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

Given,

X² + 4Y² = 4XY

⇒ X² + 4Y² − 4XY = 0

⇒ (X − 2Y)² = 0

⇒ X − 2Y = 0

⇒ X = 2Y
⇒ X/Y = 2/1

On cubing both the sides,

⇒ X³ : Y³ = 8 : 1

Hence, the correct option is (C).

In the given sentence, there is a conjunction 'but' and the second clause does not contain a verb. Therefore, the answer must contain a verb. Further, the second clause also does not contain a subject. Thus, the answer must also include a subject that matches the subject in the first sentence, that is, the answer must contain a pronoun that matches with 'the advisor' which must be a third-person singular pronoun. Also, since the tense of the first clause is present tense, the answer must also be in the present tense and since the action in both the clauses is the same, the verb must be a form of the verb 'do'.

Option (A) is incorrect. In 'he will', 'will' is in the simple future tense, and hence, is incorrect. Further, 'will' is also not a form of 'do'.

Option (B) is incorrect. In 'they do', although 'do' is present, 'they' is a third-person plural pronoun. Thus, it is incorrect

Option (C) is correct. 'he does' contains a third person pronoun 'he'. Further, the verb 'does' is the simple present form of 'do' for third-person singular pronouns. Thus, it is the correct answer.

Option (D) is incorrect. 'they are' contains a third person plural pronoun 'they', making it incorrect. Further, the verb 'are' is also not a form of the verb 'do'. Thus, it is incorrect.

Hence, the correct option is (C).

Given,

Total number of teachers = 1800

Chemistry = 23%

English = 27%

Biology = 12%

Now,

(Chemistry + English + Biology) % of Total number of teachers = (23 + 27 + 12)% of 1800

= 62% of 1800
= 62/100 × 1800

= 1116

Hence, the correct option is (B).

Given,

Word = SOFTWARE

Arrangement of n letter of a word.

Case 1: If there were no repeating letters = n!

Case 2: If there were "r" letter of one kind = n!/r!

Keeping all the vowel at one place and consider it as one vowel = SFTWR(OAE)

Here, n = 6 and number of vowel = 3

All letter can be arranged in 6! ways and vowel can be arranged in 3! ways.

Total number of ways to arrange the word,

= 6! × 3!

= 6 × 5 × 4 × 3 × 2 × 1 × 3 × 2 × 1

= 4320

∴ Total number of ways to arrange the word = 4320

Hence, the correct option is (C).

If you 'call a spade a spade', you speak honestly and directly about a subject even if it offends people.

Swimming is done in a large water body. All the other words relate to activities involving smaller volume of water.

Break, hiatus and pause, all mean a temporary halt. End is the final halt.

Suppose that 6% per annum be the rate of interest for x years, then the 4% per annum be the rate of interest for (10 - x) years.
Interest paid by Kishan Singh = Rs. 48,000
Now,

48 = 6x + 40 - 4x
2x = 8
x = 4 years
It means Kishan Singh availed the discount after 4 years.
Hence, option (3) is correct.

The first drop is 90 metres. After this, the ball will rise by 72 metres and falls by 72 metres, now this process will be continue in the form of an infinite GP with common ratio 0.8 and first term 72.
The required answer will be got by 

= 90 + 2(360)
= 810 m

Suppose that a man starts from a city. The condition is that there should be no repetition of path during his journey. So, two paths are restricted for one pass only.
Thus, the total number of possible cases = 12.
Travelling with each node in one direction only will give 6 possible routes. There are another 6 routes, when travelling in opposite direction from the first one. So, the total possible ways are 12 only.

