GATE Computer Science Engineering(CSE) 2027 Mock Test Engineering (CSE)

Given,

1764⁴⁰

log⁡6 = 0.778

log⁡7 = 0.845

As we know that,

Power Rule: log_b⁡(P^Q) = q × logb⁡P

The above property of the product rule states that the logarithm of a positive number p to the power q is equivalent to the product of q and log of p.

⇒ log⁡1764⁴⁰ = 40log⁡ (6²×7²)

= 40 (2 log⁡6+2 log⁡7)

= 40 (2 × 0.778 + 2 × 0.845)

= 40 (1.556 + 1.69)

= 40 (3.246)

= 129.84

So, number of digits in 1764⁴⁰ = [(log⁡1764_⁴⁰)+ integral part of 1]

= 129 + 1

= 130 digits

Hence, the correct option is (A).

The word 'Defiance' means refusal to obey.

The synonyms of the word 'Defiance' are "disobedience, rebellion, disrespect".

From the synonym of the given word, we can say that the word 'Disobedience' has the same meaning.

The word 'Disobedience' means failure or refusal to obey rules or someone in authority.

Hence, the correct option is (B).

Given,

Number of shake hands = 780

Let the total number of students is y.

⇒ ^_yC^₂ = 780
⇒ = 780
⇒ =780

⇒ y(y−1) = 1560

⇒ y2 − y − 1560 = 0

Using discriminant method:

D = b² − 4ac

= (−1)²−4 (1) (−1560)

= 1 + 6240

= 6241
X = 
= 
= 
= 80/2

= 40

∴ Number of students is 40.

Hence, the correct option is (B).

Recorded data are:

10, 5, 8, 16, 18, 20, 8, 10, 16, 20, 18, 11, 16, 14 and 12

Let us arrange the following data in ascending order to find out the median and mode, i.e.:

5, 8, 8, 10, 10, 11, 12, 14, 16, 16, 16, 18, 18, 20, 20

Total number of observation = 15

As we know,

Mean = ∑x/n
= 
= 202/15

Median = value of  observation

∴ Median =8^th observation = 14

From above observation 16 occurs maximum number of times.

∴ Mode =16

According to question,

One of the values is wrongly written as 16 instead of 18.

= 5, 8, 8, 10, 10, 11, 12, 14, 16, 16, 18, 18, 18, 20, 20

 Mean = ∑x/n
= 
= 204/15

Median = value of  observation

Median = 8^th  observation = 14

From the above observation 18 occurs the maximum number of times.

∴ Mode = 18

We can see that the mean and mode are changed.

Hence, the correct option is (D).

The most appropriate meaning of the given idiom/phrase 'Gordian knot' is 'A difficult problem'.

Gordian knot: A complicated problem that can only be solved with creative or unorthodox thinking.

For example: The coding problem looked like a Gordian knot until we realized we could bypass it altogether with a different approach.​

By the given meaning and example we can say that 'A difficult problem' is the appropriate meaning.

Hence, the correct option is (B).

The terms are of type x^₂ + x. (Starting from x = 1)
So, next term will be = 7^₂ + 7 = 56

'Hiatus' means 'a pause or break in continuity of a sequence or an activity'. So, options 1, 2 and 3 refer to gap or rest. But, 'end' means 'a termination of a state or situation', thus the odd one out.

% decrease in 1997-98 for ABC Engineering College = 70/150 × 100 for electronics
% decrease in 1997-98 for XYZ Engineering College = 60/180 × 100 for electronics
Difference = (70/150 - 60/180) × 100 = 13.33%

Hetero means different. It is clearly mentioned that the movement comprised of moderates, liberals, radicals, socialists and so on. Hence, option (4) is correct.

Percentage of whisky in mixture = 75%
Percentage of whisky in water = 0%

Therefore, mixture and water should be mixed in the ratio of 2 : 1.
or,
We need to replace 1/3^rd or of the mixture with water.

Let A be the event in which an item has been produced by machine A and so on.

D = The event of the item being defective.

