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

The error lies in part (A) of the sentence as the pronoun ‘him’ is incorrect and must be replaced with a reflexive pronoun ‘himself’ which is used as the object of a verb or preposition to refer to a male person or animal previously mentioned as the subject of the clause.

Total possible outcomes  = ²⁷C₂

Favourable outcomes = Selecting any two authors × selecting one book for each author=

Let us take material cost and labor cost as Rs 300 and Rs 100 respectively. Since labor cost went up by 20%, the revised labor cost is Rs 120. The revised material cost due to 15% increase is Rs (300+15% of 300) = 345. The ratio of labor cost and material cost is 120: 345 or 8: 23 i.e. (b)

You have to carefully observe the point that in the last minute, we DO NOT have to consider the sliding down of the monkey since the question states that the monkey reaches the top of the tree.Out of 58 minutes, the climb by monkey in 58th minute is 3 feet whereas for each of his earlier climbs, the monkey covers a net distance of 3-1.5=1.5  feet.We can work out the maximum possible height of the tree as 1.5×57+3=85.5+3=88.5 feet

The error lies in the first part of the sentence where the verb ‘were’ is incorrect and needs to be replaced with ‘was’ as the subject "misuse" is singular. To make the subject and the verb agree it should read as:’The misuse of drugs and alcohol was seen as..’

Let Xi denote the ith dice.

Since events are independent,

E(xi) = 7/2

V(xi) = 350/12

According to Central Limit Theorem,

P(30≤ X ≤40) = P(30 – 35 ≤ X – 35 ≤ 40-35)

Its converting to standard form, X – E[X] / std, where std is standard deviation.

Thus,

P = 2N(1.0184) – 1

= 0.69

Migration is the only word which comes to the meaning of the word ‘exodus.’

Let's understand the meanings of each phrasal verb given in the options:

Broke off = became separated/detached

Broke out = started/began suddenly

Broke in = force entry to a building

Broke down = suddenly ceased or stopped to function (of a machine or motor vehicle)

As per the context of the sentence, the correct phrasal verb is "broke down".

Since we are given time to cross a point which is 1/5th of the way and 1/3rd of the way, we can rewrite them as 3/15th and 5/15th of the way which implies that 2/15th of the way was covered in 10 minutes.
This means that the entire route was covered in 10 × 15/2 = 75 minutes (each 1/15th part is covered in 75/15 = 5 minutes) leading to time of arrival at school as 9:50 – 15 + 75 = 10:50 hours.

Literal = line for line, letter to letter
Verbatim = in exactly the same words; word for word
Formal = conventional, traditional
Idealistic = visionary, dreamy, unrealistic
Outdated = out of date, obsolete
Hence, "Verbatim" best expresses the meaning of the word "Literal".

A basic block can end in any of the following:Jump statementConditional/ unconditional branchReturn statement

The like operator is used in the string to search for characters. Eg: All names starting with ‘A’ (select name from ABC were name like ‘A%’;) while where is used to check for the condition.

The following typical process states are possible on computer systems of all kinds. In most of these states, processes are "stored" on main memory.
Created: When a process is first created, it occupies the "created" or "new" state. Ready and waiting: A "ready" or "waiting" process has been loaded into main memory and is awaiting execution on a CPU. There may be many "ready" processes at any one point of the system's execution.
Running: A process moves into the running state when it is chosen for execution. The process's instructions are executed by one of the CPUs (or cores) of the system. There is at most one running process per CPU or core. A process can run in either of the two modes, namely kernel mode or user mode.
Kernel mode : Processes in kernel mode can access both: kernel and user addresses.Kernel mode allows unrestricted access to hardware including execution of privileged instructions.
User mode : Processes in user mode can access their own instructions and data but not kernel instructions and data .access to any hardware device is allowed.
Blocked : A process that is blocked on some event (such as I/O operation completion or a signal), may be blocked due to various reasons, such as exhausting its CPU time allocation or waiting for an event to occur.
Terminated : A process may be terminated, either from the "running" state by completing its execution or by explicitly being killed

Relative: 302 + 400 = 702

A file system is said to exhibit data dependence because a change in any file’s structure, such as addition or deletion of a field, requires modifications of all programs using that file. That means, access to a file is dependent on its structure.
A file system is said to exhibit data dependence because a change in any file’s data characteristics, such as changing the filed from integer to decimal, requires change in all data access programs. That means, access to a file data is dependent on its data characteristics.