Important Points:

. ≡ ∧

+ ≡ ∨

′ ≡¬

We know that,

(A ∧ B) ∧ (¬A → B) ≡ A ⋅ B ⋅ (A′ + B) ≡ A ⋅ B ≡ A ∧ B

Statement 1: A ∧ B ≡ A . B

(A ∧ B) → (A ∧ B) ≡ ¬(A ∧ B) ∨ (A ∧ B) ≡ X′ + X ≡ 1 ≡ TRUE

Statement 2: A ∨ B ≡ A + B

(A ∧ B) → (A ∨ B) ≡ (A ⋅ B)′ + (A + B) ≡ (A′ + B′) + (A + B) ≡ A′ + A + B′ + B ≡ 1 ≡  TRUE

Statement 3: ¬(A∨ B) ≡ A′ . B′(A ∧ B) → (¬(A∨B)) ≡ (A . B)′+ A′ ⋅ B′ ≡ (A′ + B′) + A′ . B′

≡ A′ + B′ ≡ (A . B)′ ≡ FALSE (if A =TRUE and B =TRUE)

Statement 4: TRUE

(A ∧ B) → TRUE ≡ ¬(A ∧ B) ∨ 1 ≡ 1 ≡ TRUE

Statement 5: FALSE

(A ∧ B) →  FALSE  ≡¬(A ∧ B) ∨ 0 ≡ ¬(A ∧ B) ≡  FALSE (if A = TRUE and B = TRUE )

So, the statements I, II and IV are logically implied by (A ∧ B) ∧ (¬A → B).
Hence, the correct option is (D).

When an I/O device completes an I/O operation, the following sequence of hardware events occurs:

1. The device issues an interrupt signal to the processor.
2. The processor finishes the execution of the current instruction before responding to the interrupt.
3. The processor test for an interrupt and sends an acknowledgment signal to the device that issued the interrupt.
4. The processor now needs to save information needed to resume the current program at the point of an interrupt.
5. information saved to be is the status of the processor and the location of the next instruction to be executed.
6. The processor now loads the program counter with the entry location of the interrupt handling program.
7. The interrupt handler next processes the interrupt.
8. After interrupt processing, saved register values are retrieved from the stack and restored and the next instruction will be executed.

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

Given,

Input burst = 50 KB/sec for 10 sec and 10 KB/sec for 60 sec

Output rate = 5 KB/sec

Total time = 60 sec

Total Length = 500 KB + 500 KB

= 1000 KB.

Will, it is in 60 sec,

Now, output rate = 5 KB/sec

It is in 60 sec.

⇒ 5 × 60 = 300 KB

In 60 sec we can transmit 300 KB.

We should have a bucket of size = 1000 KB − 300 KB

= 700 KB

Hence, the correct answer is 700.

XOR GATE:

Symbol:

Truth Table:

Output Equation: 
 Y = A ⊕ B =
As we know,
1. If B is always high, the output is the inverted value of the other input A, i.e. 
2. The output is low when both the inputs are the same.
3. The output is high when both the inputs are different.

 

Hence, the correct answer is 1.

If minimal DFA for any regular language L has n states then minimal DFA for the complement of L will also have n states.

DFA for language represented by the regular expression Σ^∗0011Σ^∗ where, Σ = {0,1} is as following below:

The complement of the above DFA will have the same number of states but there will be non-final states to final states and final states to non-final states.

The minimum number of states in the minimal DFA that recognizes 
 
 is 5. The given regular expression matches with all strings that contain substring 0011. The complement should match with all strings except the strings with 0011 as a substring.

Hence, the correct option is (B).

Given,

Applying R₂ → R₂ - R₁

Applying R₃ → R₃ - R₁

Applying R₃ → R₃ - R₁

Now, expanding from a11.

= (m − l)(n − l)(n + l − m − l)

= (m − l)(n − l)(n − m)

Hence, the correct option is (B).

We know that,

Range of given array:

[−10…5,7……15]

Old location (Given):

A[2][10]

A[−10][7] = 250

Number of rows = 5 − ( − 10) + 1

= 16

Number of columns = 15 − 7 + 1

= 9

Array size:

A_16×9

New range:

[0…15,0…8]

New location:

A[2 + 10][10 − 7] = A[12][3]

Size of data = 4

A[0][0] = 250

Formula:

A[n][m] = A[0][0] + ( Row size  × m + n) ×  Size of cell

Now,

A[12][3] = 250 + (16 × 3 + 12) × 4

A[12][3] = 250 + 60 × 4

A[12][3] = 490

Old location of A[ − 10][7] =  New location of A[12][3] = 490.

