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Computer Science And IT (CS/IT) Mock Test - 8 For Gate


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65 Questions MCQ Test | Computer Science And IT (CS/IT) Mock Test - 8 For Gate

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Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 1

Suppose there are two teams A and B. Both of them are competing with each other on a racing track of 1 km. But team A has got the advantage of starting the race from 280 m from the starting point. If the ratio of speed of A to B is 3:4. 

Select the correct option from the options given below:

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 1

To win the race first,
A has to cover (1000 – 280) m = 720m
B covers 4 part of distance when A covers 3 part of it.
So, if A covers 720m, then B covers (4/3 * 720)m i.e. 960m
Hence, A will win the race by 40m.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 2

There are 5 brothers in a family. All were born at a gap of 3 years. If the sum total of ages of 5 brothers is 100. What is the age of 2nd most elder brother ?

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 2

Let the age of youngest brother be x
Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 100
=> 5x = 70 => x = 14
Therefore, age of 2nd most elder brother = 14 + 9 = 23.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 3

If there are 5 persons who can work individually and complete a work in 2, 4, 5, 6 and 8 hours respectively. If they work together then how much time will it take to finish the work ?

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 3

In 1 hour, ½ + ¼ + ⅕ +⅙ + ⅛ = 149/120 = 1.24 is the work done.
Hence, total time to complete the work would be = 1/1.24 = 0.80 hours

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 4

If we increase the sides of a rectangular park by 20% , then, what is the total increase in the area of that park?

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 4

Let length of park = L and breadth of park = B
Original Area = BL m2
New Length = 120L/100 = 6L/5
New Breadth = 120B/100 = 6B/5
New Area = 36LB/25 m2
so, Change in area = New – Original = 11LB/25 m2 % increase = Change / original * 100 = 44%

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 5

The sum of the digits in the unit place of all the 4 digit numbers formed with the help of 3, 4, 5 and 6 taken all at a time is:

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 5

Required sum = 3!(3 + 4 + 5 + 6) = 108
[If we fix 3 of the unit place, other three digits can be arranged in ways similarly for 4, 5, 6.]

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 6

Choose the one that best fits the sentence below. 

The state’s duty is to _______ the safety of its citizen

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 7

Cocaine will have pernicious effect on your physical fitness. 

Find synonym of the underlined word.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 8

Choose the one that best fits the sentence below.

The _______ of glory lead but to the ______

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 9

Gardener said to me, “Don’t pluck the beautiful flowers.” 

Select the appropriate Indirect Speech of the above.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 10

I haven’t had a call from the office about the meeting ________ last Monday.

Select the appropriate word to fill in the blank from the given options

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 11

Let T(n) = 2T(n/4) + 100√n 

The value of T(n) can be written as:

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 12

Consider a sorted array of n elements. Let k inversions (swapping) have been performed on the sorted array and we want to sort it again to reverse effect of inversions. If K is very small in comparison to n, then which sorting technique will prove efficient for making again the list in sorted increasing order array.

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 12

Insertion sort is directly proportional to number of inversions present in the list.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 13

What will the output of the following C code ?

#include <stdio.h>
int main()
{
int i=2, j=2;
while (i+1 ? --i : j++)
printf("%d",i);
return 0;
}

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 13

Consider the while loop condition => i + 1 ? --i : j++
In first iteration : i + 1 = 3 (true)
So ternary operator will return --i i.e. 1, condition part is 1 means true
so while condition is true.
Hence printf statement will print 1 In second iteration : i + 1 =2 (true)
So ternary operator will return --i i.e 0, condition part is 0 means false
so while condition is false.
Hence program control will come out of the while loop.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 14

A balanced tree is given below. How many nodes will become unbalanced when a node is inserted as a child of node G?

Note: A node in a tree is balanced if absolute difference between its left and right subtrees is less than or equal to 1.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 15

Consider the C code Below.

void function(int n)
{
    if (n == 1) 
      return;
    for (int i = 0; i<n; i++)
    {
        for (int j = 1; j< = n; j++)
        {
            printf("*");            
            break;
        }
    }
}

Which of the following is the tightest upper bound on time complexity of above function.

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 15

Important observation is Break statement terminates the innermost loop. So “*” is printed only n times.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 16

Consider the following statements about Bellman ford algorithm for finding shortest path in a directed connected graph G having integral edge weights. Statement I: It will always find out negative edge weight cycle in G reachable from source. Statement II: It will always give correct answer for the graph G. Choose from the options given below.

