GATE Mock Test Computer Science Engineering (CSE) - 1 - Computer Science Engineering (CSE) MCQ

Profit of A and B together in 2015 = 6000 + 16000 = Rs. 22,000

Profit of A and B together in 2017 = 8000 + 20,000 = Rs. 28,000

Required difference = Rs. 28,000 − 22,000 = Rs. 6,000

∴ The difference between the profit of A and B together in 2015 and the profit of A and B together in 2017 is Rs. 6000

 'Guinea pig' is term used to refer any person or thing used in an experiment. As first part of sentence refers the experiments as quixotic (impractical or daring), so perfect word suited to this blank is 'guinea pig', which is used in labs for testing on biological experiments.

Article 'a' is used before a word beginning with the consonant sound of  'yu' but begins with a vowel letter. For example, a university, a union, a useful thing, etc.
Such a rule doesn't apply to 'an' and 'the'. 
'few' on the other hand, is not an article.
Therefore, given the general idea of any university here, article 'a' is to be used.

Let original price = p, and original sales = s. Therefore, original gross receipts = ps.
Let new price = 0.80p, and new sales = 1.15s. Therefore, new gross receipts = 0.92ps.
Gross receipts are only 92% of what they were. Hence, gross receipts decrease by 8%.

Given,

AB = p cm and AC = q cm

∵ Area of right-angled triangle  = 1/2 ×  Base × Height

Area of triangle 

 ABC = pq/2

Applying Pythagoras theorem in △ABC,

BC² = AB² + AC²

⇒ BC² = (p² + q²)

∵ Area of semi-circle = (π/8) × (diameter)²

Area of semi-circle with diameter AB = (π/8) × (p)² = πp²/8 cm²

Area of semi-circle with diameter AC = (π/8) × (q)² = πq²/8 cm²

Area of semi-circle with diameter BC = (π/8) × (p² + q²) = π(p² + q²)/8 cm²

∴ Area of shaded region 

= (pq/2) + (πp²/8) + (πq²/8) − [π(p² + q²)/8]

=(pq/2)cm²

Given,

Total number of persons infected in Maharashtra =30,00,000

From the data given in question, we have:

Required average = (7,50,000 + 6,00,000)/2

= 13,50,000/2

= 6,75,000

∴ The average number of people infected from COVID-19 in Nagpur and Nasik is 6,75,000.

The correct answer is the first option as it is stated in the passage that most Reality TV shows centre on two common motivators: fame and money.

Choices 2 and 4 are not supported by the passage because passage is about Reality TV stars and not Reality TV.

Choice 3 is incorrect because the paragraph states that some Reality TV stars manage to parlay their fifteen minutes of fame into celebrity

Option 2: Rose said that she was very busy.

This option correctly changes the sentence from direct speech to indirect speech without changing the tense or meaning of the original sentence. It is the most appropriate and accurate representation of the given sentence.

Since fuel consumption/litre is asked and not total fuel consumed, only average speed is relevant. Maximum efficiency comes at 45 km/hr. So, least fuel was consumed per litre in lap Q.

Abelian Group: Let {G=e, a, b} where e is identity. The operation 'o' is defined by the following composition table. Then(G, o) is called Abelian if it follows the following property:

1. Closure Property
2. Associativity
3. Existence of Identity
4. Existence of Inverse
5. Commutativity

Cyclic Group: A group a is said to be cyclic if it contains an element 'a' such that every element of G can be represented as some integral power of 'a'. The element 'a' is then called a generator of G, and G is denoted by <a> (or [a]).

Theorem:

(i) All cyclic groups are Abelian, but an Abelian group is not necessarily cyclic.

(ii) The order of a cyclic group is the same as the order of its generator. 

Thus it is clear that A and B both are true.

Here, C₃ = 3 + j5
For real periodic signal,
C_-k = C_k*
Thus, C_-3 = 3 - j5

 As we know,

Since starting on disk starts from <1200,9,40> so no. of sectors left on 9^th  surface is 24.