P(A) = 1/2

P(B) = 1/4

P(D/A) = P = 2/100 (An item is defective, given that A has produced it)

P(D/B) = 2/100

P(D/C) = 4/100

P(C) = 1/4

By theorem of total productivity,

Required probability = Chosen a defective bolt from Machine A+ Chosen a defective bolt from Machine B+ Chosen a defective bolt from Machine C.

P(D) = P(A) × P(D/A) + P(B) × P(D/B) + P(C) × P(D/C)

=0.025

Hence, the correct option is (B).

According to the definition of the Algorithm,

• An algorithm is a sequence combination of finite steps to solve a particular problem. 
• It is a finite step-by-step procedure to achieve a required result.
• An algorithm always produces at least one output.
• It always terminates an infinite amount of time.
• An algorithm is independent of the programming language.

Hence, the correct option is (B).

For every packet sent the window size increases by 2. So, when

Window size [WS=1] initially

⇒ After 1 RTT, window size = 2 and 1 segment is sent

⇒ After 2 RTT window size = 4 and 3 segment sent in total

⇒ After 3 RTT, window size = 8 and 7 segment sent in total

⇒ After ‘x’ RTTS window size 2^x and 2^x−1 segment are sent

Now, 2^_x−1 = 3999

⇒ 2x = 4000

⇒ x = log2⁡(4000)

⇒ x = 12 RTT's

Alternately we can say that,

1^st RTT- 1 segment sent.

2^nd RTT- 2 new segments sent.

3^rd RTT- 4 new segment sent.

4^th RTT- 8 new segments sent.

5^th RTT- 16 new segments sent.

6^th RTT- 32 new segments sent.

7^th RTT- 64 new segments sent.

8^th RTT- 128 new segments sent.

9^th RTT- 256 new segments sent.

10^th RTT- 512 new segments sent.

11^th RTT- 1024 new segments sent.

12^th RTT- 2048 new segments sent.

Total segments sent = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024 + 2048

=4095 > 3999

Hence, the correct answer is 12.

STATEMENT- (I): TRUE

Timestamp protocol suffers from more aborts (suffer from cascading rollback).

STATEMENT- (II): TRUE

A block hole in a DFD is a data store with only inbound flows.

STATEMENT- (III): FALSE

Multivalued dependency among attributes is checked at 3 NF level because it is checked at 4 NF level.

STATEMENT- (IV): FALSE

An entity-relationship diagram is a tool to represent the event model because it is a tool to represent the data model.

Hence, the correct options are (C) and (D).

Given:

S → X|a|YXZ

X → Y∣b

Y → S∣c

Z → d

Substitute to X production in S.

S → Y|b|a|YYZ|YbZ

Y → S∣c

Z → d

Substitute to Y production in S.

S → c|b|a|SZ|ccZ|ScZ|cSZ|SbZ|cbZ

Z → d

Left recursion of above productions is,

S → SZ ⇒ S → pX′,X′ → ZX′/ε

S → ScZ ⇒ S → pY′,Y′ → cZY′/ε

S → SbZ ⇒ S → pZ′,Y′ → bZZ′/ε

Total productions are:

S → pY′,Y′→cZY′/ε

S → pX,X→ZX′/ε

S → pZ′,Y′→bZZ′/ε

S → c|b|a|ccZ|cSZ|cbZ

Z → d

So, the total productions are 13.

Hence, the correct answer is (A)

Given,

linear transformation are:

T(e₁)=7e₁−5e₃

T(e₂)=−2e₂+9e₃

T(e₃)=e₁+e₂+e₃

Let the standard matrix be A with respect to the basis e₁,e₂,e₃,

Now, T(e₁)=7e₁+0e₂−5e₃

T(e₂)=0e₁−2e₂+9e₃

T(e₃)=e₁+e₂+e₃

The standard matrix will be (transpose of linear combinations).

Hence, the correct option is (A).

It is a combinational circuit that selects binary information from one of many input lines and directs it is to single output lines. The selection of particular input lines is controlled by a set of selection lines.

