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


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30 Questions MCQ Test GATE Computer Science Engineering(CSE) 2025 Mock Test Series - GATE Mock Test Computer Science Engineering (CSE) - 5

GATE Mock Test Computer Science Engineering (CSE) - 5 for Computer Science Engineering (CSE) 2024 is part of GATE Computer Science Engineering(CSE) 2025 Mock Test Series preparation. The GATE Mock Test Computer Science Engineering (CSE) - 5 questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The GATE Mock Test Computer Science Engineering (CSE) - 5 MCQs are made for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for GATE Mock Test Computer Science Engineering (CSE) - 5 below.
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GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 1

Direction: Study the bar graph carefully to answer the following question.

The bar graph represents the profit earned by two individuals A and B over 5 years.

If the profit made by A is 50% more and that of B is 100% more in 2018, as compare to 2017, then what is the total profit made by A and B together in 2018?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 1

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.

GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 2

Direction: In the following question, one part of the sentence may have an error. Find out which part of the sentence has an error. If the sentence is free from error, select the 'No error' option.

Rarely I have seen (A)/such a scene of (B)/ violence on the streets. (C)/ No error (D)

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 2

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.

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GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 3

In the figure given below, ABC is a right-angled triangle where ∠A=90∘,AB=p cm and AC=q cm. On the three sides as diameters semicircles are drawn as shown in the figure. The area of the shaded portion, in square cm, is:

 

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 3

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).

GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 4

Direction: Study the following pie chart carefully & answer the question given below it.

The pie chart given below shows the break-up of the cost of construction of a house (in degrees). Assuming that the total cost of construction is Rs. 60000, answer the question given below.

The amount spent on labour exceeds the amount spent on steel by:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 4

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.

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

Direction: Choose the option that is the active form of the sentence.

The politician’s speech was loudly cheered.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 5

'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.
GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 6

Which of the following is the antonym of the word SILENCE?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 6

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

GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 7

Which of the following is the synonym of the word ADMONISH?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 7

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

GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 8

Which of the following statements cannot be inferred from the passage?

Not stagnating moment to moment is called achievement. Inward humility is achievement, outward courtesy is virtue. Establishing myriad truth in your nature is achievement; the mind being essentially detached from thoughts is virtue. Not departing from one's essential nature is achievement; acting adaptively without being affected is virtue. Continuity moment to moment is achievement; balance and directness of mental activity are virtue. Refining your own nature is achievement. Refining your own person is virtue.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 8

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.

GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 9

 is equal to

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 9

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,

GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 10

What will be the sum of the factors of 3129?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 10

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

GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 11

Find the value of 

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 11

Given,

As we know ,

Let sec⁡ x = u

tan⁡ x sec ⁡xdx = du

Thus, 

As we know,

Hence, the correct option is (B).

GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 12

What is the value of following recurrence?

T(n) = 1 if n = 1

T(n) = T(n/2) + n else

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 12

Given,

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

Now,

= n/2k

= 1

As we know,

⇒ log n = k

= n(2) + 1

= 2n

∴T(n) = O(n)

Hence, the correct option is (C).

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 13

A system has n = 13 resources R0,…,R12, and k processes P0,…Pk−1. The implementation of the resource request logic of each process Pi is as follows:

if (i%2 == 0)
{
if (i < n) request Ri
if (i+2 < n) request Ri+2
}
else
{
if(i < n) request Rn - i
if (i+2 < n) request Rn-i-2
}
Minimum number of the process so deadlock is possible _________.


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 13

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.

GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 14

If  A =  then find a + b:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 14

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).

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 15

A circuit outputs a digit in the form of 4 bits. 0 is represented by 0000, 1 by 0001,…, 9 by a combinational circuit is to be designed which takes these 4 bits as input and outputs 1 if the digit ≥ 5, and 0 otherwise. If only AND, OR and NOT gates may be used, what is the minimum number of gates required?


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 15

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.

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 16

The circuit shown below acts as:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 16

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).

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 17

Consider the equation (213)6 =  (a3)b with a & b values to be unknown. Then what are the number of possible solutions for the values a & b?


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 17

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.

GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 18

Consider a three-level page table to translate a 39-bit virtual address to a physical address as shown below:

 

The page size is 4 KB(1 KB = 210 bytes ) and page table entry size at every level is 8 bytes. A process P is currently using 2 GB(1 GB = 230 bytes) virtual memory which is mapped to 2 GB of physical memory. The minimum amount of memory required for the page table of P across all levels is _____________ KB.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 18

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).

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 19

In X = , how many one-address instructions are required to evaluate it?


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 19

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.

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 20

Which of the following statements is/are correct about the 3NF and BCNF normal forms?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 20

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).

GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 21

Consider the following statements:

S1: The relation ≤ is reflexive on any set of real numbers.

S2: If R1 and R2 are reflexive relations, R1∪R2 is also reflexive.

S3: Every superset of a symmetric relation is symmetric.

S4: Smallest relation which is irreflexive contains 1 element.

Which of the above statements are false?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 21

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).

GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 22

A micro-instruction format has micro-ops field which is divided into three subfields F1, F2, F3 each  having seven distinct micro-operations, condition field CD for four status bits, branch field BR having four options used in conjunction with address field ADF. The address space is of 128 memory locations. The size of micro-instruction is:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 22

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).

GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 23

Consider a network having 6 routers P to U connected with links having weights as shown below:

All routers are using distance vector routing algorithm to update their routing table. Then what will be the shortest distance between router P and router U?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 23

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).

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 24

Which among the following is the only difference between Turing Machine and Finite Automaton?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 24

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

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

GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 25

The lexical analysis for a modern computer language such as Java needs the power of which one of the following machine models in a necessary and sufficient sense?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 25

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).

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 26

Which of the following is/are false about deterministic context free language (DCFL)?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 26

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 = pnqnrm ∪  pmqnrn 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 = wr is a CFL but not a DCFL. It can be derived from the following grammar.

S → aSa | bSb | ∈

Option (D): False

L = pnqnrm ∩ pmqnrn ∣ 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).

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 27

Consider the following memory values and a single address machine with an accumulator:


Which of the following instructions will load 230 into the accumulator?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 27

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).

GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 28

The least number of computers required to connect 10 computers to 5 routers to guarantee 5 computers can directly access 5 routers is _________.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 28

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).

GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 29

A 4:1 multiplexer is to be used for generating the output carry of a full adder. A and B are the bits to be added while Cin  is the input carry and Cout  is the output carry. A and B are to be used as the select bits with A being the more significant select bit.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 29

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

Cout = ∑ m(1, 2, 4, 7)

Another equation for Cout  is:

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

The equation for Cout  is:

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

Hence, the correct option is (A).

GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 30

Which one of the following array represents a binary max-heap?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 5 - Question 30

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).

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