Practice Test: Computer Science Engineering (CSE) - 4 - Computer Science Engineering (CSE) MCQ

The correct spelling of the word is 'definitive' which means '(of a conclusion or agreement) done or reached decisively and with authority.' Thus option 2 is the correct answer.

The preposition 'about' is mandatory here thus option 1 and 4 are eliminated. Option 2 is correct as the tense present perfect continuous fits here. It conveys the meaning that the person usually talked about the place. 

Options 1 and 2 are incorrect as they change the meaning of what is mentioned. Option 3 is incorrect as with 'less' we use the adjective in positive form and not comparative form. The correct word here should be 'brave.' Option 4 is thus the correct answer.

Let number is N

N = 65 k +29

The word 'here' and 'there' both can be used here but should be followed with 'waiting' as no other word can fit here. The word 'caught' is correct here as 'catch a sight' means 'to see something.' Option 2 thus has the correct combination of words. The word 'cot' means 'a small bed with high barred sides for a baby or very young child.'

For a number to be greater than 5000, d₁ should be filled with either 5 or 7

∴ Total numbers formed when the digits are repeated = 2 × 5 × 5 × 5 = 250

total cases = 250 -1 = 249 ( case of 5000 is not included)

Now, For the number to be divisible by 5, unit digit d₄ should be either 0 or 5.

∴ Total no. of ways = 2 × 5 × 5 × 2 = 100

favorable cases = 100 - 1=9 ( 5000 is not included))

∴Required Probability = 99/249 = 0.397

The question asks which of the statements given in the options can weaken the argument put by the author that all bacteria from Venus must have died out.

The passage states that since there is a single strain of bacteria which exists on the Earth, they all must be belonging to the Earth or let's say a single planet. But here the author does not take into consideration (as can be argued from his theory) the fact that may be all the bacteria came from Venus and there are none which originally belong to the Earth. So this criticism as mentioned in option 3 makes the argument of the author vulnerable. 

Options 1, 2 and 4 are completely irrelevant criticisms as they do not address the main argument. The argument claims that if there were Venusian bacteria on Earth, then they must have died out by now. Whether there are bacteria originally from Earth that have also disappeared from Earth is irrelevant to the question and has no effect on the given argument.

Let a, b and c be the cost prices of the three articles A, B and C.

SP = CP + Profit (or) SP = CP – Loss

⇒ SP of A = 1.1a; SP of B = 1.2b; SP of C = 0.9c

By question,

1.1a + 0.9c = a + c ⇒ 0.1a = 0.1c ⇒ a = c

1.2b + 0.9c = 1.05(b + c) ⇒ 0.15b = 0.15c ⇒ b = c = a

Gain% = {(SP – CP)/CP} × 100

Net gain on the sale of all the articles = 

Net gain on the sale of all the articles = 6.66%

Option 1 is false as graph says there is decrease in students qualifying in Physics in 2015 compared to 2014.

Option 2

Let no. of students qualifying in Biology in 2013 be 100

⇒ No. of students qualifying in Biology in 2014 = 100 – 10% of 100 = 90

⇒ No. of students qualifying in Biology in 2015 = 90 + 10% of 100 = 99

∴ The number of students qualifying in Biology in 2015 is less than that in 2013

Option 3 and option 4 are incorrect since no detail is given regarding how many students qualified the subject in 2013.

Area of triangle PAB = Area of triangle RCB (By symmetry)

∴ Area of Δ PAB = ½ × PA × AB

= ½ × 2A/3 × A = A²/3

Area of ΔRCB = A²/3

Now, Area of ABCD = Area of DRQP + Area of PAB + Area of RCB + Area of PBRQ

A² = a² + A²/3 + A²/3 + Area of PBRQ

As, A/a = 3 ⇒ a = A/3

If 'k' is the number of states of the NFA, it has 2^k subsets of states. Each subset corresponds to one of the possibilities that the DFA must remember, so the DFA simulating the NFA will have 2^k states.

The chromatic number of a complete graph is the number of vertices it has. 

Therefore, the chromatic number of a complete graph having 100 vertices is 100.

No. of bits required for 128 instructions = log₂ 128 = 7

No. of bits required for operand field = 7 

No. of bits required for register field = 24 – 7 – 7 = 10

Maximum no. of registers = 2¹⁰ = 1024

i.e. from 0 to 1023. Hence, maximum value = 1023

Let 

The given expression can be solved as:

1 – P (x) ≥ 50%

In a circular queue, the new element is always inserted at the rear position. If queue is not full then, check if (rear == SIZE – 1 && front!= 0) if it is true then set rear=0 and insert new element.

The inorder traversal of the binary search tree gives the sequence in ascending order.

Out of the given sequences, only 2 and 3 are in ascending order.

Therefore, option 2 is the correct answer.

Transfer time = propagation delay + copy time

= 35 + 18 

= 53 ns

Statements S2 and S3 are true.

Statement S1 can be corrected as: Given a context-free language, there is no Turing machine which will always halt in the finite amount of time and give answer whether language is ambiguous or not.

A relation  on a set  is irreflexive provided that no element is related to itself; in other words,  for no  in .

An irreflexive relation contains (n² - n) elements.

Therefore, number of irreflexive relations =

In class B, size of net id = 16 bits 

Size of host id = 16 bits

In network id first two bits are reserved for leading bits i.e. 10

Hence total number of networks possible = 2^{16 – 2} = 2¹⁴

In C, when preprocessor sees the #define directive, it goes through the entire program in search of the macro templates; wherever it finds one, it replaces the macro template with the appropriate macro expansion. In this program wherever the preprocessor finds the phrase MUL(x) it expands it into the statement (x * x).

p = 4 x 5 = 20

now i = 6

q = 8 x 8 = 64

p + q = 84

Number of sectors  = Number of surfaces*Number of tracks per surface*Number of sectors per track

= 2³ * 2⁷ * 2⁷ = 2¹⁷

Therefore, number of bits required = 17

Dijkstra’s algorithm solves the single-source shortest-paths problem on a weighted, directed graph G for the case in which all edge weights are nonnegative. The time complexity of Dijkstra’s algorithm is O(E log V).

One can implement Dijkstra’s shortest path algorithm to find all pair shortest path by running it for every vertex. The time complexity of this Dijkstra’s algorithm is O(V³ log V).

A receiver and network can dictate to the sender the size of the sender's window. If the network cannot deliver the data as fast as they are created by the sender, it must tell the sender to slow down. In addition to the receiver, the network is a second entity that determines the size of the sender's window.

Actual window size = minimum (rwnd, cwnd)

rwnd = size of the receiver window

cwnd = size of the congestion window

2 = minimum(8, cwnd)

cwnd = 2