Y = S¯₁S¯₀I₀ + S¯₁S₀I₁ + S₁S¯₀I₂ + S₁S₀I₃

Now,

From the given multiplexer,
f = 

Given that the output is,
f = 

Now, we need to express every term in terms of w1 and w4 as these are the inputs of selection lines. 

  represents NOR gate.

 represents AND gate.

represents OR gate.

Hence, the correct options are (A), (B) and (C).

Only if part should be consider because if part contains the statement printf(“#”);

12 times # is printed which can be seen from the above given tree

Important points:

# will be printed if a ≠ 0 and b ≠ 0

log₂ 22048+1

=11+1

=12

Hence, the correct answer is 12.

Assume: Computation time of P2 is less than or equal to computation time of  P3.

Gantt chart:

Process table:

Average Turnaround time =13/2
 =13/2
3X + 17 = 26
∴ X = 3 ms
As we know that,
TAT = CT − AT
ms → milliseconds
Explore: If computation time of P2 is more than computation time of P3.
Hence, the correct answer is 3.

A foreign key is a set of attributes references to the primary key or alternative key of the same relation or some other relation. The correct statement regarding the foreign key is that a referencing relation record whose foreign key value is NULL is out of referential integrity constraints. Between referenced and referencing is (1:M) mapping. Each referenced record can be related to many [0 or more] referencing records and each referencing record can relate to at most 1 referenced records.

Referential integrity is the state of a database in which all values of all foreign keys are valid. A foreign key is a column or a set of columns in a table whose values are required to match at least one primary key or unique key value of a row in its parent table. A referential constraint is a rule that the values of the foreign key are valid only if one of the following conditions is true:

• They appear as values of a parent key.
• Some component of the foreign key is null.

To map (1:M) relationships, the primary key on the `one side' of the relationship is added to the `many sides' as a foreign key.

Hence, the correct options are (B), (C) and (D).

We know that:

Using precedence graph method we can find out Schedule is conflict serializable or not.

If precedence graph has any loop then that schedule is going to be not conflict-serializable.

If precedence graph doesn’t have any loop then that schedule is going to be conflict-serializable.

If one transaction  (T₁) is writing some value × and that value is read by some other transaction (T₂) then T₁ should commit first then only  (T₂) should read that value and commit in order to become recoverable schedule.

There is no loop in this graph so schedule is conflict serializable.

T₂: T₃: T₁: T₄ is the only serial schedule that is equivalent to given schedule. (Using Topological Sorting)

There are two write read dependency.

1. T₁ write (x) and T₄ read (x)

2. T₂ write (y) and T₄ read (y)

In first case T₁ is committed first then read by T₄ then T₄ is committed so this is recoverable.

In second case T₂ is committed first then read by T₄ then T₄ is committed so this is also recoverable so schedule S is recoverable.

Hence, the correct answer is (C).

The clock rate or clock speed typically refers to the frequency at which the clock generator of a processor can generate pulses, which are used to synchronize the operations of its components, and is used as an indicator of the processor's speed.

For P₁:

Pipeline time =max(1.5,2,1 and 0.5) n\sec

=2 n\sec

Clock rate =1/2n\sec

=500 MHz

For P₂:

Pipeline time =max(1, 2.5, 1.5, 2 and 1) n\sec

= 2.5 n\sec

Clock rate = 1/2.5 n\sec

= 400 MHz

So, processor P₁ faster than processor P₂.

Hence, the correct option is (A).

As we know that,

For a general 4 to 1 multiplier as shown, with S₁ as MSB and S₂ as LSB, the output expression is written as: 
F= 

i.e. I₀ will be the output when the select inputs are 00,I₁ will be the output when the select Inputs are 01, and so on

We get it as,

OR

Hence, the correct options are (A) and (D).

Correct option: B

The production S → aSd | aAd adds one a at the left and one d at the right for each application of the recursion, so it enforces an equal number of a's and d's. The base production aAd ensures at least one pair of a and d, so n >= 1.

