Test : CSE Past Year Paper 2018 - GATE MCQ

The series starts with 2. We multiply first term with 6 to get second term 12 Then we multiply second term with 5 to get third term 60 Then we multiply third term with 4 to get fourth term 340 Then we multiply fourth term with 3 to get fifth term 720 Then we multiply fifth term with 2 to get fifth term 1440. Then we need to multiply with 1 and we get 1440 again.

We need a number that can be written as 20x + 7, 42y + 7, 76z + 7 for three integers x, y and z. We basically need to find least common multiple of 20 (2 * 2 *5), 42(2 * 3 * 7) and 76(2 * 2 * 19) which is 2 * 2 * 5 * 3 * 7 * 19 = 20 * 21 * 19 = 420 * 19 = 7980 So our number is 7980 + 7 = 7987. 
Note - You can also verify options using virtual calculator.

"From where are they bringing their books? Who bringing Whose books from Where." Who - A group as subject = They are Whose - of that group = their Where - from a place = there "From where are they bringing their books? They are bringing their books from there." So, option (B) is correct choice.

Meandering (as adjective) = proceeding in a convoluted or undirected fashion. Timely (as adjective) = done or occurring at a favourable or useful time; opportune. Consistent (as adjective) = acting or done in the same way over time, especially so as to be fair or accurate. Systematic (as adjective) = done or acting according to a fixed plan or system; methodical. Therefore, meandering is correct choice, all other options are similar to the word organized.

One important observation to solve the question :
Diagonal of Square = Diameter of Circle.
Let side of square be x.
From Pythogorous theorem. Diagonal = √(2*x*x)
We know area of square = x * x = d
Diameter = Diagonal = √(2*d)
Radius = √(d/2)
Area of Circle = π * √(d/2) * √(d/2) = 1/2 * π * d

Let total males and females be 60x and 40x respectively.
Total number of people = (60x + 40x)
Total number of people who attended : 0.8(60x + 40x) = 80x
Let y males attended. It is given all 1females attended
40x + y = 80x
y = 40x which is same as females.
Alternative Approach - Lets total number of people = 100. Therefore, 60 are male and and 40 are female. But total 80 guests are attended and all 40 female attended the party. So, there remaining (80 - 40 = 40) attendees should be male. Then the ration of male to female among attendees is 40 : 40 = 1 : 1

Considering uniformly distributed outcomes, we get 4 greens and 2 reds in six throws. Then in one more throw, green is more likely. So, option (C) is correct.

Taking logs of given three values, we get
1/q = p^-x -------(1)
1/r = q^-y -------(2)
1/p = r^-z -------(3)
1/q = p^-x
= r^-xz [Putting value of p from (3)]
= q^-xyz [Putting value of r from (2)]
= 1 / q^xyz
On comparing power of q both sides, we get xyz = 1
So, option (C) is correct.

Let there be x total men.
Total amount paid = x * 750 * 8/9 + (300 - x)*1000*2/3
= x*2000/3 + 300*1000*2/3 - x*2000/3
= 200000

For Enqueue operation, performs in constant amount of time (i.e., Θ(1)), because it modifies only two pointers, i.e.,
Create a Node P.
P-->Data = Data
P-->Next = Head
Head = P
For Dequeue operation, we need address of second last node of single linked list to make NULL of its next pointer. Since we can not access its previous node in singly linked list, so need to traverse entire linked list to get second last node of linked list, i.e.,
temp = head;
While( temp-Next-->Next != NULL){
temp = temp-Next;
}
temp-->next = NULL;
Tail = temp;
Since, we are traversing entire linked for each Dequeue, so time complexity will be Θ(n). Option (B) is correct.

Chromatic number of given graph is 3.

Note that graph is Planar so Chromatic number should be less than or equal to 4 and can not be less than 3 because of odd length cycle. In other words if a graph is planar and has odd length cycle then Chromatic number can be either 3 or 4 only.

Given, 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. But 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.

