Practice Test: Computer Science Engineering (CSE) - 5 - Computer Science Engineering (CSE) MCQ

The sentence talks about a past incident and a condition has been mentioned. Thus option 2 fits here correctly. 'Would have' is the correct tense to be used here. Option 1 is illogical. 'Would' must be used here and not 'will' as it is in the past tense. Past perfect tense is incorrect here.

The word that fits here is a noun thus options 3 and 4 are eliminated. The form should be singular as for abstract nouns we do not use the plural form. Option 1 is thus the correct answer.

The word powerful is an adjective and 'empowering' is a verb.

The radius of outer circle = r

∴ Diagonal of square = 2r

Side of square (a)=√2r

Average speed = Total distance/Total time

The sentence mentions a person doing something as per her wishes thus the first word must be a positive reaction. 'Restive' which means 'restless' and 'jinxed' which means 'cursed' are incorrect here. Between options 1 and 3,the word 'elated' which means 'too happy' and the word 'venture' which means 'undertake a risky or daring journey or course of action' fit here correctly. The word 'cease' means 'stop' and does not convey a proper meaning. Thus option 3 is the correct answer.

As, the faster runner is twice as fast as the slowest runner, the faster runner would have complete two rounds by the time the slowest runner completes one round.

Their first meeting takes place after the fastest runner takes 6 min to complete two rounds and the slowest runner completes one round.

Time taken to complete one round by fastest runner = 6/2 = 3 minutes

Rounds needed to complete the race = 1600/400 = 4

Time taken to complete the race = 4 × 3 = 12 min

The young one of a horse is called a foal. Similarly, the young one of a goose is known as gosling. Thus, option 2 contains the pair that best expresses a relationship similar to that expressed in the given pair. In all the other pairs, the first word expresses the male form of the latter.

Since teenagers here have tender age and commit crimes unknown to its consequences, clearly they should not be treated as any criminal. They need to be taught so that they can make difference between right and wrong and do not commit these crimes in future. Hence inference in option 1 is correct.

Previous

450 y = n³ implies that 450 y is the cube of an integer. 

When we prime-factorize the cube of an integer, we get 3 (or a multiple of 3) of every prime factor.

8 is the cube of an integer because 8 = 2³ = 2*2*2.

Thus, when we prime-factorize 450 y, we need to get at least 3 of every prime factor. 

Here's the prime factorization of 450 y:

450 y = 2 * 3² * 5² * y 

Since 450 provides only one 2, two 3's, and two 5's, and we need at least 3 of every prime factor, the missing prime factors must be provided by y. Thus, y must provide at at least two more 2's, one more 3, and one more 5. 

Smallest possible case: 

y = 2² * 3 * 5

In this problem, cumulative frequency is given, so converting into frequency

From the table, it is clear that option b, i.e.

60% of the scholars published at least 2 papers is correct.

Augmented matrix will be

Rank of A ≠ Rank of Augmented matrix

Hence given system of equations has no solution.

Concept:

Keep the n leftmost bits of given address in the block and set the 32 − n rightmost bits to 0s to find the first address where n is the prefix length. 

Calculation:

Here, prefix length is 20.

135.16.27.116 ⇒⇒ 10000111 00010000 00011011 01110100 

The first 20 bits of this octal are fixed as 10000111 00010000 0001

The remaining 32 – 20 = 12 bits can get a maximum value as 0000 00000000.

So the possible first octal IP address is 10000111 00010000 00010000 00000000 which is 135.16.16.0

The expression retrieves name of all employees in the Marketing department along with their manager ID.

In case of continuous random variable probability at a point = 0

The lexical analyzer reads the stream of character making up the source program and groups the characters into meaningful sequences called as lexemes. For each lexeme, the lexical analyzer produces as output a token of the form <token name, attribute value>

Strings of length not more than 250 means all those strings whose length is less than or equal to 250.

Note: If |w| <= n, the number of states in minimal DFA is n+2.

Therefore, the minimum number of states in a minimal DFA that accepts all the strings of length not more than 250 is 252.

fork() can be used to create a new process, known as a child process. This child is initially a copy of the parent but can be used to run a different branch of the program or even execute a completely different program. After forking, child and parent processes run in parallel. fork() returns 0 in child process and non-zero value in parent process.

CA is a federal or state organization that binds a public key to an entity and issues a certificate. The CA has a well-known public key itself that cannot be forged. The CA checks Bob's identification. It then asks for Bob's public key and writes it on the certificate. To prevent the certificate itself from being forged, the CA signs the certificate with its private key.

Therefore, the number of functions possible = 2² = 4

t1 = i + 1

i = t1

t2 = i * 8 // refers memory allocation for given data type

t3 = a[t2]

if t3 < b

Hence, 3 variables are required in three address code.

Therefore, maximal elements are {12, 20, 25} and the minimal elements are {2, 5}.

Post-order traversal: 25, 30, 20, 45, 60, 50, 40, 35

In order traversal is the arrangement of keys in ascending order. Therefore,

In order traversal: 20, 25, 30, 35, 40, 45, 50, 60

The tree is given by:

Concept:

The client TCP after receiving close command sends the first segment, FIN segment. In which only FIN flag is set.

Let the sequence no. be x and acknowledge no. be y.

The server TCP sends FIN + ACK segment after receiving the FIN segment.

Here, sequence no will be y and ack no will be x + 1.

The client TCP sends the last segment an ACK segment. This segment contains an acknowledgment number which is plus 1 the sequence number received in the FIN + ACK segment from the server.

Calculation:

Acknowledgment number in last segment i.e. ACK segment is 2164. Hence, sequence number in FIN + ACK segment is 2164 – 1 = 2163

I, III and IV are CFL because one comparison can be done by a push down automata at one time.

Total number of entries = 2 ×× 10⁷ 

Space occupied by overhead = 2 x 10⁷ x 6 = 12 x 10⁷

Maximum file size = total file system size – space consumed by overhead

= 90000 x 10⁷ – 12 x 10⁷

= 899880 MB

The chromatic number of a planar graph is no greater than four.

First(S) = {p, q, r} 

First(P) = {q, r, ε} 

First(Q) = {q, ε} 

First(R) = {r, ε}

Follow(S) = {$} 

Follow(P) = {p} 

Follow(Q) = {p, r} 

Follow(R) = {p}