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 16

Statement 1 is true as in the nth iteration of Bellman Ford, if the length of the path of any node reachable from the source is decreased, that means we are having negative edge weight cycle in the graph. Statement 2 is false as no algorithm can give give correct answer if the Graph is having negative edge weight cycle.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 17

Consider the case: f(n) = O(g(n)). Then, following two statements are claimed to be inferred from the above case. 
Statement I: 2f(n) = O(2g(n))
Statement II: 2g(n) = O(2f(n)) 

Choose the correct option from the given.

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 17

if f(n) = n and g(n) = 2n. then f(n) = O(g(n))
here, 2^n = O(2^(2n)) = O(4^n), but not vice versa.
Hence, I is true. II is false. --------------
Now, if f(n) = 2n and g(n) = n then also f(n) = O(g(n)) because we can ignore constant but, 2^(2n) != O(2^n),
hence I is false, but II is true.
In both of the above cases, f(n) = O(g(n)).
But both the cases are counter of each other. Hence both I and II are wrong.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 18

Let X and Y be the integers representing the number of simple graphs possible with 3 labeled vertices and 3 unlabeled vertices respectively. Let X - Y = N. Then, find the number of spanning trees possible with N labeled vertices complete graph.

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 18

Number of simple graphs possible with n labeled vertices is 2^(n(n-1)/2).
Number of simple graphs possible with n unlabeled vertices is n+1.
Number of spanning tree possible with n vertices complete graph n^(n-2) X =8 Y = 4 X-Y=4
Therefore required answer is 42=16.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 19

Consider the following pseudo code.

x = 0;
for (i = 1 to n)
  for (j = 1 to 4i – 3)
    x = x + 1

Find the value of x, when the above code is executed in a function.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 20

Consider the following statements: 
S1 : DFS of a directed graph always produces the same number of edges in the traversal, irrespective of the starting vertex.
S2 : If all of the back edges that are found while DFS traversal on directed graph are removed, the resulting graph is acyclic. 

Which of the following statements above are valid ?

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 20

Statement S1 : consider the graph

 

A (source vertex ) we will get 2 edges Starting with B will get only 1 edge
Starting with C we will get no edge
Therefore DFS on directed graph may not give same number of edges.
Statement S2 : Back edges are those edges (u,v) connecting a vertex u to an ancestor u in a depth-first tree. Self-loops are considered to be back edges. Back edges describe descendant-to-ancestor relations, as they lead from “high” to “low” nodes.
Suppose that there is a back edge (u, v). Then vertex v is an ancestor of vertex u in the depth-first forest.
There is thus a path from v to u in G, and the back edge (u,v) completes a cycle.
Removing the back edge will break the cycle. Therefore removing all the back edges will make the graph acyclic.
So the statement is true.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 21

Which data structure would be the most appropriate to implement a collection of values with the following three characteristics? 
i) Items are retrieved and removed from the collection in FIFO order.
ii) There is no priori limit to the number of items in the collection.
iii) The size of an item is large relative to storage required for a memory address.

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 21

Head and tail pointers in singly link list will make the insertion and deletion in O(1) time complexity if we are accessing the elements in FIFO order. In doubly link list since only head pointer is given then for insertion we have to traverse the complete link list so insertion will be O(n) so not appropriate. In binary tree we have only a pointer to the root. Insertion and deletion in binary tree will be O(log n) so not appropriate. In hash table accessing the data in FIFO order will not be possible.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 22

Stack A has the entries as following sequence a, b, c (with ‘a’ on top), stack B is empty, as shown in the diagram below.

 An entry popped

out of stack A can be printed or pushed to stack B. An entry popped out of stack B can only be printed. In this arrangement which of the following permutation of a, b, c are not possible to print?

(i)   bac
(ii)  bca
(iii) cab
(iv) abc

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 22

Follow these steps to print bac. 1) POP element ‘a’ from stack A, push ‘a’ to Stack B. 2) POP element ‘b’ from A, print it. 3) POP element ‘a’ from B, and print it. 4) POP element ‘c’ from A, and print it. Now, perumtation bac has been printed. Similarly, we can print bca and abc, but can’t print cab.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 23

Considering the data given in the previous question, if the stack A had 4 entries, then the number of possible permutations that can be printed will be:

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 24

Let f(n) = Σ [(log(n/2i) +100] where i limits from 0 to k, and n = 2k. 
Find the time complexity of f(n).

