Practice Test: Computer Science Engineering (CSE) - 6 - Computer Science Engineering (CSE) MCQ

The spokes are units which radiate of a wheel and make the wheel a complete entity. Similarly, fingers, tentacles and petals are integral units of hand, octopus and flower respectively. On the other hand, roots and leaves do not share this kind of relationship. Thus option 4 does not expresses a relationship similar to that expressed in the given pair.

Let a = 3x and b = 4x

Similarly b = 2y and c = y

∴ 4x = 2y ⇒ y = 2x

∴ c = 2x

Now a + b + c = 3x + 4x + 2x = 9x

So, the minimum integer value = 9

Friday → 2 days earlier

Therefore, scheduled day = Friday + 2 = Sunday

Sunday + 3 = Wednesday

Therefore, I would have been late by 3 days

The sentence implies that it was hoped that that place would become the point from where the African civilization would proceed, but the belief that it would happen was not fulfilled.

Therefore, the correct word to fill in the blank is expectation as it means a strong belief that something will happen or be the case.

Mean = 4

k ≠ m so k = m = 4 is out.

{k, m} = {1, 7} or {2, 6} or {3, 5}

for median of {3, k, 2, 8, m, 3}

{1, 2, 3, 3, 7, 8} or {2, 2, 3, 3, 6, 8} or {2, 3, 3, 3, 5, 8}

Median: (3 + 3) /2 = 3

A person can take 1^st step in any direction independently

Suppose he moves to A

Now to return to O ( initial point) in 2 steps he can move in 2 directions. Either B or f

Thus probability he will move to either B or F will be = 2/6

Suppose he moves to B

Now to return back to O

he has only 1 option

⇒ to move in BO

Probability he will move in BO direction  = 1/6

Total probability he will return

The nth line drawn will add n more regions to the circle. 

1st line adds 1 region to the circle for a total of 2 regions. 

2nd line adds 2 more regions to the circle, bringing the total number of regions to 2 + 2 = 4. 

3rd line adds 3 more regions to the circle, bringing the total number of regions to 4 + 3 = 7. 

4th line adds 4 more regions to the circle, bringing the total number of regions to 7 + 4 = 11.

5th line adds 5 more regions to the circle, bringing the total number of regions to 11 + 5 = 16. 

6th line adds 6 more regions to the circle, bringing the total number of regions to 16 + 6 = 22. 

7th line adds 7 more regions to the circle, bringing the total number of regions to 22 + 7 = 29. 

8th line adds 8 more regions to the circle, bringing the total number of regions to 29 + 8 = 37. 

9th line adds 9 more regions to the circle, bringing the total number of regions to 37 + 9 = 46. 

10th line adds 10 more regions to the circle, bringing the total number of regions to 46 + 10 = 56. 

11th line adds 11 more regions to the circle, bringing the total number of regions to 56 + 11 = 67. 

12th line adds 12 more regions to the circle, bringing the total number of regions to 67 + 12 = 79. 

The correct answer is option 2 i.e. To implement hardware protocols to minimize risks and reduce electromagnetic radiation production significantly. 

The passage states about the electromagnetic radiations released from the devices and how they affect the individuals. Out of the given options, only option 2 states the possible solution which can be implemented to reduce electromagnetic radiation.

Let the number of students be X who cleared cut off in all sections.

The number of students who cleared the cutoff in section 1 and section 2, only = 20 – X

The number of students who cleared the cutoff in section 2 and section 3, only = 16 – X

The number of students who cleared the cutoff in section 1 and section 3, only = 26 – X

Now, consider section 1:

24 + 20 – x + x + 26 – x = 60

70 – x = 60

x = 10

If w is in L, then either

(a) w does not contain any 0, or

(b) it contains a 0 followed by 11. So, w can be written as w₁w₂.........w_n, where each w_i is either 0 or 011.

So, L is represented by the regular expression (1 + 011)*.

Therefore, option 2 is the correct answer.

Concept:

Contention period is minimum time a host must transmit for before it can be sure that no other host's packet has collided with its transmission. It takes minimum of RTT to detect collision.

