Computer Science And Information Technology - CS 2017 GATE Paper - 1 (Practice Test) - GATE MCQ

Explanation: 

162 = 3 * 3 * 3 * 3 * 2 To make it a perfect cube, we need at least 2 * 2 * 3 * 3 which is 36. 

Alternate Solution 

Take the prime factor of 162: 162 = 2 * 81 = 2 * 9 * 9 = 2 * 3 * 3 * 3 * 3, now if we add minimum two 2 and two 3, then it will be perfect cube of a number. Therefore, the value of y will be = 2 * 2 * 3 * 3 = 4 * 9 = 36 Note that you can multiple 162 with values in given options, then take cube root with the help of calculator for each option, the answer should be an integer.

Assume, they are looking to the center of circular table, then the possible arrangement is:

Grammatically, besides is an adverb or a preposition, and beside a preposition. Beside means next to. As a preposition, besides means in addition to or apart from. As an adverb, besides means as well or furthermore. 
Here, we use "besides" as a preposition meaning "in addition to". 
"besides money" is correct choice.

For 10 digits(0 to 9) total possible cases = (10)^k 
Excluding digits 0, 5, 9, total possible cases = (7)^k with numbers 1,2,3,4,6,7,8 
Required probability = (7)^k/(10)^k= (0.7)^k 
Therefore option C is correct 

Alternate Solution 
Case(1): When 0 can not be as left most digit: 
Then sample space = 9 * 10^{k-1} 
Favorable event = 7^{k} 
Required probability = (7^{k}) / (9 * 10^{k-1}). 
None option matches. 
Case(2): When 0 can be as left most digit: 
Then sample space = 10^{k} 
Favorable event = 7^{k} 
Required probability = (7^{k}) / (10^{k}) = (7/10)^{k} = (0.7)^{k}. 
Option (c) matches. 

According to the rule : when the main clause is in the past or past perfect tense, the subordinate clause must be in the past or past perfect tense.

Let's draw what these statements say:

• There are atleast Two Men and Two Women.
• There are atleast 3 Right Handed People
• No women are Right Handed - infers there are three right handed men
• Every woman has a left-handed person to her immediate right - Means women circled in red needs to have a left handed person on her right and that too a man as on right side of ?? there is a right handed man.
• So,  ?? - will be a left handed man

​

Therefore, option A (only 2 women) is correct

The author is trying to say that everything related to colonial past has a hold of nationalism in people’s imagination. It is not necessary that anything which is represented inadequately is the history. History and nationalism has a strong connection, but sometimes history is viewed with nationalism in mind.

As we know that, if x > y, then |x - y| = x - y and 
if x < y then |x - y| = y - x , because value of |x - y| is always non-negative. 
Therefore,

• Case 1: If x > y : (x + y) - |x - y|] / 2 =  (x + y) - (x - y)] / 2 = 2y / 2 = y (Minimum of x , y)
• Case 2: If x < y : (x + y) - |x - y|] / 2 = (x + y) - (y - x)] / 2 = 2x / 2 = x (Minimum of x , y)

Therefore in both the case we get minimum of (x,y). So, option B 
Note that you can take some random values of x and y, then verify given options.

Given data - height of all villages

• P = 575
• Q = 525
• R = 475
• S = 425
• T = 500

Villages  below the flood level - 525m are

1. R <500m
2. 425m<= S<=450m
3. 500m <=T<= 525m

Therefore option C is correct.

Total number of Shirts = 4 , Total candidates =4

• So total combination of possibilities = 4 X 4 = 16
• No.of ways  'Arun chooses red' + No.of ways Shwetha chooses white '=  2

Therefore, Answer is 16-2 = 14

Here, half of the values of Y are to the left of the mean X = 0 and the remaining half of the values of Y lies to the right of the mean X = 0. hence,The median of Y = 0. 

