contains 10(no. of columns) elements and in 12^{th} row we have to cross 5 elements

total elements to cross= $11*10+5=115$

each element occupies 4 words space

total space = $4*115=460$

address of $A[11][5] = base address + space occupied by elements before it$

$=100+460=560$

option (A)