Practice Test: Computer Science Engineering (CSE) - 2 - Computer Science Engineering (CSE) MCQ

Let the cost price = 100 Rs.

From the options:

If profit % = 60%

Then SP = 160 Rs.

New SP = 160/4 = 40 Rs.

Then,

Percentage loss= (100-40)/100 = 60%

Hence Verified

ALTERNATE:-

Let the selling price is 100

New selling price 100/4 = 25

He suffers a loss of 60%.

CP*0.4 = 25

CP = 25/0.4 = 62.5

Actual Profit = 100 - 62.5 = 37.5

profit% = 37.5/62.5 *100 = 60%

the given series is prime numbers from 23

a. Jab as a noun refers to a quick, sharp blow, especially with the fist.

b. A joke is a thing that someone says to cause amusement or laughter, especially a story with a funny punchline.

c. Satire is the use of humour, irony, exaggeration, or ridicule to expose and criticize people's stupidity or vices, particularly in the context of contemporary politics and other topical issues.

d. A gag is a joke or an amusing story, especially one forming part of a comedian's act, or in a film.

‘Hilarious satires’ and ‘hilarious jabs’ seem inappropriate and ambiguous. Among all the options, ‘jokes’ and ‘gags’ fit perfectly in the blank. Therefore, option D is the apt answer.

speed of car = distance covered / time taken

720 / 9 = 80 km/h

Now, speed of bus = 3 / 4 x 80 = 60 km/h

Speed of train = 27 / 15 x 60 = 108 km/h

Distance covered by train in 7 h = 108 x 7 = 756 km

Kaur Singh, former heavyweight Boxer is struggling with the treatment for heart disease and admitted at a private hospital in Mohali. Under such circumstances, the Union Sports Ministry has approved five lakh rupees from the National Welfare Fund for Sportspersons for Kaur Singh.

Selling price of T.V = 5000 x 100 / 10 x 90 / 100 = 45000

Marked price of laptop = 45000 + 15000 = 60000

Required % = 45000/60000 x 100 = 75%

Statements

All oils are sands

Some clays are oils

All clays are rocks

Combining all three statements, we get

Conclusions

I. At least some clays are sands

II. Some oil is not rock

Conclusion I

Some clays are oils + All oils are sands = Some clays are sands

Hence, Thus, the conclusion I follow.

Conclusion II

Some clays are oils → conversion → some oils are clays + All clays are rocks = Some oils are rock. Hence, conclusion II does not follow.

The first sentence 1 introduced a situation, and the rest of the sentences should be arranged in the order that they happened. Clearly A is the first in sequence as it uses the word 'first' indicating the first step of these six men. The next step is described in D. Now, the twist comes in sentence C as the suspicions rise. People cannot get suspicious of something unless something actually happened. So, we place A first, followed by D and B. This leaves C for the last, which describes the result of those suspicions. Hence, the correct answer is d.

• Let the inputs to JK flip flop are J = K = 1 and Q = 0. After the propagation delay Δt, the output will change to Q = 1.
• Now, we have J = K = 1 and Q = 1 and after another time interval of Δt, the output will change back to Q = 0.
• Hence, for the duration tp of the clock pulse, the output will oscillate back and forth between 0 and 1.
• At the end of the clock pulse, the value of Q is uncertain. This situation is referred to as the race around the condition.

Therefore, the race around the condition can be avoided by keeping the propagation delay as tp < Δt < T. 

Note:

For JK flip-flop if J, K, and Clock are equal to 1 the state of flip-flop keeps on toggling which leads to uncertainty in determining the output of the flip-flop. This problem is called Race around the condition.

This can be eliminated by using the following methods.

• Increasing the delay of flip-flop
• Use of edge-triggered flip-flop
• Use of master-slave JK flip flop

It contains cycles so it is not serializable.

It contains no cycle so is serializable.

So option (C) is correct.

+ and – are both right associative and + has higher precedence than*. So the correct option is (c).

In symmetric ⇒ Asymmetric ⇒ 2k

Explanation for asymmetric:

Every user has a (public key, private key) pair.

1 user ⇒ 2 keys

k user ⇒ 2k keys

Hence option (C) is the right answer.

(x⊙y)=(y⊙x)

x² + y² + 2xy = y² + x² + 2yx

Since LHS = RHS

So ⊙ is commutative.

