GATE CSE 2007 | GATE CSE Year Wise Previous Years Questions

GATE CSE 2007

MCQ (Single Correct Answer)

-0.6

Consider the B⁺ tree in the adjoining figure, where each node has at most two keys and three links. GATE CSE 2007 Database Management System - File Structures and Indexing Question 16 English 1

GATE CSE 2007 Database Management System - File Structures and Indexing Question 16 English 1

Keys K 15 and then K 25 are inserted into this tree in that order. Exactly how many of the following nodes (disregarding the links) will be present in the tree after the two insertions?

GATE CSE 2007

MCQ (Single Correct Answer)

-0.6

The order of a leaf node in a B⁺- tree is the maximum number of (value, data record pointer) pairs it can hold. Given that the block size is 1K bytes, data record pointer is 7 bytes long, the value field is 9 bytes long and a block pointer is 6 bytes long, what is the order of the leaf node?

GATE CSE 2007

MCQ (Single Correct Answer)

-0.6

Consider the following schedules involving two transactions. Which one of the following statements is TRUE?

S₁: r₁(X); r₁(Y); r₂(Y); r₂(X); w₂(Y); w₁(X);

S₂: r₁(X); r₂(X); r₂(Y); w₂(Y); r₁(Y); w₁(X);

Both S₁ and S₂ are conflict serializable.

S₁ is conflict serializable and S₂ is not conflict serializable.

S₁ is not conflict serializable and S₂ is conflict serializable.

Both S₁ and S₂ are not conflict serializable.

GATE CSE 2007

MCQ (Single Correct Answer)

-0.6

Consider the following two transactions: T₁ and T₂.

T₁: read (A);                  T₂: read (B);
    read (B);                      read (A);
    if A = 0 then B ← B + 1;       if B ≠ 0 then A ← A - 1;
    write (B);                     write (A);

Which of the following schemes, using shared and exclusive locks, satisfy the requirements for strict two phase locking for the above transactions?

S1 : lockS(A); 	        S2 : lockS(B);
     read (A);               read (B);
     lockS(B);               lockS(A);
     read (B);               read (A);
     if A = 0                if B ≠ 0
     then B ← B + 1;        then A ← A - 1;
     write (B);              write (A);
     commit;                 commit;
     unlock (A);             unlock (B);
     unlock (B);             unlock (A);

S1 : lockX(A);          S2 : lockX(B);
     read (A);               read (B);
     lockX(B);               lockX(A);
     read (B);               read (A);
     if A = 0                if B ≠ 0
     then B ← B + 1;        then A ← A - 1;
     write (B);              write (A);
     unlock (A);             unlock (A);
     commit;                 commit;
     unlock (B);             unlock (A);

S1 : lockS(A);           S2 : lockS(B);
     read (A);                read (B);
     lockX(B);                lockX(A);
     read (B);                read (A);
     if A = 0                 if B ≠ 0
     then B ← B + 1;         then A ← A - 1;
     write (B);               write (A);
     unlock (A);              unlock (B);
     commit;                  commit;
     unlock (B);              unlock (A);

S1 : lockS(A);           S2 : lockS(B);
     read (A);                read (B);
     lockX(B);                lockX(A);
     read (B);                read (A);
     if A = 0                 if B ≠ 0
    then B ← B + 1;        then A ← A - 1;
     write (B);               write (A);                                                                                                     
     unlock (A);              unlock (A);
     unlock (B);              unlock (B);
     commit;                  commit;

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