So, on 9^th  surface total storage of "12288 B" is possible.

Now, a part from 9^th  surface, on cylinder no. 1200 only 6 surface is left.

To storage possible on these 6 surfaces are = 6 × 2⁶ × 2⁹→ Storage on each sector.

No. of sectors on each surface

= 196608 B

So, total on cylinder no. 1200, storage possible is:

⇒ 196608 + 12288

= 208896 B

So, since the file size is 42797 kB and out of which 208896 B are stored on cylinder no. 1200. So, we are left only with only 43615232 B.

Since in 1 cylinder, storage possible is = 2⁴ × 2⁶ × 2⁹ B

= 524288/B

So, we need about = 43615232 B/524288 B

= 83.189 more cylinders

So, we'll need the [1284^th Cylinder] to completely store the file. C₀₂ after 1283^rd  cylinder we will left with data which will need 189.

Sum of eigen values = Sum of diagonal elements (Trace) = 12
Product of eigen values = Value of determinant = 40
Now check by options, only (4) satisfies these conditions.
Hence, eigenvalues are 1, 2, 4 and 5.

Given,

The values of a node of a tree are:

1,3,7,15,29

As we know,

In worst case, tree might be skewed tree and in that case, there will be 5 levels and root will be at 0. So, the levels can be like this.

1 - Level 0

3 - Level 1

7 - Level 2

15 - Level 3

29 - Level 4

The maximum value of level number possible is 4

We must undo log record 6 to set B to 10000 and then redo log records 2 and 3 because system fails before commit operation. So, we need to undo active transactions (T2) and redo committed transactions (T1).
Note: Here we are not using checkpoints.

Checkpoint : Checkpoint is a mechanism where all the previous logs are removed from the system and stored permanently in a storage disk. Checkpoint declares a point before which the DBMS was in consistent state, and all the transactions were committed.

Recovery: When a system with concurrent transactions crashes and recovers, it behaves in the following manner:
 The recovery system reads the logs backwards from the end to the last checkpoint.
 It maintains two lists, an undo-list and a redo-list.
 If the recovery system sees a log with <Tn, start> and < Tn, Commit> or just <Tn, Commit >, it puts the transaction in the redo-list.
 If the recovery system sees a log with <Tn, start> but no commit or abort log found, it puts the transaction in undo-list.All the transactions in the undo-list are then undone and their logs are removed. All the transactions in the redo-list and their previous logs are removed and then redone before saving their logs.

Multiple threads of the same process share other resources of process except register, stack and stack pointer. In particular, a process is generally considered to consist of a set of threads sharing an address space, heap, static data, code segments and file descriptors.

Thread of the same process doesn't share program counter (register), stack, registers

On the basis of computational power, DFA and NFA are both equally powerful. Since we have an algorithm to convert NFA to DFA, and DFA is by default an NFA, both of them have equal computational power.

DFA is said to be a specific case of NFA and for every NFA that exists for a given language, an equivalent DFA also exists.

The computational power of other mechanisms:

• Power of DFA = Power of NFA.
• Power of NPDA (Non-deterministic Push-down Automata) > Power of DPDA (Deterministic Push-down Automata) i.e. language accepted by DPDA is a subset of the language accepted by NPDA.
• Power of NTM (Non-deterministic Turing Machine), Multi-Tape TM and DTM (Deterministic TM) are the same since every NTM can be converted to corresponding DTM.

