ISRO CSE 2014 PYQ

The fourth-smallest key in a binary search tree is found by an in-order traversal (left subtree, node, right subtree), which visits keys in ascending order. Count the fourth visited node. The supplied answer V depends on the omitted BST diagram and cannot be independently verified without that image; retain V provisionally.

A candidate key is a minimal superkey: it uniquely identifies a tuple, and no proper subset does. Among the listed superkeys, {x} is already minimal, so {x,y} and {x,y,z} are not minimal. Likewise {a,b} is minimal, while {a,b,c} is not. Therefore the candidate keys are {x} and {a,b}.

After pushing A,B,C,D,E, the top is E. Four pops enqueue E,D,C,B in that order; A remains in the stack. Deleting two queue items removes E then D. Pushing those deleted items onto the stack in that order makes D the top (A, then E, then D). The next pop therefore returns D.

Apply LRU with four initially empty frames. References 0, 2, 4, and 5 cause four faults. 2 and 4 are hits. Reference 3 faults and replaces least-recently-used page 0. Reference 11 faults and replaces page 5. Reference 2 is a hit. Reference 10 faults and replaces page 4. Faults occur for 0, 2, 4, 5, 3, 11, and 10: seven faults total.

Frame transmission time is L/R = 1000 bits ÷ 50,000 bits/$s = 0.02 s = 20 ms$. The given 500 ms is round-trip propagation delay, so one-way propagation is 250 ms (assuming a symmetric path). The receiver gets the final bit after transmission plus one-way propagation: 20+250 = 270 ms. This excludes processing and queuing delays.