The given scenario can be illustrated as:

According to Static Single Assignment

1. A variable cannot be used more than once in the LHS
2. A variable should be initialized atmost once.

Now looking at the given options

• a - code violates condition 1 as p1 is initialized again in this statement: p1 = u * v
• c- code is not valid as  q₁ = p₂ * c , q₂ = p₄ + q₃ - In these statements p₂, p₄, q₃ are not initialized anywhere
• d- code is invalid as  q₂ = p + q is incorrect without moving it to register

Therefore, option B is only correct option.

Given

V -> W
VW -> X
Y -> VX
Y -> Z

We need to find the minimal cover of these FDs 
Option B. W->X can  not implied by the given FDs, so incorrect 
Option C. Y->X can be implied from Y->V and V->X, hence redundant 
Option D. W->X can  not implied by the given FDs, so incorrect 
Option A. Minimal cover of dependencies that can be extracted out as

1. V -> W
2. V -> X
3. Y -> V
4. Y -> Z

Therefore, option A is most appropriate. 

Alternate Solution

Irreducible equivalent functional dependency is minimal cover. Given,

{V → W, VW → X, Y → VX, Y → Z}

W extraneous from VW, since we have V → W,

= {V → W, V → X, Y → VX, Y → Z}

X extraneous from VX, since we have V → X,

= {V → W, V → X, Y → V, Y → Z}

Turn Around Time

P1  = 12-0 = 12
P2 = 6-3 = 3
P3 = 17-5 = 12
P4 = 8 - 6 = 2

Waiting Time

P1  = 12-7 = 5
P2 = 3-3 = 0
P3 = 12-5 = 7
P4 = 2 - 2 = 0

Average Waiting time = (7+0+5+0)/4 = 3.0 Therefore, option C is correct 

Alternate Solution

Given, with arrival time and burst time:

Using (preemptive) shortest remaining time first algorithm, gantt chart is:

Therefore, Average waiting time = (5 + 0 + 7 + 0)/4 = 12/4 = 3 

The minimum height of a binary search tree will be when tree is full complete tree:

Now, let h be the height of the binary tree, then, 2^{0}+2^{1}+2^{2}+2^{3}+...+2^{h}=2^{h+1}-1 <= n 
So, Min height of a binary search tree = log₂(n+1) - 1 = log₂(15+1) - 1 = 4 - 1 = 3

The maximum height of a binary search tree will be when the tree is fully skewed: (like below)

Max height of the binary search tree = n-1 = 15 - 1 = 14, where the tree is Skewed tree

Therefore, option B 

(¬ p) => (¬ q) is logically equivalent  ¬ q + p 
I. p => q ≡  ¬ p +  q 
II. q => p ≡ ¬ q + p 
III. (¬ q) ∨ p ≡ ¬ q + p 
IV. (¬ p) ∨ q ≡ ¬ p + q 
Therefore option D, II and III only

Output of Query will be:

Thread share all other  resources of process except local data like  - register , stack.

The vectors a1, a2, …, an are linearly dependent. 
For the system AX = B, 
Rank of coefficient matrix A = Rank of augmented matrix (A / B) 
= k (k< n) 
Hence, the system has infinitely many solutions. 

Given, first order logic sentence 
F: ∀x (∃y R(x, y)) with following sentences: 
(i) ∃y (∃x R(x, y)) is true. Because we have ∀x (∃y R(x, y)) → ∃x (∃y R(x, y)) → ∃y (∃x R(x, y)). 
(ii) ∃y (∀x R(x, y)) is false. Because we have ∀x (∃y R(x, y)) ← ∃y (∀x R(x, y)). 
(iii) ∀y (∃x R(x, y)) is false. Because for ∃y can not imply ∀y.
(iv) ∼∃x (∀y ∼R(x, y)) is true. Because ∼∃x (∀y ∼R(x, y)) = ∀x (∼ ∃y ∼R(x, y)) = ∀x (∃y ∼∼R(x, y)) = ∀x (∃y R(x, y)). 

