Practice Test: Computer Science Engineering (CSE) - 13 - Computer Science Engineering (CSE) MCQ

The distance travelled by Ram in February = 950 km

Distance travelled by him in month of May = 1200

Total distance travelled by Ram in Feb and May = 950+1200 = 2150 kms

The distance travelled by Amar in February = 1200 km

Distance travelled by him in month of May = 800

Total distance travelled by Amar in Feb and May = 1200+800 = 2000 kms

The distance travelled by Ram in Feb and May is 7.5% more than distance travelled by Amar in Feb and May.

Hence, option (c) is correct.

‘Flake out’ means to fall asleep from exhaustion.

Fall back on - begin to use something held in reserve

Fritter away - To waste or squander time or money.

Ferret out - To discover something after searching.

It is given that the total number of students are = 1000

Only Hockey = 380

Only Cricket = 170

Only Football = 240

Hockey + Cricket + Football = 40 {all the three sports}

Now, let X be the number of students who play Cricket and Football.

And Y is the number of students who play football and hockey.

Now total number of students,

170 +60 +40 +380+ 240+ X + Y= 1000

X + Y = 110

And 60 students play only hockey and cricket, and that number we get after removing students who play all three sports from students who play hockey and cricket.

X + Y + 60 = 110+ 60 = 170

There are 170 students who play exactly two sports.

Let Bholu’s share = X Rs.

Arun’s Share = X + 30 Rs

Chetan’s Share = X + 30 + 20 = X+50

Now, Arun+ Bholu+ Chetan = (X + 30) + (X) + (X + 50) = 200

3X = 200 – 80

X = 120/3

X = 40

Bholu’s share is 40 Rs.

Hence, (b) is the correct option.

According to the question let m = 1

∴ 5 consecutive integers are = 1, 2, 3, 4, 5

n = 15/5 = 3

6 consecutive integers starting with (m+2) are = 3, 4, 5, 6, 7, 8

Now check from option to put n = 3

Option : (a)

(Satisfied)

Hence, the correct option is (a).

Cost price of the television = = Rs. 20000

To earn 16% profit

Selling price = = Rs. 23200

Amount after 1^st year serves as principal for 2^nd year. So,

Dividing (?) by (??) we get,

P=Rs.562.50

Present age of Kanika = 16 years

Age of Kanika one year ago =16-1=15 years

Age of Monika after four years = 15×2=30 years

Present age of Monika = 30-4=26 years=26 years

e present ages of Kanika, Monika and Shweta =22×3=66 years

Present age of Shweta =66-16-26 = 24 years

Present average age of Kanika and Shweta =x=(24 + 16)/2=20 years

So, the value of ‘x’ is 20 years

So option (d) is the correct answer.

Let the two numbers be 9X and 13X

So, (9x) × (13x)= 5 × 585

117x² = 25 x 117

x²=25

X = 5

Required sum =9 × 5 + 13 × 5 = 45 + 65 = 110

So option (a) is the correct answer.

Here 60 is given as the first element.

After the first pass, the pivot element goes to it’s exact location.

Here 60 goes to 6th place.

All the elements less than 60 go to the left of 60 and all the elements greater than 60 go to the right of 60.

After 1^st pass

⇒ 5! × 3!

⇒720 possible arrangements.

The way of removing left recursion is-

A→ AB / C is replaced by A → CA'

A' → BA' / epsilon

So the answer is - E → +E'

E' → S+E'/+E'

E' → EPSILON

S → *S'

S' → *S'/EPSILON

int i,j,x,y,m,n;

N = 20;

for(i = 0; i < n; i++)

{

for(j = 0;j < n; j++)

{

if(i%2)

{

x + = 4*j[4<

y + = (7+4*j);

}

}

}

M = x + y;

Given 8 – bit delimiter pattern of 01111110.

Output Bit string after stuffing is 01111100101

StuffedBit

Now, Input String is 0111110101.

Therefore, maximum frame size = 2 × L × Bw / v

= 20 bits = 2.5 bytes.

With the above-given data, after allocating 2 units of tape to each process, with 1 available unit any of the 3 processes can be satisfied in such a way that no deadlock will be there. So the answer is 7 tape units.

F= {A →BC, CD → E, E →C, D → AEH, ABH →BD, DH → BC} then a nonredundant cover for F { A → BC, E →C, D → AEH, ABH →BD }. Then FD ABH → BD can be decomposed into FDs ABH → B and ABH →D

Now since, the FD A →B and AH → D. We also notice that AH →B is redundant since the FD A → B is already in F

That gives us the canonical cover as {A → B, A→C, E → C, D →A, D → E, D →H, AH → D}.

Therefore X - Y = 29040

Max-Heap after all insertions

Clearly, no element has the same index.

Hence, answer = 0

No. of candidate keys at R = {A.C} i.e. 2

Therefore for a relation related to 7 entities

No. of candidate keys = 7 {A, C, E, G, I, K, M}

We need to apply Optional Elimination

Taking A = 6 AND B = 4

Case it’s GCD when GCD = 2

Now dry run the code

1 → A = 2, B = 4

2→ A = 2, B = 2

Prints A which is 2 which means it’s a GCD

From the given heap, it is obvious that 3 and 9 need to be exchanged for 17 and 3, but there are also exchanges needed for 2 and 6. So, 2 should be at the root as it is the minimum. Since the minimum number of exchanges is asked, it is clear that if exchanges are carried out in this manner, 2 – 6, 2 – 17, 6 – 17 and 3 – 9 separately, then the number of exchanges required = 4. But in other way, exchanges would be like:

3 – 9, 3 – 17, 9 – 17, (3 is at root now) and on the right side:

2 – 6, 2 – 3. Thus 5 exchanges. So P = 4.

But we have to find the minimum number of exchanges.

Now the elements in BST with height 4 is 4+1(root) = 5

S1: Referencing relation is the relation referring to a primary key of some other relation. Deletion in that table will not cause any violation. Hence, no deletion anomaly hence this is true.

So, this statement is true.

S2: This statement is true, since foreign key can be created within a relation

C is a foreign key referring to the primary key "A'' of the same relation R.

S3: S3 is true since count (*) will count the no. of tuples in a relation, on the basis of ID value.

S4: It is true. Since it is possible to decompose every relation into BCNF with lossless join decomposition property but it may or may not preserve the dependencies of the relation.