Sample GATE Mock Test - Computer Science And IT (CS/IT) - Computer Science Engineering (CSE) MCQ

Past Tense is used

In this question, we need to find the HCF of 374 and 578 which is 34. We can look at 374 = 34 x 11 and 578 = 34 x 17. This gives the number of square tiles required to cover the entire room is 11 x 17 = 187 and area of each such tile is 34² = 1156 cm².

In Q part, 'the' is not required.

Let us assume the capacity of the tank to be 48 units (48 being LCM of 8, 12, 16 and 24). This leads us to get the rate of filing of T1 to 4 as 6, 4, 3 and 2 units per hour.

Taking pairs of taps, we find that T1 + T2 will take 48/10 whereas the remaining pair will take 48/5 hours leading to the ratio of time taken to fill up 2 tanks as 5:10 which is not the required ratio.

Taking T1 and T3, the time taken is 48/9 and the time taken for the other tank is 48/6 leading to the required ratio as 2:3.

Total installed capacity = 33.3 million kW and total power generated = 119.3 billion kWh

Percentage utilization   which gives nearest integer value as 41%.

Let the ages of father and son 12 years ago were F and S respectively. Then A/Q,

Present age of son  x+ 12 = 9+12 = 21 yesrs

Let x and y be the time for which tap ‘A’ and ‘B’ are opened

We have x + y = 90 … (i)

90x + 60y = 6000 … (ii)

Solving (i) and (ii), gives x = 20 and y = 70.

 ⇒  Difference in the duration for which taps ‘A' and ‘B’ are opened = 70 – 20 = 50 minutes.

Let them meet at X after t hours and speed of train 2 be s kmph as shown in figure below

Train 1 is shown above the line and Train 2 is shown below the line

Distance travelled by train 1 before meeting = Distance travelled by train 2 after meeting

45*t = 9*s   .... i

Also Distance travelled by train 2 before meeting = Distance travelled by train 1 after meeting

st = 45*16 .... ii

Using eq i ;

t= s/5

Putting in eq ii ;

s^2 / 5 = 45*16

s = 60 kmph

OR

Short trick:

Since we are given time to cross a point which is 1/5th of the way and 1/3rd of the way, we can rewrite them as 3/15th and 5/15th of the way which implies that 2/15th of the way was covered in 10 minutes.

This means that the entire route was covered in 10 × 15/2 = 75 minutes (each 1/15th part is covered in 75/15 = 5 minutes) leading to time of arrival at school as 9:50 – 15 + 75 = 10: 50 hours.

Total number of boys opting for craft and dancing = 365 + 370 = 735

Total number of boys opting for drawing and swimming = 140 + 235 = 375

Required difference = 735 - 375 =  360

In operating systems, seek time is very important. Since all device requests are linked in queues, the seek time is increased causing the system to slow down. Disk Scheduling Algorithms are used to reduce the total seek time of any request.

TYPES OF DISK SCHEDULING ALGORITHMS

Although there are other algorithms that reduce the seek time of all requests, I will only concentrate on the following disk scheduling algorithms:

First Come-First Serve (FCFS)

Shortest Seek Time First (SSTF)

Elevator (SCAN) 

Circular SCAN (C-SCAN)

LOOK

C-LOOK.

Vectored interrupts are the fastest. Slow and inefficient peripherals are not able to provide device identification number or ISR address. Best option is to have ISR at a fixed address. So for a n-bit CPU, ISR of peripheral attached to vector x will be at x*8.

For each node in preorder linear search in the order traversal gives left subtree and right subtree.

Time complexity =O(n²)

For daisy chaining, there is a single bus request line and bus grant line which is passed through each of the devices. In polling, there is log2(n) bit polling number for the sequence maintaining I/O devices. Bus is granted using a poll count. Thus log2(n) lines needed to maintain poll count. Bus request line is single. So total no of control lines = log2(n)+1. For independent requesting separate bus request and bus grant line for each I/o device. Thus 2n control lines needed.

Deadlock occur if each and every process holds one resource and request resource of other process such that circular wait condition get satisfied. Let A, B, C be these processes holding one process each.

At source , packet visits all the layer once before getting transmitted.

At router R1 , it will go upto the network layer because router is a networking device.

At gateway G1 , it will go upto transport layer because gateway is an all layer switch who works on transport layer.

At router R2 , it will go upto network layer.

At destination , packet will once again visit all the layers.

Hence

A - Application Layer

P - Presentation Layer

S - Session Layer

T - Transport Layer

N - Network Layer

D - Data Link Layer

P - Physical Layer

Hence , from figure

x = 3 y= 6 z = 8

so, x + 2y + 3z = 3 + 12 + 24 = 39

gradeup ≤ gaterank and

gaterank is decidable.

If gaterank is decidable, then gradeup is also decidable.

So option (a) is correct.

NOT and OR can implement NOR (universal gate). Therfore {NOT, OR} is functionally complete. NOR and NAND are designated as universal logic gates because using any one of them we can implement all the logic gates.

D →A A → E

N →R R → N

C →  C →I

(R, C) →G

From the given relation

R( R, N, D, A)

The functional dependencies which this particular relation hold:

D →A N →R R → N

The candidate key for the relation will be: RD, DN

Hence partial dependency exists in the given relation R:

D →A

So relation is in 1NF but not in 2NF. 

For regular language

1) Equivalence: it can be checked by developed product automate for XOR function.

if XOR = O ⇒ equal else not.

2) Ambiguity: Ambiguity is decidable for regular languages & grammar since they are deterministic in nature.

3) Regularity: it is the trivial problem.

For context free language

⇒ it is not closed in any of those problems.

We know Subnet mask is needed by routers (stored in routing table) inorder to direct the packet to intended network. But how will host know whether the packet belongs to its own network, whether to send the packet to router or not. For this reason, even host needs the Subnet mask.

(A) is the correct answer.

A Pre-order traversal visits nodes in the following order:

25, 15, 10, 4, 12, 22, 18, 24, 50, 35, 31, 44, 70, 66, 90

A Post-order traversal visits nodes in the following order:

4, 12, 10, 18, 24, 22, 15, 31, 44, 35, 66, 90, 70, 50, 25

Hence 4 elements are having same index numbers in both the arrays.

Explanation:

In a Spanning Tree where a subgraph is freely connecting to vertices. The cost of a spanning tree will be the sum of costs on its edges. The minimum spanning tree is tree with minimum cost.

So MST will be 3 + 6 + 4 + 5 + 5 = 23

The correct statement are.

1.  Recovering lost packet between two directly connected nodes is done by data link layer.

2. Recovering lost packet between two nodes separated by multiple hops is done by transport layer.

The function is returning i++ which is post incremental, so first the value of i which is 10 will be returned and then it will be incremented. so the value of i in the main function will be 10 and printing --i will print 9.

FIRST() is calculated by the following rules-

i. FIRST(TERMINAL)=TERMINAL

ii. FIRST(EPSILON)=EPSILON

iii. If we have to calculate FIRST(ABC) and FIRST(A) contain epsilon then also include FIRST(BC)

FOLLOW() is calculated by the following rules-

i. If A is start symbol then FOLLOW(A) also contain $

ii. Eg- B -> DACE

C -> a/b

FOLLOW(A)=FOLLOW(CE) = FIRST(C) = {a,b}

iii. Eg- B ->DA or B->DAC

C-> epsilon

FOLLOW(A)=FOLLOW(B)

Now coming to question , answer is - FIRST(S) = d,e,f,a,c

FOLLOW(S)= $

FOLLOW(A)= c,f,a