According to Lagrange's theorem, states that for any finite group G, the order (number of elements) of every subgroup H of G divides the order of G. Therefore, possible subgroups of group on 84 elements are : 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84. Subgroups with element 1 and 84 are trivial groups. The size of a largest possible proper subgroup of G is 42.

Given, average time for a memory access = M units if page hits, and average time for a memory access = D units if page fault occurred. And total/experimental average time taken for a memory access = X units. Let page fault rate is p. Therefore, Average memory access time = ( 1 - page fault rate) * memory access time when no page fault + Page fault rate * Memory access time when page fault → X = (1 - p)*M + p*D = M - M*p + p*D → X = M + p(D - M) → (X - M) = p(D - M) → p = (X - M) / (D - M) So, option (B) is correct.

Given a_n = 2n + 3 Generating function G(x) for the sequence a_n is G(x) =

Type checking is done at semantic analysis phase and parsing is done at syntax analysis phase. And we know Syntax analysis phase comes before semantic analysis. So Option (B) is False. All other options seems Correct.

Given, post-order - 8, 9, 6, 7, 4, 5, 2, 3, 1 and in-order - 8, 6, 9, 4, 7, 2, 5, 1, 3 Construct a binary tree from postorder and inorder traversal :

The height of the binary tree above is 4.

(A) (p⊕q)' = (pq' + p'q)' = (p'+q).(p+q') = (pp' +p'q' + qp + qq') = pq + p'q' = (p⊙q) (B) (p')⊕q = (p')q' + (p')'q = pq + p'q' = (p⊙q) (C) (p'⊕q') = (p')'.(q') + (p').(q')' = p.(q') + (p').q = p⊕q. (D) (p⊕(p'))⊕q =(pp''+ p'p')⊕q =(p+p')⊕q = 1⊕q = 1.q' + 0.q = q' And (p⊙(p'))⊙(q') = (pp' + p'.p'')⊙(q') = 0⊙(q') = 0(q') + 1.(q')' = q Please note that (p⊕q) = (p'⊙q) = (p⊙q') = (p'⊙q')' but not (p'⊙q'). Similarly, (p⊙q) = (p'⊕q) = (p⊕q') = (p'⊕q')' but not (p'⊕q'). So, option (D) is false.

Slow-start begins initially with a congestion window size (cwnd) of 1, 2, 4 or 10 MSS. The value of the Congestion Window will be increased by one with each acknowledgement (ACK) received, effectively doubling the window size each round-trip time (approximately exponential). The cwnd increase by 1 MSS on every successful acknowledgement. Therefore, only statement (iv) is correct. Option (C) is true.

Since given relation is many to one :

Therefore, no entity in E1 can be related to more than one entity in E2 and an entity in E2 can be related to more than one entity in E1. Only option (A) is correct.

The minimum number of state in DFA will be 1 when we have only starting state and all other states will not be reachable from stating state. The minimum number of state in DFA will be 2ⁿ as the number of subsets of a set with n elements in the worst case. Therefore, k ≤ 2ⁿ Option (D) is correct. Related question - GATE | Gate IT 2008 | Question 6

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.

Recursive Enumerable Language are closed under Union, Intersection, Concatenation and Kleene Closure (but not Complementation). So, we can easily rule out option (A) and option (B) comes to be TRUE. Recursive Language are subset of REL, but option (C) is saying opposite, so Option (C) is FALSE. REL are countable, since set of all Turing Machines are countable. So option (D) is also FALSE. Only Option (B) is CORRECT.

In SQL the FULL OUTER JOIN combines the results of both left and right outer joins and returns all (matched or unmatched) rows from the tables on both sides of the join clause. So, option (D) is correct. See : Join (Inner, Left, Right and Full Joins), Inner Join vs Outer Join

Given, Number of processes (P) = 3 Number of resources (R) = 4 Since deadlock-free condition is:
R ≥ P(N − 1) + 1

Where R is total number of resources, P is the number of processes, and N is the max need for each resource.
4 ≥ 3(N − 1) + 1
3 ≥ 3(N − 1)
1 ≥ (N − 1)
N ≤ 2

Therefore, the largest value of K that will always avoid deadlock is 2. Option (B) is correct.