Calculation:

Contention Period = RTT = 2 × Tp

600 = 2 × Tp

Tp = 300 μs

The minterm y₁y₂ ...y_n is 1 if and only if each y_i is 1, and this occurs if and only if x_i = 1 when y_i = x_i and x_i=0 when y_i = x_i.

Therefore, the minterm will be x₁'x₂x₃'x₄x₅ i.e. 01011. 

The answer will be 11.

The buddy system allocates memory from a fixed-size segment consisting of physically contiguous pages. Memory is allocated from this segment using a power-of-2 allocator, which satisfies requests in units sized as a power of 2.

In a double-ended queue, elements can be inserted and deleted from both the front and back of the queue.

Time complexity of all operations like insert element at front, insert at rear, delete element from front, and delete element from rear is O(1).

For all values of a and b,     is an Eigen vector of A.

⇒ ab + b² = 2a²

⇒ 2a² - ab - b² = 0

⇒ 2a² - 2ab + ab - b² = 0

⇒ 2a (a - b) + b (a - b) = 0

⇒ b = a, b = -2a

For b = a, (or) b = -2a, X is an eigen vector of matrix B.

In indexing addressing mode, the address field references a main memory address, and the referenced register contains a positive displacement from that address.

Address of operand = base register + index register + displacement

Address of operand = 3456 + 4 + 1500 = 4960

The number of available hosts on a subnet is 2^h−2, where h is the number of bits used for the host portion of the address.

h² = 14

h = 4

Option 4 is incorrect. It can be corrected as:

The cartesian product of two countable sets is countable.

Proof: There are three cases to consider:

Case 1: If both A and B are finite with |A|=m and |B|=n, then it is easy to show that |A×B|=mn and hence A×B is finite and so it is countable.

Case 2: If A is finite with |A|=n and B is countably infinite, there exist bijective functions f: A→{1,2,...,n} and g: B→N. We can define h: A×B→N for all (a,b)∈A×B.

Then clearly h is injective. So A×B is countable.

Case 3: If A and B are both countable, there exist bijective functions f: A→N and g: B→N and defining h: A×B→N gives us an injective function, and so A×B is countable. 

In resource allocation graph, a claim edge Pi → Rj indicates that process Pi may request resource Rj at some time in the future. An assignment edge Rj → Pi indicates that process Pi is holding resource Rj. The above system is deadlock-free although there is a cycle in a graph because there are more than one instance per resource type.

The sizeof operator returns the size of the struct in bytes. The size of a struct is not always the sum of the size of the individual members, hence use sizeof rather than a literal.

struct Student *sptr;

sptr = (struct Student*)malloc( sizeof(struct Student) );

We know that,

Concept:

The measured round-trip time for a segment is the time required for the segment to reach the destination and be acknowledged, although the acknowledgment may include other segments. In TCP, there can be only one RTT measurement in progress at any time.

Calculation:

RTTM = 3.2

RTTS = 3.2

RTTD = RTTS/2 = 1.6

RTO = RTTS + 4 × RTTD = 9.6

The first rule defines the inherited attribute T’.val using F’.val and F appears to the left of T’ in the production. The second rule define T’₁.val using the inherited attribute T’.val associated with head and F.val, where F appears to the left of T’ in the production

Probability (1 faulty in box) 

The chances of box having other than 1 defective is 5/8

The probability exactly 1 one fault in exactly 1 box

Alternate:

The given situation is possible if

(1) 1 faulty among 4 components

(2) 3 faulty among 4 components

For case (1)

For case (2)

Out of all the options, option 3 is false.

It can be corrected as: It is desirable to maximize CPU utilization and throughput and to minimize turnaround time, waiting time, and response time. 

In IPv4, the maximum length of packet size is 65,536 and length of IP header is 20 bytes. So, maximum data size (including UDP header) = 65,535 – 20 = 65515

Size of UDP header = 8 bytes

Maximum data size = 65,515 – 8 = 65507

Number of balanced parenthesis = Number of binary search trees

Number of binary search trees = 

When n = 4

Number of binary search trees =

Therefore, value of n = 4

Both the given statements are true.

It is possible to have a weak entity set with more than one identifying entity set. A particular weak entity would then be identified by a combination of entities, one from each identifying entity set. The primary key of the weak entity set would consist of the union of the primary keys of the identifying entity sets, plus the discriminator of the weak entity set.