If variables are declared previously, then semantic analyzer is responsible for generation of symbol table.

If variables are not declared, then lexical analyzer will generate the symbol table.

Hence option b is correct.

Gantt chart for the given situation is:

Now, the waiting time is 0 milliseconds for process P1, 22 unit for process P2 and 27 unit for process P3

Thus, the average waiting time = (0 + 27 + 22)/ 3 = 49/3 = 16.33 = 16 units.

So, the insertion is followed in the same order, hence, we will insert 10 first which doesn’t effect the tree. Now, we insert 19. After insertion of 19 max_heapify founds an error in the node 17 → 11, 17 → 19. So, 19 is send up and 17 is brought down. After that the level order traversal changes to 30, 18, 19, 12, 11, 10, 17

Following is the syntax for creating views.

CREATEVIEW view_name AS

SELECT column_name(s)

FROM table_name

WHERE condition

Backslash actually acts an escape character and helps to print double quotes, where % can be printed only with double % i.e. %%

For this purpose, we can ignore the below forty numbers and have an array of 60 index to store the individual frequencies from 40 to 100.

given

now, the random variable X does represent number of throws required for getting 3. So,

The SUBI instruction, means the subtraction is in immediate addressing mode.

The given sequence is <1, a¹, a², a³, a⁴, ………..>.

The generating function for this sequence will be;

The only generating sequence that matching the above sequence is of option D

since , we know that 

Since we know that

As it is clearly mentioned that each team at least require one player so one team can acquire n number of athletes where n>=1. And also one athlete can play for more than one team i.e for m teams. So maximum cardinality is m:n

To check LL(1) we have to use some rules :
i. First(A1) ∩ First(A2) ∩ ...... ∩ First(An) = Ø
Where , A -> A1 | A2 | A3 ...... | An
ii. If any production contains null then do as follow
First(Ai) ∩ Follow(Ai) = Ø
If the above two rules satisfy then given grammar is LL(1) , but as per grammar given in question it is not satisfying above rule in production of B [ First(B) ∩ Follow(B) != Ø ].
To check LR(0) when we draw DFA for each item in Goto and Clousre , there is shift and reduce conflict.
Same case is with SLR(1) , in production B , there is shift reduce conflict.

We don’t need to calculate the final result, just calculate the unit digit raised to the power of unit digit of power of each term and add it. So,  1+7-3=5

In the above example food items are generalized on the basis that they all belong to fast food.

It is the property of Optimal and LRU page replacement that S and SR will have same number of page fault.

Also, FIFO suffer from Belady’s anomaly but LRU and Optimal do not suffer from belady’s anomaly.

{K MOD N = i}

4-M: 4 MOD 4 = 0

5-M: 5 MOD 4 = 1

7-M: 7 MOD 4 = 3

12-M: 12 MOD 4 = 0

4-M; 4 MOD 4 = 0

5-H

13-M; 13 MOD 4 = 1

4-H

5-M; 5 MOD 4 = 1

7-H

H = 3/10 = 0.3

41, 11

Class without name is allowed. In this case object of class is created along with definition.

cnct(x,y) x##y ,## is a preprocessor macro used for concatenation.

x##y= xy

Call this creates a single terms named xy,here in main pq; cnct(p,q) places value of pq in its position that is 15.

cout<<p+q+pq+cnct(p,q)

5+6+15+15=41

P+++q is interpreted here as p++ + q(1st two for post increment operator,3rd one operator for adding )

Plus operator(+) has higher precedence than post increment operator. So, Value of p & q (5 & 6) get added before p increments. So, p+++q->11

Height is defined as path length i.e (height+1) will give number of levels. In a binary tree of height h = 4;

Maximum number of nodes = 2^(h+1)-1

= 31

Minimum number of nodes = 5

Therefore (d) is correct answer.