Hence, the correct answer is 490.

Sequence: Y, Z, X

Gantt chart:

Processs Table:

Average waiting time 
= 8/3
Sequence: X, Y, Z
Gantt chart:

Average waiting time 
 = 13/3
Sequence: X, Y, Z
Gantt chart:

Average waiting time 
 = 11/3
Sequence: X, Y, Z
Gantt chart:

Average waiting time 
 = 13/3
{Y, Z, X} is an optimal non-preemptive scheduling sequence that gives 8/3 average waiting time.
Formula used:
TAT=CT-AT
WT=TAT-BT

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

FCFS is the simplest disk scheduling algorithm. As the name suggests, this algorithm entertains requests in the order they arrive in the disk queue. The algorithm looks very fair and there is no starvation (all requests are serviced sequentially) but generally, it does not provide the fastest service.

Algorithm:

1. Let Request array represents an array storing indexes of tracks that have been requested in ascending order of their time of arrival. ‘head’ is the position of disk head.

2. Let us one by one take the tracks in default order and calculate the absolute distance of the track from the head.

3. Increment the total seek count with this distance.

4. Currently serviced track position now becomes the new head position.

5. Go to step 2 until all tracks in the request array have not been serviced.

Given,

A disk drive has cylinders = 200

They are numbered 0 to 199.

Disk requests come to the disk driver for cylinders 82, 170, 43, 140, 24, 16 and 190.

The driver is currently serving a request at a cylinder 50.

So, total seek time:

= (82 − 50) + (170 − 82) + (170 − 43) + (140 − 43) + (140 − 24) + (24 − 16) + (190 − 16)

= 642

Hence, the correct answer is 642.

Correct options: B and D.

Z implements (A ∧ B) ∨ (C ∧ D). By De Morgan's law, this can be written as ¬(¬(A ∧ B) ∧ ¬(C ∧ D)).

Using only 2-input NAND gates, compute N1 = NAND(A, B) and N2 = NAND(C, D) with two NAND gates (these produce ¬(A ∧ B) and ¬(C ∧ D) respectively).

Then compute Z = NAND(N1, N2) with a third NAND gate. This realizes ¬(¬(A ∧ B) ∧ ¬(C ∧ D)) = (A ∧ B) ∨ (C ∧ D), so 3 NAND gates suffice.

Two NAND gates are insufficient because with only two 2-input NAND gates you cannot both form the two intermediate signals N1 and N2 and also combine them; a third gate is required to combine the two intermediate outputs.

For timing, the first-level NANDs produce N1 and N2 after 10 ns. The final NAND produces Z another 10 ns after its inputs are stable. Therefore the total propagation delay is 20 ns, so option D is correct and option C is false.

In a carry look-ahead adder if we ignore the input size of a gate, then to add two n bit numbers using CLA the total no of AND and OR gate required. In a look-ahead carry generator, the carry generate function G_i and the carry propagate function P_i for inputs A_i and B_i are given by:

P_i = A_i ⊕ B_i and G_i = A_i × B_i

In the carry generator function first thing is that either the carry C_i is generated by the current bits i.e. by A_i and B_i therefore by G_i or it is generated on lower level and propagated to the higher level.

Therefore,

C₅ = G₅ + P₅G₄ + P₅P₄G₃ + P₅P₄P₃G₂ + P₅P₄P₃P₂G₁ + P₅P₄P₃P₂P₁G₀ + P₅P₄P₃P₂P₁P₀C_{− 1}

Hence, the correct option is (A).

The main advantage of Static RAM over Dynamic RAM is that SRAM is faster than DRAM which means it is faster in operation. SRAM can be used to create a speed-sensitive cache. SRAM only has medium power consumption. SRAM has a shorter cycle time since it does not require pausing between accesses.

Hence, the correct option is (D).