Alternate Solution

To compute FOLLOW(A) for all non terminals A, apply the following rules until nothing can be added to any FOLLOW set:

• Place $ in FOLLOW(S), where S is the start symbol and $ is the input right end marker.
• If there is a production A → αBβ, then everything in FIRST(β), except for ∈, is placed in FOLLOW(B)
• If there is a production A → αB, or a production A → αBβ where FIRST(β) contains ∈ (i.e., b → ∈), then everything in FOLLOW(A) is in FOLLOW(B).

Therefore, follow(Q) = First(RS) = ( w ) ∪ First(S) = { w } ∪ { y } = { w,y }. 
Note that here, first(S) is not ∈. 

Given, Memory reference/instruction = 1.4 
So, total number of memory references = 1000 * 1.4 = 1400 
No. of  7 misses in L2 cache = 7/1400 = 0.005 
Miss rate of L2=No. of misses in L2/No. of misses in L1 =0.005/0.1 =0.05 
Therefore, option A is correct. 

Alternate Solution 

On Average, 1.4 memory accesses are required for one instruction execution. 
So, for 1000 instructions, 1400 accesses are needed. 
Number of misses occurred in cache L2 for 1000 instruction = 7/1400 = 0.005 
Miss rate of L2 cache = misses occured in L2 cache / miss rate in L1 cache 
= 0.005 / 0.1 = 0.05 (A)

For the large number, value of inverse of number is less than a constant and value of constant is less than value of square root. 
Therefore, 100/n < 10< log₂n< √n< n 
It can be proven by substituting any big number in all the functions.  

The address of student grade can only be accessed by using the base address of Student.

• Post-increment addressing mode. (R1)+ - Will give the next address and not the desired grade address
• Pre-decrement addressing mode, -(R1) - Will give the previous address and not the desired grade address
• Register direct addressing mode, R1 : In this mode, the address of the operand is embedded in the instruction code.
• Index addressing mode: It is the only mode which gives access to next address by adding displacement in base address using displacement mode. . Base register contains a pointer to a memory location. An integer (constant) is also referred to as a displacement. The address of the operand is obtained by adding the contents of the base register plus the constant.

Therefore, option D is correct

The code will run and give output = 10, so option A and B are discarded.

int * x= malloc (sizeof(int));

This statement assigns a location to x. Now,

(int*)malloc(sizeof(int));

again assigns a new location to x , previous memory location is lost because now we have no reference to that location resulting in memory leak. Therefore, option D is correct. Memory leak occurs when programmers create a memory in heap and forget to delete it. Memory leaks are particularly serious issues for programs like daemons and servers which by definition never terminate. Detailed article on Memory LeakRunning code of the question: [sourcecode language="C"] #include int *assignval (int *x, int val) { *x = val; return x; } void main () { int *x = malloc(sizeof(int)); if (NULL == x) return; x = assignval (x,0); if (x) { x = (int *)malloc(sizeof(int)); if (NULL == x) return; x = assignval (x,10); } printf("%d ", *x); free(x); } [/sourcecode]

Solving the above k- map, = ca'(b+b') = ca' This can be implemented using one NOR gate. Therefore, option A is correct  

• Kruskal’s  is a greedy technique of Minimum Spanning Tree Algorithm to find an edge of the least possible weight that connects any two trees in the forest.

• QuickSort is a Divide and Conquer algorithm. It picks an element as pivot and partitions the given array around the picked pivot.
• Floyd Warshall Algorithm is for solving the All Pairs Shortest Path problem using Dynamic Programming. The problem is to find shortest distances between every pair of vertices in a given edge weighted directed Graph.

As it is stated in the question, that m and n are valid Lists but not explicitly specified if the lists are empty or not. We can have two cases:

1. Case 1: If lists are not NULL : Invocation of join will append list m to the end of list n if the lists are not NULL. For Example:Before join operation : m =1->2->3->4->5->null n =6->7->8->9->nullAfter join operation : 6->7->8->9->1->2->3->4->5->null
2. Case 2: If lists are NULL : If the list n is empty and itself NULL, then joining and referencing would obviously create NULL pointer issue.

Therefore option B is correct