The nonterminal A → bAc | bc generates strings with equal numbers of b's and c's and at least one such pair, so it enforces b^mc^{m with m >= 1. Combining these gives strings of the form anbmcmdn, so this grammar exactly generates L.}

Option C is incorrect because S → A and A → aBd fix exactly one a and one d around the b^mc^m part; that forces n = 1, so it cannot generate all strings of L where n may be greater than 1.

Option A is incorrect because A → aAb | ε and B → cBd | cd produce aⁿbⁿc^md^m with possible n = 0, tying the number of b's to the number of a's rather than to the number of c's; thus it does not match the required pattern.

Option D is incorrect because the alternative A → ad (in addition to bAc and bc) allows generation of strings without the required b^mc^m block or with mismatched structure, producing strings not in L.

Therefore, only option B generates the language aⁿb^mc^mdⁿ for n, m >= 1.

According to Church Turing Thesis, every problem which can be logically solved using some algorithm can be implemented on a Turing machine.

So, now these statements can be analyzed:

S1: For calculating GCD, we have an algorithm, so it can be implemented on a Turing machine.

S2: Copying string have algorithm, so it can be implemented on a Turing machine.

S3: Predicting a tomorrow's match result is undecidable and it has no algorithm. So, it cannot be implemented on a Turing machine.

S4: Calculating surface area of a cone has also logic behind it, a formula has to be used, so it can be implemented on a Turing machine.

Hence, the correct option is (B).

Option (A): TRUE

In the polling method, the CPU polls each device to check status bits to find out if the device has raised any interrupt. It is a software method.

Option (B): FALSE

In the vectored interrupt, a vector address is given to the CPU to identify the source of the interrupt. For example, in 8085μP, RST 4.5 is a vectored interrupt. Here, 4.5 is the interrupt vector which gives the vector address as 4.5 × 8 = (36)₁₀ = (0024)₁₆.

Option (C): FALSE

In DMA mode, either the CPU or the DMA controller would gain control of the system bus time, but not both. Accordingly, the CPU would be either in a busy state or a Hold state. It would be in a busy state until the I/O device prepares the data and would go to Hold state when the I/O device starts transferring data to the main memory via the DMA controller.

Option (D): TRUE

In the daisy-chaining method of interrupt handling, the devices are connected serially in such a manner that the nearest device to the CPU has the highest priority, followed by the next device and so on.

Hence, the correct options are (B) and (C).

Given,

Simplify the function, we get

We know that,

( a + b )² = a² + 2ab + b²

As we know,

Differentiating with respect to x, we get

f′(x) = 2^4x × 4 × ln⁡2 + 2^3x × 3 × ln⁡2

f′(x) = ( 4 × 2^4x + 3 × 2^3x) ln⁡2

Hence, the correct option is (D).

The simplified SOP (Sum of Product) form of the boolean expression by the two methods:

Method ( 1 ): By Minimization

Method ( 2 ): By Truth Table

Hence, the correct option is (B).

The largest equivalence relation on S is a relation that contains all pairs (x, y), where x and y are elements of S. The number of such (ordered) pairs is n x n = n²
The smallest equivalence relation on S is such a relation that every element x of S is only equivalent to itself. Thus, this relation will have n ordered pairs.
Note that any equivalence relation must be reflexive (i.e. each element must be equivalent to itself), so we cannot have 0 or 1 pair in this case.
So, the smallest equivalence relation for set of n elements has n ordered pair and largest equivalence relation has n² ordered pairs.

Pipelining: Pipelining is a process of arrangement of hardware elements of the CPU such that its overall performance is increased. Simultaneous execution of more than one instruction takes place in a pipelined processor.

In a Non-Pipelining system, processes like decoding, fetching, execution and writing memory are merged into a single unit or a single step.

For non-pipelining system,

Average execution time = (20+10+30+25+40+15) = 140 ns

For pipelining system,

Average execution time = max[ (20+5), (10+5), (30+5), (25+5), (40+5), (15+5) ]

Cycle (ideal condition) = 45 ns

Speed up = 

Speed up factor 
 
= 140/45

Hence, the correct answer is 3.11.