SQL uses the five aggregations operators SUM, AVG, MIN, MAX and COUNT. These operators are used by applying them to a scalar-valued expressions, typically a column name, in a SELECT clause.

Dijkstra's single source shortest path is not guaranteed to work for graphs with negative weight edges, but it works for the given graph.

Let us see...

Let us run the 1st pass b 1 b is minimum, so shortest distance to b is 1

After 1st pass, distances are c 3. e -2. e is minimum, so shortest distance toe is -2 After 2nd pass, distances are c 3, f 0. f is minimum, so shortest distance t of is 0 After 3rd pass, distances are c 3, g 3. Both are same, let us take g. so shortest distance to g is 3. After 4th pass, distances are c 3, h 5 c is minimum, so shortest distance to c is 3 After 5th pass, distances are h -2 h is minimum, so shortest distance to h is -2.

As seen in the above image, best job is already assign to the best employee. Now we have 6 remaining job, that we need to assign to 4 employee. why 4?
Because best employee still can do other jobs.
Let N(e2) denotes no. of function where e2 is not assigned to any job.
 this will represent the no. of functions where e2, e3 and e4 is assign to any job.

why we are not counting best employee for this counting because best employee is always assign to the job

This program finds a repeated number (duplicate) in the list. This program adds all values in the variable “ value” and substract “k(k-1)2” from it.Array has 0 to (k-1) locations and numbers are from 1 to (k-1).so one number will be repeated in the array.

When you dereference a pointer, it actually gives out the type of object being pointed. Hence, in this case it is int and the answer is 10 * 2 * 2 (integer size ) = 40

Inference A, is Modus Ponens.

Inference B, is false.

Let P = Sita go to school.

Q = Ram go to school.

For P →Q to be true.

If P is true then Q has to be true. (There is no other choice for Q).

But if P is false then Q can be anything (True or False). Still P →Q is true.

This is also known as "The fallacy of denying the antecedent".

P →Q

~P

∴(Nothing can be concluded).

The given expression can be written in a+ib form as :

where r, θ are the magnitude and argument of z respectively.

The function prints first n fibbonacci numbers note that 0 and 1 are initially there in the queue. In every iteration of loop, sum of two queue items is enqueued and the front item is dequeued.

Total number of bit strings that start with two zeros (2⁶) or end in a one 

2⁶+2⁷−25

Loc[-2] [50] = Base address

+[(-2+5)*(50-5+1)+(50+5)]*1

= 0+3*46+45=183

Number of BST with ‘n’ nodes is given by ²ⁿC_n/(n+1)

=>¹²C₆/7 =132.

if we get A² = I →then it's know as  Involuntary Matrix

if we get A² = A → then matrix is Idempotent Matrix

and

If A is a symmetric matrix, A^-1 will also be a symmetric matrix

In this, we see that on b state, state 2 will go to state 1 with state 3 going to state 4 which further will go as state 1 and state 4 are in different set. Further, states 2 and 3 are different from each other. As state 4 goes to state 4 with state 3 goes to state 5 and state 4 which lies in different sets, hence state 3 and state 4 gets separated from each other. Further, state 2 goes to state 1 while state 4 goes to state 4 and to state 1. Further it is seen that state 2 and state 4 are separated from each other.

Further it is observed that as state 1 goes to state 3 placed on node a and further to state 2 which lies on node b as shown, hence the minimized DFA transitions are added from state 1 to state 3 on a while state 1 to state 2 on node b. Further as state 2 goes to state 1 on node b and state 3 goes to state 1 on node a, so minimized DFA transitions adds from state 2 to state 1 on node b and further state 3 to state 1 on node a.