Given,

(2x)^2y = 4e^{2x − 2y}

log_b⁡(M^p) = plog_b⁡(M)

log_b⁡(MN) = log_b⁡(M) + log_b⁡(N)

Taking log both sides, we get

ln⁡ (2x)^2y = ln⁡ 4e^{2x − 2y}

⇒ ln⁡ (2x)^2y = ln ⁡4+ln ⁡(2x − 2y)

⇒ 2yln ⁡(2x) = ln ⁡2² + (2x−2y)

⇒ 2yln ⁡(2x) = 2ln ⁡2+(2x−2y)… (1) 

⇒ yln⁡ (2x) = ln⁡ 2+(x−y)

⇒ y(1+ln⁡ 2x) = x+ln ⁡2

∴ y = x + ln⁡ 2/1 + ln⁡ (2x)…(2)

As we know,

d/dx(ln⁡x) = 1/x

If f and g are both differentiable, then,

Differentiating equation (1) with respect to x, we get,

Putting equation (2) in equation (3), we get,

Taking the first 22 bits of 135.46.63.10 as network address, we have 135.46.60.0. The bit pattern of 135.46.63.10 is 10000111.00101110.00111111.00001010.
When we perform the bit and operation with 22 leading bit 1s and 10 bit 0s, it is equivalent of making the last 10 bit zero.
We get the following network address bit pattern: 10000111.00101110.00111100.00000000.
The first two bytes are not changed. The 3^rd type changes from 63 to 60 while the 4^th byte become zero. Match with network address in the routing table.
The 2^nd row matches. The router will forward the packet to Interface 1.
Taking the first 23 bits of the IP address 192.53.40.7 as network address, we have 192.53.40.0.
The packet will be forwarded to Router 1.

As we know,

{PT}⁺ = {PTR}

Therefore, option (A) is false.

{ST}⁺ = {STUPR}

Since Q is not present in {ST}⁺, so, option (B) is false.

{QT}⁺ = {PQRSTU}

Therefore, option (C) is true.

{PQ}⁺ = {PQRSTU}

In option (D), S is not present.

∴ It is false.

Now,

You might feel that option (D) is also correct but in question, it is already defined that it should contain all are derived attributes.

An operating system is built to directly communicate with the hardware of the system. An OS instructs the hardware to perform intended operations and gives commands for the same.

Given,

Packet Size = 2 KB = 2¹¹ bytes = 2¹⁴ bits

Bandwidth = 1 Gbps = 10⁹ bps

Lenght of link 1= Lenght of link 2 = 500 m

Propagation speed = 2.5 × 10⁵ km/s = 2.5 × 10⁸ m/s

Queuing delay at R = 10 microseconds

Diagram:

As we know,

Total transfer delay = transmission delay from A to link 1+ propogation delay from A to R+ queuing delay at R+ transmission delay from R to link 2+ propagation delay from R to B

Now,

T_trans  (from A to link 1)

 = T_trans  (from R to link 2)

= 2¹⁴ ÷ 10⁹ bps

= 16.384 microseconds

T_prop  (from A to R)

= T_prop  (form R to B)

= (500 m) ÷ (2.5 × 10⁸ m/s)

= 2 microseconds

Total transfer delay = 16.384+0.2+10+16.384+2

= 46.768 microseconds

In indirect addressing:
Opcode 200

In relative addressing,
effective address = (200 + value of PC) = 200 + 300 = 500

Cache Memory is a special very high-speed memory. It is used to speed up and synchronize with a high-speed CPU. Cache memory is costlier than main memory or disk memory but more economical than CPU registers. Cache memory is an extremely fast memory type that acts as a buffer between RAM and the CPU. It holds frequently requested data and instructions so that they are immediately available to the CPU when needed.

The performance of cache memory is frequently measured in terms of a quantity called hit ratio.

{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

L₁ → recursive

L₂,L₃ → recursively enumerable but not  recursive.

So L₁ can be recursive enumerable.

RE - RE = RE

So L₁ → L₃ can be recursive enumerable.

Given,

Number of processes = 3

The available instance of resources = N

Max needs of process (X,Y,Z) = (5,11,9)

Now,

Deadlock can occur: needed (requested) resource > available resources

Max resource per process to be in deadlock = needed −1

Nmax = (5 − 1) + (11− 1) + (9 − 1) = 4 + 10 + 8

= 22

∴ ≤ 22 will leads to deadlock.

∴ 20,21,22 will leads to deadlock.

Number of possible strings = 2 x 2 x 2 x 2…..n times = 2ⁿ.