Practice Test: Computer Science Engineering (CSE) - 9 - Computer Science Engineering (CSE) MCQ

Exulted- to express great pleasure or happiness, especially at someone else's defeat or failure

In the sentence a word that is complementary to 'freedom' is required. Hence, 'piece' which means a portion can be eliminated. This leaves us with options 2 and 3. It can be gathered from the sentence that invasion resulted in violence and hardships though it promised positive things like democracy, peace and freedom. So all these promises were false ideas. Thus, 'illusion' best fits here, making option 2 the correct answer.

An 'allusion' is an indirect or passing reference.
E.G. The Renaissance writers alluded to literature written during the Classical Greek period.

The relation among the given numbers is

Hence, the next numbers are 58, 77, 100.

By similar triangles ΔADB and ΔABC, we get

⇒ AD/AB = DB/BC=AB/AC

⇒ AD/6=DB/8=6/12

⇒ DB/8=6/12

⇒ DB = 48/12 = 4 cm

∴ BD = 4 cm

Given expression is,

Using: 1 + tan²θ = sec²θ and 1 + cot²θ = cosec²θ

Using: cosec θ = 1/sin θ and sec θ = 1/cos θ

= sin²θ + 3 cos²θ + 2 sin²θ

= 3sin²θ + 3cos²θ

= 3(sin²θ + cos²θ)

Using: sin²θ + cos²θ = 1

= 3 × 1 = 3

The statement provided to us in the question states the fact that an initial team was sent for the reconnaissance of the target area before the main strike team. It is to be noted here that the word ‘reconnaissance’ means observation of an area to get information about it.

Now, nowhere in the sentence is there any mention of the relative training or the equipments of the two teams, neither is there any indication to asses this fact. Therefore, it is safe to eliminate options 1 and 4.

Now, the third statement indicates that the strike team could not work without the main team. There is no evidence in the sentence to point out to any such fact. It is simply stated that the initial team was sent to make reconnaissance before the main team, in order to assist them. It cannot be assumed that the job of the initial team was absolutely essential for the second team.

The second inference states that the team did not posses sufficient and complete information about the target area. This is an inference that can be safely drawn from the given statement, since if they would have had enough information, there would have been no need for an initial team.

Hence, option 2 is the correct answer.

From graph it is evident that option 1 is true.

From graph, Of the surface area above sea - level, larger proportion lies below 2 km elevation

From graph, Of the surface area below sea - level, larger proportion lies below 4 km depth.

Option 4 is true as maximum depth is 12 km and maximum elevation is 8 km.

∴ Option 3 is FALSE

Given, Ananth takes 6 hours and Bharath takes 4 hours to read a book.

Let the number of pages of the book be n.

∴ In 1 hour, Ananth reads n/6 pages of the book while Bharath reads n/4 pages of the book.

Let after ‘t’ hours the number of pages remaining to be read by Ananth is twice that number of pages remaining to read by Bharath.

⇒ 1 – t/6 = 2 – t/2

⇒ t/3 = 1

⇒ t = 3 hours

Number of ways 5 people can sit around round table = (5-1)! = 4!

Now 5 spots are created for other 5 people wearing black white shirt.

∴ Total ways = 4! × 5!

Now for each arrangement 10 spots are created for the person wearing Red shirt.

∴ Total ways = 4!5!10 = 28800

The First 500 Natural Odd numbers are given as 1,3,5,7,…………….999. 

The Numbers divisible by 3 from the above list are given as 3,9,15,21…………999 

The above series is in Arithmetic progression,

T_n = a+(n−1)d

999 = 3+(n₃−1)6 ⇒ n₃ = 167 

So total Numbers which are divisible by 3 in the given list is, n₃ = 167 

The Numbers divisible by 5 from the above list are given as 55,15,25,35…………99 

The above series is in Arithmetic progression,

T_n = a+(n−1)d

995 = 5+(n₅−1)10 ⇒ n₅ = 100 

So total Numbers which are divisible by 5 in the given list is, n₃ = 100.

The Numbers divisible by 15 from the above list are given as 15,45,75,105…………975 

The above series is in Arithmetic progression,

T_n = a+(n−1)d

975 = 15+(n₁₅−1)30 ⇒ n₁₅ = 33 

So total Numbers which are divisible by 3 in the given list is, n₃ = 33 

So Required Probability p = p(n₃)+p(n₅)−p(n₁₅) = = 0.468

≡ 1

Hence it is a tautology.

Total number of functions = 4⁴ = 256

Number of functions which are either one -one or onto = 4! = 24

Number of functions which are neither one -one nor onto​ = 256 - 24 = 232

baabbabb can not be generated using above grammar.

BST has sorted in-order traversal, 10-1-3-5-6 will be the longest path from root to leaf. Hence height of BST will be 4.

S ≥ 2 × bandwidth × t_d ≥ 2 × 10⁹ × 1000/2 × 10⁸ ≥ 10000bits

FIRST/FIRST Conflict for LL(1)

S -> E | E 'a'

E -> 'b' | ε

FIRST(E) = {'b', ε} and FIRST(E 'a') = {'b', 'a'}

Thus in LL(1) table, there is conflict under terminal 'b' of production rule S.

FIRST/FOLLOW Conflict for LL(1)

S -> A 'a' 'b'

A -> 'a' | ε

The FIRST set of A now is {'a', ε} and the FOLLOW set {'a'}.

Go through this Wikipedia page.

To sort a linked list, Merge sort is the best choice and Heap sort is impractical.

p(p−1) < 0, because product of roots is a negative number.

Thus p must be less than 1 and greater than 0

If pointer to the node to be deleted is given, delete operation is more efficient in doubly linked list O(1) than singly linked list O(n), because to delete a node in singly listed list, pointer to the previous node is needed. To get this previous node, we have to traverse the list. But in doubly linked list we can get the previous node using previous pointer.

In slow-start phase, for each ACK, the sender increases the current transmit window by Maximum Segment Size (MSS). As we are given a packet consists of 2000 bytes and that can be taken as MSS. So, after two ACKs, current transmit window:

= 4000 + 2000 + 2000

= 8000

For making sure that the graph is having every vertex degree greater than 1 have to add the edge that will repeat at least one vertex causing cycle.

Multiplication will not form a group because inverse of zero does not exist.

Division operation will not form a group because n/0 is not defined.

Total number of rows in the bank is 2¹⁴. Total number of banks is 8. And in 1.024 sec total number of refresh operation performed by the controller is 1024/64 =16.

Total number of refresh operation performed by the controller is

2¹⁴ × 8 × 16 = 2²¹

Whenever a number is in 2ⁿ form then the minimal 2’s complement representation is 1 followed by n number of zeros.

So, – 64 = – 2⁶ = 1000000

A broadcast address is a network address at which all devices connected to a multiple-access communication network are enabled to receive datagrams. A message sent to a broadcast address may be received by all network-attached hosts.

negate subnet mask and perform logical or operation with given IP address.

The function returns minimum value from the list 

I is context sensitive language

II and III are CFLs.

III can not generate L rest can.

To avoid spilling we have to use 3 register

a = 1;

b = 2;

c = 3;

c = a + b;  r1 = c, r1 = b, r2 = a

b = a + c r2 = a, r1 = b, r3 = c, because we will using the value a,b,c in future   

d = b + c;  

e = d + a;

return d + b;

As the flag is 01111110, we have to delete 0 after every five consecutive 1’s.