GATE Mock Test Computer Science Engineering (CSE) - 5 - Computer Science Engineering (CSE) MCQ

Given,

Profit of A in 2017=8000

Profit of B in 2017=20000

According to question:

Profit of A in 
 2018=8000×(150100)

= Rs. 12000

Profit of B in 
 2018=20000×(200100)

= Rs. 40000

Total profit made by A and B in 2018= Rs. (12000+40000)

= Rs. 52000

∴ Total profit made by A and B in 2018 is Rs. 52000.

The error lies in part (A) of the given sentence.

Meaning of 'rarely': not often.

Inversion of sentences: sentences structure becomes: Verb+ subject.

Inversion of sentences takes place following negative adverbs and adverb phrases: hardly, never, rarely, seldom, scarcely, only later, on no account etc.

Therefore, we can see that the above-given sentence starts with 'rarely'.

So, inversion of sentence structure will take place and 'I have' will be replaced by 'have I'.

Thus, part (A) is having grammatical errors.

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) x diameter²

Area of semi-circle with diameter 

Area of semi-circle with diameter 

Area of semi-circle with diameter 

Hence, the correct option is (D).

Given,

Total cost = Rs. 60000

Amount spent on labour 
= 90/360 × 60000 = Rs. 15000

Amount spent on steel 
= 54/360 × 60000 = Rs. 9000

Difference = 15000 − 9000 = Rs. 6000

Let the amount spent on labour exceeds the amount spent on steel by x% of the total cost.

Then,

⇒ x × 600 = 6000

 ⇒ x = 6000/600

x = 10

The amount spent on labour exceeds the amount spent on steel by 10% of the total cost.

'The audience loudly cheered the politician’s speech.'

In the active form, the subject and the object will get interchanged. So, 'the politician's speech' will become the object. The passive form contains 'was' which means the active form must be in the past tense. 

Option (A) is in the present tense.

Option (B) changes the meaning.

Option (D) is in the past perfect tense.

Rules of Conversion from Passive to Active Voice:

1. Identify the subject, the verb and the object: S+V+O.
2. Change the subject into an object.
3. Omit the suitable helping verb or auxiliary verb.
4. Change the past participle to a simple past tense form of the verb.
5. Omit the preposition "by“.
6. Change the object into a subject.

Silence refers to 'no speech'. Babble means continuous, and sometimes incoherent speech.

Admonish means taking someone to task for wrong doing. This is similar to warning or cautioning someone for some wrong doing.

Going through the passage,
(1) is eliminated as it is true as per line,' Inward humility is achievement, outward courtesy is virtue' of the passage.
(2) is true as per the line ,'Refining your own nature is achievement. Refining your own person is virtue'. Hence, eliminated.
(3) can also be inferred in the last lines.
(4) is the only statement that can not be inferred.

Remember that you can only add fractions with the same denominator.
Rearrange 1/3⁸ so that it can be added to 1/3⁹ . That is, try and turn  into a fraction with 3⁹ in the denominator.
Multiply 1/3⁸ by 3/3 to get

Solving,

Canceling out a factor of 3 gives,

N is a number such that N = a^p x b^q x c^r ... where a, b, c are prime number and p, q, r are positive integers.
So, sum = ;
Here a = 3, p = 129,
So required sum =.

Given,

As we know ,

Let sec⁡ x = u

tan⁡ x sec ⁡xdx = du

Thus, 

As we know,

Hence, the correct option is (B).

Given,

T(n) = 1 if n = 1
T(n) = T(n/2) + n

Now,

= n/2^k

= 1

As we know,

⇒ log n = k

= n(2) + 1

= 2n

∴T(n) = O(n)

Hence, the correct option is (C).

As we know,

n should be odd then deadlock will be possible if k should be = (n+2)/2.

Now,

n =13 then k will be floor (13+2)/2 = floor 15/2 = 7

(n is taken as 13 because as mentioned in the question Resources are R₀ to R₁₂)

So, n = 13 and k = 7 then deadlock is possible.

Case I:

Even processes request even resources and odd processes request odd resources. i.e.

Even Processes:

P₀ requests R₀ and R₂

P₂ requests R₂ and R₄ and so on.

Odd Processes:

P₁ requests R₉ and R₁₁

P₃ requests R₇ and R₉ and so on.

So, there would not be any scenario of deadlock.

Case II:

All process request even resources:

Even Processes:

P₀ requests R₀ and R₂

P₂ requests R₂ and R₄

P₄ requests R₄ and R₆

P6 requests R₆ and R₈ (Deadlock arise)

Odd Processes:

P₁ requests R₁₂ and R₁₀

P₃ requests R₁₀ and R₈

P₅ requests R₈ and R₆ (Deadlock arise)

Hence, the correct answer is 7.

