GATE | GATE-CS-2015 (Set 3) | Question 56

Consider B+ tree in which the search key is 12 bytes long, block size is 1024 bytes, record pointer is 10 bytes long and block pointer is 8 bytes long. The maximum number of keys that can be accommodated in each non-leaf node of the tree is
(A) 49
(B) 50
(C) 51
(D) 52

Answer: (B)


Let m be the order of B+ tree

m(8)+(m-1)12 <= 1024  
[Note that record pointer is not needed in non-leaf nodes]

m <= 51

Since maximum order is 51, maximum number of keys is 50. 

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