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 25

Given a hash table with n keys and m slots with simple uniform hashing. If collisions are resolved by chaining then what is the probability that the first slot ends up empty ?

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 25

​Probability of one particular slot = 1/m ( because total m slots)
Probability that a value should not go in one particular slot = 1 – (1/m)
For n values (keys) probability = [1 – (1/m)]n

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 26

Consider the following instance of knapsack problem:

The maximum weight of 12 is allowed in the knapsack. Find the value of maximum profit with the optimal solution of the fractional knapsack problem.

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 26

Decreasing order of P i /W i is X1, X4, X3, X5, X2 X1 → profit = 15 and weight = 2 Including X4 → profit = 15 + 16 and weight = 2 + 4 = 6 Including X3 → profit = 40 and weight = 9 Now weight left = 3 Weight of X5 = 6 → half of X5 can be included. Profit = 40 + 17/2 = 48.5.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 27

Find the maximum value of the expression (x+y+k) where (x,y) satisfies the equation (x-2)2 + (y-3)2 = 25

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 27

Since (X,Y) is a point on circle, the general form of the point is X = 2 + 5*cost, y = 3 + 5*sint
We need to maximise the value of x+y+k x+y+k = 2 + 5*cost + 3 + 5*sint + k = (5+k) + 5*(cost+sint)
Here, k is a constant. The maximum value of c + acost + bsint is equals to c + sqrt (a*a +b*b).
Maximum value of (5+k) + 5*(cost+sint) (5+k) + 5*sqrt(2)
The result is (5+k) + 5*sqrt(2).

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 28

Which of the following argument is invalid ?

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 28

“⟺” mean equal. ifA⟺B means A is equal to B.
Option (a); (p→(q⋁r) )⟺(~p⋁(q⋁r) ) ⟺((~p⋁q)⋁r) ⟺(~(p⋀~q)⋁r) [∵De morgan’s law] ⟺(p⋀~q)→r ∴(p→(q⋁r) )⟺((p⋀~q)→r)
Option (b); (p→r)⋀(q→r)⟺(~p⋁r)⋀(~q⋁r) ⟺(~p⋀~q)⋁r [∵ De morgan’s law] ⟺~(p⋁q)⋁r ⟺(p⋁q)→r ∴(p→r)⋀(q→r)⟺(p⋁q)→r
Option (c): p→(q→r)⟺~p⋁(q→r) ⟺ ~p⋁(~q⋁r) ⟺~q⋁(~p⋁r) ⟺~q⋁(p→r) ⟺q→(p→r)
Option (d): (p↔q)⟺(p→q)⋀(q→p) ⟺(~p⋁q)⋀(~q⋁p) ⟺(~p⋁q)⋀(p⋁~q) ⟺(~p⋀~q)⋁(p⋀q) ⟺ ~(p⋁q)⋁(p⋀q) (p↔q)⟺(p⋁q)⋁(p⋀q) is wrong.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 29

The value of the following Integral is:

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 29

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 30

consider a Matrix   . The number of linearly independhent Eigen Vector for the Eigen value 1 is ______________.

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 30

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 31

The language L = {WbcWR | W ∈ (a+b)*} is _____.

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 31

Any language for which we can have a Deterministic PDA is always a DCFL.
Here for language L= {WbcWR | W ∈ (a+b)*} we can have a PDA with following transitions in which PDA accepts a string when it halts on a final state.
q0 in initial state and qf is final state. 1. (q0, a , Z) -> (q0, aZ) 2. (q0, b , Z) -> (q0, bZ) 3. (q0, a , a) ->(q0, aa) 4. (q0, b , a) -> (q0, ba) 5. (q0, a, b) -> (q0, ab) 6. (q0, b , b) -> (q0, bb) 7. (q0, c , b) -> (q1, null) 8. (q1, a , a) ->(q1, null) 9. (q1, b , b) -> (q1, null) 10. (q1, null, Z ) -> (qf, Z)
Here all a’s and b’s are initially pushed onto the stack for W. As soon as a c occurs after b , B is popped from the stack after which W^R is checked. If the further string alphabets match the alphabet of the stack the alphabet is popped from the stack and finally the string reaches the final state if language is of the form L= {WbcW^R | W ∈ (a+b)*}. Hence the above language is a DCFL.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 32

Given a graph G (V, E) is Bipartite, what is the chromatic number of G ?