If the queue is implemented with a linked list, keeping track of a front pointer, Only rear pointers will change during an insertion into a non-empty queue.

A queue data structure can be used to implement the least recently used (LRU) page fault algorithm and Quicksort algorithm.

When the queue is empty both the pointers are initialized to NULL front ptp = rear ptp = NULL

Now, when a new node is inserted into the linked list both the front and a rear pointer will point to the new node.

After that, whenever a new node will be inserted the rear ptp will change and point to the new node.

As we know that ptp is a pointer.

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

ICMP error message should not be sent after receiving when:

• After another ICMP error message.
• After an IGMP message of any kind.
• After any fragment other than the first fragment from a fragmented IP datagram.

ICMP error message should be sent after receiving when:

• If DF ( do not fragment) bit is set and a router needs to fragment the datagram to send it further then the ICMP error message must be sent to the sender.
• If TTL is exceeded then also it should report to the sender by sending the ICMP error message.

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

Shortest remaining time, also known as shortest remaining time first (SRTF), is a scheduling method that is a pre-emptive version of the shortest job next schedule. In this scheduling algorithm, the process with the smallest amount of time remaining until completion is selected to execute. Since the currently executing process is the one with the shortest amount of time remaining by definition, and since that time should only reduce as execution progresses, processes will always run until they are complete or a new process is added that requires a smaller amount of time.
SRTF: Gantt chart:

Here, we know that
At time 0, P₁ is the only process, P₁ runs for 15 time units.
At time 15, P₂ arrives, but P₁ has the shortest remaining time. So, P₁ continues for 5 more time units.
At time 20,P₂ is the only process. So, it runs for 10 time units. at time 30, P₃ is the shortest remaining time process. So, it runs for 10 time units.
At time 40,P₂ runs as it is the only process. P₂ runs for 5 time units. At time 45, P₃ arrives, but P₂ has the shortest remaining time. So, P₂ continues for 10 more time units. P₂ completes its execution at time 55 .
So, order is P₀, P₂, P₁ and P₃.
LRTF: Gantt chart:

So, order is P₀, P₁, P₂ and P₃.
Hence, the correct option is (C).

The not null specification prohibits the insertion of a null value for the attribute.

The unique specification says that no two tuples in the relation have the same values on all the listed attributes (Here name is the listed attribute). Rita is repeated and here it is given UNIQUE(Name).

So, INSERT INTO Student VALUES(1005,Rita,EE ); will generate error.

Hence, the correct option is (D).

To count the no of strings we have to check for every length string:

For 0 length string − No string possible

For 1 length string − No string possible

For 2 length string − No string possible

For 3 length string − S → 0A1 → 0 B1 → {001,011}

For 4 length string − No string possible

For 5 length string − S → 0 A1 → 0BB11 → {00011, 00111, 01011, 01111}

For 6 length string − No string possible

Hence, the correct answer is 6.

Given,

Differentiating with respect to x, we get

Hence, the correct option is (C).

Output expression Q is equivalent to NAND gate.

NAND GATE:

Symbol:

As we know,
1. If A is always high, the output is the inverted value of the other input B,i.e. 
2. The output is low only when both the inputs are high.
3. It is a universal gate.
Hence, the correct option is (B).

For regular grammar and CFL grammar:

1. Equivalence: It can be checked by developed product automate for XOR function. If XOR = 0 ⇒ equal else not. Two context-free grammars are defined as being structurally equivalent if they generate the same sentences and assign similar parse trees (differing only in the labeling of the nodes) to each. It is argued that this type of equivalence is more significant than weak equivalence, which requires only that the same sentences be generated.

2. Ambiguity: Ambiguity is decidable for regular languages & grammar since they are deterministic in nature. A context-free language is called ambiguous if there is no unambiguous grammar to define that language and it is also called inherently ambiguous.

3. Regularity: It is the trivial problem. For context-free language. It is not closed in any of those problems.

So, Regular grammar is closed in all three problems given but CFL is not closed in anyone.

Hence, the correct option is (C).