Given,

Using the formula,

AA^-1 = 1

On comparing both matrix,

⇒ 3b = 1
⇒ b = 1/3

⇒ 2a − 0.1b = 0

⇒ 2a = 0.1b
⇒ a = 0.1b/2 …… (i)

Putting value of b in equation (i), we get
a = 1/60

Therefore, a = 1/60,b = 1/3

Hence, the correct option is (B).

According to the question,

Let the 4 inputs be (w,x,y,z) and output be F.

Min-terms: 5, 6, 7, 8, 9

Don't care: 10, 11, 12, 13, 14, 15

Now, the K- map:

So, F = w + xz + xy
F = x(y + z) + w
Now, realizing this expression using min number of gates:

So, the minimum number of gates required is 3.
Hence, the correct answer is 3.

According to the question,

If control input is '0', then

If control input is '1', then

So, the circuit given in the question acts as both half adder and half subtractor.

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

Given,

(213)₆ = (a3)_b

Converting the both in decimal base to form an equation, we get

⇒ 2 × 6 × 6 + 1 ×  6 + 3 = ab + 3

⇒ 72 + 6 + 3 = ab + 3

⇒ ab = 78

Now, by finding the factors of 78, we get

= 2 × 3 × 13

Now,

We have two equation which should be followed while deciding the pairs of a & b:

(I) a < b

(II) b > 3

So, the possible solutions for the answer is only 4 l.e.

= (1, 78), (2, 39), (3, 26), (6, 13)

Hence, the correct answer is 4.

Given,

• Virtual address (VA) = 39 bits
• Page size = 4 KB
• Physical address (PA) = 2 GB
• Page table entry size (PTE) = 8 B
• Three level pages tables with address division (9, 9, 9, 12)

Three level pages tables with address division (9, 9, 9, 12) means:

• 9 most significant bits for indexing into the level−1 (outer level).
• 9 bits for the level−2 index.
• 9 bits for the level−3 index.
• 12 bits for the offset within a page.

The entries of the level−1 page table are pointers to a level−2 page table, the entries of the level−2 page table are pointers to a level−3 page table, and the entries of the level−3 page table is PTEs that contain actual frame number where our desired word resides.

9 bits for a level means 2⁹ entries in the one-page table of that level.

For our process P:

P is using 2 GB of its VM. The rest of its VM is unused.

2 GB VM will have 2 GB/4 KB = 2¹⁹ Pages.

But level −3 page table has only 2⁹ entries. So, one-page table of level −3 can point to 2⁹ pages of VM only. So, we need 2¹⁰ level−3 page tables of process P.

So, at level−3, we have 2¹⁰ page tables. So, we need 2¹⁰ entries in level−2. But level −2 page table has only 2⁹ entries, so, one-page table of level −2 can only point to 2⁹ page tables of level−3. So, we need 2 level−2 page tables.

So, we need 1 level−1 page table to point to level−2 page tables.

So, for process P, we need only 1 level−1 page table, 2 level−2 page tables, and 2¹⁰ level−3 page tables.

Note that, all the page tables, at every level, have the same size which is 2⁹ × 8 B = 2¹² B = 4 KB (because every page table at every level has 2⁹ entries and page table entry size at every level is 8 B ).

So, in total, we need 1 + 2 + 2¹⁰ page tables (1 level−1,2 level−2,2¹⁰ level−3, and each page table size is 4 KB.

So, total page tables size,

= 1027 × 4 KB

= 4108 KB

Hence, the correct option is (A).

Given,

As we know,

To obtain operand value from memory, the address field of an instruction is used by the CPU. In single address instruction, one of the operands is stored in the accumulator and the other operand may be either in register or memory. 

All the operations will be performed in the Accumulator register (AC).

The load operation is used to fetch the value from the register or memory to the accumulator.

The store operation is used to store the value from the accumulator to the register or memory.

The one address instructions for the given equations are:​

Hence, the correct answer is 8.

Lossless join: It guarantees that the spurious tuple generation problem does not occur with respect to the relation schemas created after decomposition.

Dependency preservation: It ensures that each functional dependency is represented in some individual relation resulting after decomposition.

3NF:

• Third normal form is based on the concept of transitive dependency. A functional dependency X→Y in a relation schema R is a transitive dependency if there exists a set of attributed Z in R that is neither a candidate key nor a subset of any key of = R.
• A relation is in 3NF if it satisfies 2NF and no prime attribute of R is transitively dependent on the primary key.
• If X → A is a functional dependency, then A should be prime attribute or X should be a candidate key.
• Loss less join and dependency preservation is always possible in 3NF. 3NF decomposition is always lossless join and dependency preserving.

BCNF:

• It stands for boyce codd normal form.
• A relation R is in BCNF if whenever a non-trivial functional dependency X → A holds in R, then X is a superkey of R. Any relation with two attributes is always in BCNF. Because when a relation contains only two attributes than one attributes determines another and left side of the functional dependency will always be a candidate key in that case.
• BCNF is not always dependency preserving.

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

As we know,

The statement S₁ is true because ∀i,(i ≤ i).

The statement S₂ is true because R₁∪R₂ will contain each element (i,i) which is present in R₁ and R₂.