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 32

Since the graph G is bipartite, the vertices set V can be partitioned into two disjoint sets. This shows that we can color the graph with 2 colors such that no two adjacent vertices will have same color.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 33

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 33

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 34

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 34

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 35

Print the maximum of the two Eigen values.

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 35

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 36

If P, Q are the roots of the quadratic equation x2 + ax + b. Find the value of P3 +Q3.

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 36

For any quadratic equation which is off the form x2+ax+b=0 1. Sum of the roots P+Q = -a2.
Product of the roots PQ = b => P3 + Q3 = (P+Q)(P2+Q2-PQ) => (P+Q)((P+Q)2-2PQ-PQ) => (P+Q)((P+Q)2-3PQ) => (-a)(a2-3b) => 3ab - a3

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 37

Given F is an expression of P,Q. Derive the expression F(P,Q) from the truth table given below.

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 37

From the truth table, the value of F can be written as: => F = PQ + PQ' + P'Q' => P(Q+Q')+P'Q' => P+P'Q' => (P+Q')(P+P') => (P+Q')T => (P+Q') Answer :- Option (B)

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 38

In a random experiment, suppose 3 unbiased coins are tossed simultaneously, what is the probability of getting at least 2 Heads ?

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 39

Consider a DFA accepting all strings over {a, b} where the number of a’s mod 3 = 2 and number of b’s are odd. What is the minimum number of states of such DFA ?

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 39

​The above question is an example of product automata. If we have two cases for which we can have separate DFA’s we can merge the two by product automata.
The resulting DFA has number of states equal to the product of the states of the separate DFA’s.
Here DFA for accepting all strings over {a, b} where the number of a’s mod 3 = 2 would have 3 states.
( number of a’s mod 3 would give remainder either 0, 1, 2 so 3 states to depict each). Similarly DFA accepting all strings over {a, b} where the number of b’s are odd ( number of b’s mod 2 = 0) would have 2 states.
Hence resulting DFA for both the conditions would have 2*3 = 6 states.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 40

Which of the following languages are regular over ∑ = {a, b} ? 

L1 = { anbn | 0 < n ≤ 10}
L2 = { a2n | n ≥ 1}
L3 = { an! | n > 0}

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 40

Any language for which we have a finite set or we can possibly have a DFA drawn for it is regular. L1: It gives a finite set generating string with equal number of a’s and equal number of b’s for range 0

Hence L2 is also regular. L3: It gives all possible strings with a’s such that number of a’s are in factorial values of 1,2,3,4 ……. i.e. {a1, a2, a6, a24……………}. The set is neither finite nor a DFA is possible for such set. Hence L3 is not regular.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 41

The language L = {WbcWR | W ∈ (a+b)*} is _____.

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 41

Any language for which we can have a Deterministic PDA is always a DCFL. Here for language L= {WbcWR | W ∈ (a+b)*} we can have a PDA with following transitions in which PDA accepts a string when it halts on a final state. q0 in initial state and qf is final state. 1. (q0, a , Z) → (q0, aZ) 2. (q0, b , Z) → (q0, bZ) 3. (q0, a , a) →(q0, aa) 4. (q0, b , a) -> (q0, ba) 5. (q0, a, b) → (q0, ab) 6. (q0, b , b) → (q0, bb) 7. (q0, c , b) → (q1, null) 8. (q1, a , a) →(q1, null) 9. (q1, b , b) -> (q1, null) 10. (q1, null, Z ) -> (qf, Z) Here all a’s and b’s are initially pushed onto the stack for W. As soon as a c occurs after b , B is popped from the stack after which W^R is checked. If the further string alphabets match the alphabet of the stack the alphabet is popped from the stack and finally the string reaches the final state if language is of the form L= {WbcW^R | W ∈ (a+b)*}. Hence the above language is a DCFL.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 42

Which of the following languages are closed under complementation ? 

A) Context free
B) Recursive
C) Recursive Enumerable

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 42

According to closure properties of languages, context free and recursive enumerable languages are not closed under complementation while recursive language is.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 43

Which of the following statements is incorrect ?

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 43

Statement 1 is correct because union of two DCFL may or may not be DCFL but surely CFL. Statement 2 is correct because regular languages are closed under complementation. Statement 3 is correct because Turing recognizable languages or recursive enumerable languages are closed under intersection.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 44

Which of the following is not a CFL? 