Ternary operator has the following syntax:

exp 1 ? exp 2 : exp 3

if exp 1 is TRUE, then exp 2 is evaluated else exp 3 would be evaluated. Do(30,90)would return 0 and Do(30,0) would return 30.

Hence A) is the correct answer.

An attribute or set of attributes which can be used to distinguish every tuple in a table is called a super key.A super key for which no proper subset is a super key is called candidate key.There can be many candidate keys for a table and the candidate key selected by the database designer to identify the tuples uniquely is called a primary key.

From the above cardinalities, we see that ratios like 1:1, 1:N, N:1 and N:M gives cardinality constraint or numeric restriction on possible relationships. From this, we see that the ratio 1:N relation is most comfortable to describe the constant of ER team.

Max heap is stored in the array such that

Arr[k]>=Arr[2k] and Arr[k]>=Arr[2k+1]

Where k is the index of the array such that 1<=k<=n/2, Arr[2k] is left child and Arr[2k+1] is right child. Option (c) is the only array which satisfies above condition.
 

1 is false because for P →Q to be true.

If P is true then Q has to be true. (There is no other choice for Q).

But if P is false then Q can be anything (True or False). Still P → Q is true.

This is also known as "The fallacy of denying the antecedent".

P → Q

~P

∴ (Nothing can be concluded).

2 is modus ponens so it is valid, similarly 3 is hypothetical syllogism, it is also valid.

4 is false because 

The edge (d-e) cannot be considered before (d-c) in Kruskal’s minimum spanning tree algorithm because Kruskal’s algorithm picks the edge with minimum weight from the current set of edges at each step.

a) p is a function that takes an argument as an array of pointers to characters and returns a pointer to an integer.

b) p is a function that takes an argument as a pointer to a character array and returns an integer.

c) p is a pointer to a function that takes an argument as a pointer to a character and returns an integer.

d) This is an illegal statement.

Hence (d) is the correct answer

Here, According to FD set F,

A+= {ABC}, B+= {ABC}, C+= {ABC}

According to FD set G,

A+= {ABC}, B+= {ABC}, C+= {C}

So, F covers G but G doesn’t covers F. Hence, option A is correct.

(a)f(n)=n^2 logn(true)

(b) If the graph has no edges (True)

(c) In a DAG the strongly-connected components consist of single nodes. To compute them just

scan the array of nodes. (True)

(d) They form a tree but the nodes in this tree are only those nodes reachable by a path from ‘s’and there may be strictly less than ‘n’ such nodes, therefore the tree may have strictly less

than (n – 1) edges. (False)

As there have to be at least k balls in each bag, so firstly put k balls in each bag i.e k*y balls. Balls remaining: x-ky

We can now apply balls and sticks method.

y bags= y variables, they need to equal to x-k*y, no restrictions on how many balls in each bag. We can make a equation out of this:

a1 + a2 + ... + ay = x- k*y

On solving it we get our answer:  (x-ky+y-1) C (y-1)

12 bits is the answer

as it is given that page size is 2¹² Bytes, so 12 bits are needed to specify that offset. The number of bits specify the second level index is:

36 - (12 + 8)

= 36 - 20

= 16

Returning from x = 1 to 5 “f” value is updated being a static variable = 40 + 5 × 40 = 240.

Most efficient solution would be to traverse given BST in inorder sequence and store in an array which takes O(n). Building a balanced BST from an array takes O(n). So complexity = O(n).

For the system not to be in safe state, values should be distributed in such a way that remaining resources should not be able to satisfy the need of any of the process.

After Observing, it is clear that when X= 40 and Y=20 the remaining need of the processes can be listed as:

Remaining Resources= 150-140=10

And we cannot satisfy the need of any of the given process.

Hence option d is the correct answer.