PREVIOUS YEAR SOLVED PAPERS MCQ - GATE 2018

46 The postorder traversal of a binary tree is 8, 9, 6, 7, 4, 5, 2, 3,1. The inorder traversal of the same tree is 8, 6, 9, 4, 7, 2, 5, 1, 3. The height of a tree is the length of longest path from the root to any leaf. The height of the binary tree above is __________.

Note – Numerical Type question

Given, post-order – 8, 9, 6, 7, 4, 5, 2, 3, 1 and in-order – 8, 6, 9, 4, 7, 2, 5, 1, 3
So the height of tree is 4.

47 Two people, P and Q, decide to independently roll two identical dice, each with 6 faces. Numbered 1 to The person with the lower number wins. In case of a tie, they roll the dice repeatedly until there is no tie. Define a trial as a throw of the dice by P and Q. Assume that all 6 numbers on each dice are equiprobable and that all trials are independent. The probability (rounded to 3 decimal places) that one of them wins in the third trial is ___________.

Note – Numerical Type question

Given there are two identical dices, each having 6 faces.
Favorable events for tie = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
Since two dice are thrown, total sample space will be 6 x 6 = 36

Therefore, P(tie) = 6/36 = 1/6
and Probability of not tie = (1 – 1/6)
To one of them win on third trial, previous two trials should be tie.

= 1/6 * 1/6 * (1 – 1/6)
= 1/36 * 5/6
= 5/216 = 0.023
So, option (C) is correct.

48 The value of

correct to three decimal places (assuming that π = 3.14) is __________.

Note – Numerical Type question

put x² = t
2xdx = dt or x dx = 1/2 dt
if x = 0 then t = 0 and if x = π/4 then t = (π/4)²
Gate2018

49 The chromatic number of the following graph is __________.

Note – Numerical Type question

Sub graph is K₃ chromatic number is at least 3. We can try for a chromatic number of 3 by using 3 colors, as follows :
Gate2018 cs

All vertices with only 3 colors, the chromatic number of this graph is 3.

50 A 32-bit wide main memory unit with a capacity of 1 GB is built using 256M x 4-bit DRAM chips. The number of rows of memory cells in the DRAM chip is 2¹⁴. The time taken to perform one refresh operation is 50 nanoseconds. The refresh period is 2 milliseconds. The percentage (rounded to the closest integer) of the time available for performing the memory read/write operations in the main memory unit is ____________.

Note – Numerical Type question

Total number of rows is 2¹⁴ and time taken to perform one refresh operation is 50 nanoseconds. So, total time taken to perform refresh operation = 2¹⁴*50 nanoseconds = 819200 nanoseconds = 0.819200 milliseconds.
Refresh period is 2 milliseconds. So, time spent in refresh period in percentage = (0.819200 milliseconds) / (2 milliseconds) = 0.4096 = 40.96%
Hence, time spent in read/write operation = 100% – 40.96% = 59.04% = 59 (in percentage and rounded to the closet integer).
So, answer is 59.