A) L = {albmcn where l=m or l=n}
B) L = {albmcn where l=m and l=n}
C) L = {albmcn where l = m + n}

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 44

Language in A is CFL because CFL can make one comparison which is either l and m must be equal or l and n must be equal. Similarly language in C is also CFL. But B is not because it has to make two comparisons. So language in B is CSL not CFL.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 45

Identify the circuit shown below:

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 45

In the circuit shown there are two tristate buffers.
One with active low enable and second one with active high enable. As per the operation of these buffers truth table is given below.
It is equivalent to 2 x 1 MUX B is selection line, A & C are the input lines, ‘D’ is the output.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 46

The 32 bit floating point representation of (-12) is ___

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 46

To convert the floating point into decimal, we have 3 elements in a 32-bit floating point representation:

  • Sign (MSB)
  • Exponent (8 bits after MSB)
  • Mantissa (Remaining 23 bits)

Sign bit is the first bit of the binary representation. '1' implies negative number and '0' implies positive number. Sign bit=1 Exponent is decided by the next 8 bits of binary representation. Hence the exponent of 2 will be 3. i.e 23=8. 127 is the unique number for 32 bit floating point representation. It is known as bias. It is determined by 2k-1-1 where 'k' is the number of bits in exponent field. Thus bias = 127 for 32 bit. (28-1-1 = 128-1=127) 127+3=130 i.e. 10000010 in binary representation. Mantissa: 12 in binary = 1100 Move the binary point so that there is only one bit from the left. Adjust the exponent of 2 so that the value does not change. This is normalizing the number. 1.100 x 23 10000000000000000000000 Thus the floating point representation of -12 is 1 10000010 10000000000000000000000

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 47

The following is the AB flip flop:

What are values of A, B when change in the flip flop state is from 0 to 1 ? 

Note: x in the options below represents that it can have any value out of 0 and 1.

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 47

Characteristic equation => Qn = A′.Q′ + B′.Q To change from 0->1 means Q = 0 and Qn = 1 Thus 1 = A’(1) + B’(0) = A’ Hence , A = 0 and B can have 0 or 1 not matters. 

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 48

An instruction pipeline has 4 stages Instruction Fetch(IF), Instruction Decode(ID), Execute instruction (Ex), Write Back(WB). All instructions take all stages and takes 4 clock cycles. Branch instructions are not overlapped, i.e. the instructions after the branch are not fetched till branch is known. Branch is known in the execute phase. Suppose 20 % instructions are conditional and 80 % unconditional. Calculate speed up for 100 instructions (upto 2 decimal place). Ignore the case that the branch may not be taken.

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 48

Suppose each stage takes 1 s. 20 conditional – 20*3 s (Cycles per instruction for conditional instructions is 3 as branch is known at the third stage) 80 unconditional – 80 s (Cycles per instruction for unconditional is 1) So total time taken with pipeline = 20*3 + 80 = 140 s Time taken without pipeline = 4 * 100 (Cycle per instruction for all is 4) Speed up = 400 / 140 = 2.86

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 49

A digital computer has a memory unit of 256k x 16 and a cache memory of 4k words. The cache uses direct mapping with a block size of 16 words. How many bits are there in index, tag, block and words fields of address format ?

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 49

Main Memory has 256k = 2^8 x 2^10 = 2^18. i.e.
We need 18 bits to address main memory. Cache Memory has 4k = 2^2 x 2^10 = 2^12. i.e. We need 12 bits to address cache memory. Cache consists of Index and tag which together are used to address main memory location.
Here Index is 12 bits and tag is 6 bits. (18 - 12 = 6). Index is divided into block part and word part.
Block part is used to address blocks in cache and word part addresses individual word in a block.
Here a block size is 16 words. i.e. 2^4, we need 4 bits to address a word and 12 - 4 = 8 bits to address a block in cache memory.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 50

Using the data from Question 49, How many blocks can cache accommodate ?