2. Check for associative:

(x⊙y)⊙z=x⊙(y⊙z)

(x² + y² + 2xy) Η z = x Η ( y² + z² + 2yz)

x⁴ + y. + 4x²y² + 2x2y2 + 4xy³ + 4x³y + z² + 2x²z + 2y²z + 4xyz ≠ x² + y⁴ + z⁴ + 4y²z² + 2y²z² + 4y³z + 4yz³ + 2xy² + 4xyz

So A is not associative.

C + R × S = M × S

Where, C = Initial capacity

R = Token rate

M = Output rate

S = Bursty traffic

Given, C = 10 Mbits

Token arrives at an interval of 10 μsec.

Then in 1 sec 100000 token arrives.

So, 100000 × 16 bit

= 16 × 105 bits/sec

= 1.6 Mbps

Hence R = 1.6 Mbps

Substituting we get

S = = 1.5625

38 × 2¹⁰ = [21⁶ – [Group1 + Group2 + Group3]]

38 × 2¹⁰ = [2¹⁶ – [2⁷ × 2⁶ + 2⁶ × 2⁸ + 2^m × 2ⁿ]]

2^x × 2^y = 2¹⁶ – 2¹³ – 2¹⁴ – 38 × 2¹⁰

= 2¹⁰[2⁶ – 2³ – 2⁴ – 38]

= 2¹⁰[64 – 8 – 16 – 38] = 2¹⁰ × 2¹

2^x × 2^y = 211

Option (d), which 26 × 25 = 211

Insertion Sort ⇒ O(n²)

Selection Sort ⇒ O(n)

Bubble Sort ⇒ O(n²)

Quick Sort ⇒ O(n²)

Amongst all these algorithms, selection sort requires the least number of swaps.

Selection sort, which in many respects has incredibly poor performance (not adaptive; quadratic performance re: num of comparisons), actually requires the minimum number of swaps. It is based on the concept of doing a significant number of comparisons before moving each element directly to its eventual sorted resting place. At most the algorithm requires N swaps; once you swap an element into place, you never touch it again.

And then overwrite the array by first filling the array with 1's - the number of 1's being equal to one_count value, and do the same for 0's also.

All this will take O(n) time and (A) is the answer.

Number of lines =

Number of sets =

Cache representation

Physical address size = 16 MB = 24 bit

24 = x + 7 + 5

x = 12 bit

Now, to map physical memory at 13th set bit (b₁₁ b₁₀ b₉ b₈ b₇ b₆ b,) MM address should be 13.

In option (d)

Hence option (d) is correct.

SELECT S.Name FROM Student S WHERE

S.snum NOT IN (SELECT E.snum FROM Enroll E)

↓

It return snum of all students who is enrolled in any course.

Query return the sname of a student who is not enrolled in any course.

SQL does not eliminates duplicate so relation given by SQL query is

Total 2 tuple returns.

L₁ . L₂ will be equal to {aⁿ b^m| m , n > = 0}, which is same as {aⁿ b^m| m = n, n > –1}.

Number of binary trees possible are = 10C5 / 6 = 10!/ 5! * 5! * 6 → 42

Hence option D is the correct answer

Total number of different Binary tree of size

5 : 14 + 5 + 4 + 5 + 14 = 42

For a 2 × 1 MUX is shown above, the output function F is expressed as:

F = S̅₁ I₀ + S₁ I₁

i.e. when S₁ = 0, I₀ is transmitted to the output.

And when S₁ = 1, I₁ is transmitted to the output.

Application:

The output of the 1^st MUX will be:

E = B̅C + BC̅ 

The final output will be:

f = E̅.0 + E.A = E.A

f = (B̅C + BC̅) A

f = AB̅C + ABC̅ 

Hence, Option (C) 4 is the correct choice.

f = ( C’ + BC ) D + A D’

f = C’ D + BCD + AD’

Min-terms that represent f are

C’ D - 0 0 0 1, 0 1 0 1, 1 0 0 1, 1 1 0 1

B C D - 0 1 1 1, 1 1 1 1

A D’ - 1 0 0 0, 1 0 1 0, 1 1 0 0, 1 1 1 0

f = min-terms (1 , 5 , 7 , 8 , 9 , 10, 12 , 13 , 14 ,15)

Now

1 + 5 + 7 + 8 + 9 + 10 + 12 + 13 + 14 + 15 = 94.