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A block can hold either $12$ records or $42$ key pointers .A database contains $96$ records , then how many blocks are required to hold the data file and the dense Index?

a). $10$

b). $12$

c). $11$

d). $13$

No of blocks for record=$96/12=8$

No of block for index=$96/24=4$

So total no of blocks= $12$

Don't we have to consider one more block for indexing the 4 blocks?
how is answer calsulated? any formula for that? I am not getting it.

No of blocks for record = 96/12 = 8

For indexing = ceil(96/42 ) = 3

Hence,

So total no of blocks = 8+3 = 11

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Why aren't you considering one more block that holds the pointers for 3 index blocks?