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 50

Since cache is 4k words and block size is 16 words i.e. No of blocks in cache = 4k/16 = 256.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 51

On a system using round-robin scheduling let 's' represent the time needed to perform a process switch, 'q' represents the round-robin time quantum, and 'r' represents the average time a process runs before blocking on I/O. The CPU efficiency when s = q < r is:

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 51

‘r’ is the average time a process runs before I/O block and ‘s’ is the time needed for switch. 'q' represents the round-robin time quantum. The efficiency of the CPU when s < r and q < r. = r/(r/q)*s+r = q/(q+s) Here s = q. Hence efficiency is ½.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 52

A counting semaphore was initialized to 0, then 20 V operations were successfully completed on this semaphore, followed with 18 P operations, the resulting value of the semaphore is:

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 52

20V => increments the semaphore 20 times. Hence , the value becomes 20. 18P => decrements the semaphore 18 times. Hence , the value becomes 2.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 53

Consider all the processes are arriving at large time intervals. Let t be the time interval between two processes Pi and Pi+1 for any i and service time of Pi is Si. If t > Si for every i then, what should be the strategy to schedule the processes ?

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 54

The following graph shows the schedule of five transactions. The schedule is

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 55

A student relation with 4 attributes is defines as R(ABCD). Which of the following query will execute without error if all attributes are numeric. 
A) ∏B,C(σA>20) 
B) σA>20(∏B,C)

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 55

Query in A is correct. It will first select required rows and then project required columns. But query in B is not correct as after projecting B and C columns, it can't have row selection based on condition on attribute A. So correct option is 1.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 56

Using the EMPLOYEE table, the following query is issued to generate the name, salary and the salary increased after an appraisal by 25 %. The increased salary for all the employee should be above 25000.

SELECT fname, salary, salary + (salary *0.25) AS "INCREASED_SALARY" 
FROM employee 
WHERE increased_salary > 25000; 

The above query throws an error. What is the reason for the error ?

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 57

Consider the following CFG

S -> AaAb | Bb
A -> ԑ
B -> ԑ

The above grammar is:

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 57

The grammar is unambiguous (only one parse tree) , not left recursive ( Non-terminal not present in second and third grammar rule) . It is left factored . It is LL(1).

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 58

Consider the following SDT
S -> 1A23    {print “GA”}
A -> 4S        {print “TE”}
A -> 5          {print “C”}
A -> B          {print “SE”}
B -> 6B       {print “TE”}
B -> 2         {print “ST”}
What will be output by the SDT for the input string “14122323”?

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 59

A directed acyclic graph represents one form of intermediate representation. The number of non terminal nodes in DAG of a = (b+c)*(b+c) expression is:

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 60

Consider the following grammar
S -> Aa | bAc | dc | bda
A -> d
The above grammar is:

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 61

Let 5 stations are connected to CSMA/CD network . Every station wants to transmit data with probability p = 0.6. It is given that there will be no collision if only one station will transmit. What is the probability of successful transmission ?

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 61

Success will be for one station and failure for other four Thus Psucc =5 c1 *(0.6)*(0.4)^4 =0.0768.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 62

Consider the following two statements: 
Statement 1: Stop and Wait protocol is preferred for LANs compared to WANs

Statement 2: Stop and Wait protocol is good for bursty data transmission.
Which of the above two statements are true?

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 62

Efficiency of Stop and Wait is given by, E= 1/(1 +(2*d*B)/v*L), link speed(v) and Bandwith (B) are fixed . Thus only variable terms are packet length(L) and distance(d) Cleary, E decreases with increase in d . And E will increase with increase in L . Hence Stop and Wait is suitable for LANs and bursty data.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 63

A broadcast channel has 10 nodes and total capacity of 12 Mbps. It uses polling for medium access. Once a node finishes transmission, there is a polling delay of 50 μseconds to poll the next node. Whenever a node is polled, it is allowed to transmit a maximum of 1000 Bytes. The maximum throughput of broadcast channel is:

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 64

Match the protocol with the characteristics: 

1. Mails are stored on the computer client use.
2. Stateless protocol
3. Converts MAC address to IP address.
4. Used to send error messages

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 64

In POP3 protocol mails are stored on client’s computer while in IMAP mails are stored on the server. HTTP is a stateless protocol, RARP is used to convert MAC to IP while ARP is used to convert IP to MAC, ICMP is used for sending error messages.

Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 65

In public key and private key cryptography, host ‘A’ has PuA and PrA, host ‘B’ has PuB and PrB as public and private keys respectively. Now, if ‘A’ wants to send a message to ‘B’ securely, which key will be used by ‘A’ for encryption ?

Detailed Solution for Computer Science And IT (CS/IT) Mock Test - 8 For Gate - Question 65

If the message is encrypted using PuB, then it can be decrypted using only PrB, which will be known only to ‘B’.

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