The statement S₃ is false because there can be some pair which is present in a relation R whose symmetric pair is not present in its superset.

The statement S₄ is false because the smallest relation which is irreflexive contains 0 element. A null set is the smallest relation which is irreflexive.

Therefore, both S₃ and S₄ are false.

Hence, the correct option is (A).

The size of the micro-instruction is 20.

The given micro-instruction consists of the following three subfields− F₁, F₂, F₃, one condition field − CD, one branch field − BR, one address field ADF. Since each sub-field has seven distinct operations, a minimum of 3 bits is required to indicate all of them CD has four bits.

So, 2 bits are sufficient to indicate all four status bits BR has four options. So, 2 bits are required here. To address all the 128 memory locations, 7 bits are required. 

Total number of bits required for this micro-instruction,

= 3 + 3 + 3 + 2 + 2 + 7

= 20

Hence, the correct option is (B).

There are various routes possible from P to U.

First route from P to U is:

1. P → S → U: COST = 4 + 2 = 6

2. P → R → U:  COST = 1 + 2 = 3

3. P → R → S → U: COST = 1 + 1 + 2 =  4

4. P → R → T → U: COST = 1+ 8 + 3 = 12

5. P → Q → T → U: COST = 3 + 6 + 3 = 12

6. P → Q → R → U: COST = 3 + 7 + 2 = 12

7. P → Q → R → T → U: COST = 3 + 7 + 8 + 3 = 21

So, minimum cost is 3. Shortest route from P to U is 3.

Hence, the correct option is (A).

TM has read/write head, can go in both directions, can have multiple tapes.

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

Finite Automata(FA):

The first phase of the compiler is called lexical analysis or scanning. The lexical analyzer reads the stream of characters making up the source program and groups the characters into a meaningful sequence called lexemes.

For each lexeme, the lexical analyzer produces tokens as output for the parser and tokens are expressed in regular expressions.

So, a simple finite automaton is sufficient for it.

Phases in a compiler:

Hence, the correct option is (A).

As we know,

NPDA → Non-deterministic pushdown automaton.

DPDA → Deterministic pushdown automaton.

DCFL → Deterministic context free language.

CFL → Context free language.

Option (A): False

L = pⁿqⁿr^m ∪  p^mqⁿrⁿ is accepted by NDPA and so it is a CFL, but it is not accepted DPDA and so it is not a DCFL.

Option (B): True

DCFL is closed under complementation. Therefore if L is DCFL then L is also DCFL.

Option (C): False

L = w^r is a CFL but not a DCFL. It can be derived from the following grammar.

S → aSa | bSb | ∈

Option (D): False

L = pⁿqⁿr^m ∩ p^mqⁿrⁿ ∣ m,n > 0.L = pnqnrn it is not accepted by a deterministic pushdown automaton and so it is not a DCFL.

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

Indirect addressing mode refers to the address field refer to the address of a word in memory.

LOAD DIRECT 150:

Above instruction points to memory location 150 which contains 230.

LOAD INDIRECT 108:

Above instruction points to memory location 150 which contains 230.

LOAD INDIRECT 150:

Above instruction points to memory location 230 which contains 405.

LOAD DIRECT 108:

Above instruction points to memory location 108 which contains 150.

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

It is given that, each 5 computer needs directly connected with each router. 

Let c1 to c5 be the 5 computer needs directly connected with each router r1 to r5. Now,

c1 → r1, r2, r3, r4, r5

c2 → r1, r2, r3, r4, r5

c3 → r1, r2, r3, r4, r5

c4 → r1, r2, r3, r4, r5

c5 → 1, r2, r3,r 4, r5

c6 → 1

c7 → r2

c8 → r3

c9 → r4

c10 → r5

So, 25 connections + now remaining 5 computer, each connected to 5 different routers, so 5 connections =30 connections.

Now, any pick of 5 computers will have a direct connection to all the 5 routers.

Hence, the correct option is (C).

As we know that,

The functional table is mentioned below:

The general output equation is:

The truth table for a full adder is as shown:

The expression for the sum bit and the output carry will be:

S = ∑ m(3, 5, 6, 7) and

C_out = ∑ m(1, 2, 4, 7)

Another equation for C_out  is:

Now,
Given S₁ = A and S₀ = B. The output expression becomes:

The equation for C_out  is:

Comparing equations (iii) and (iv), we get:
I₀ = 0, I₁ = C_in, I₂ = C_in, I₃ = 1

Hence, the correct option is (A).

A tree is max-heap if data at every node in the tree is greater than or equal to its children’s data.

For max heap we will compare parent node (i) with its left-child (2 × i) and rightchild (2 × i + 1) :

• In first option node (2) < node (5) which is violating the max-heap property.
• In second option node node (2) <  node (5) which is violating the max-heap property.
• In the third option, there is no violation.
• In fourth option node (3) < node (7) which is violating the maxheap property.

Hence, the correct option is (C).