Direction: Study the following piechart and tables carefully and answer the questions given below.
Total Number of email received by the organization = 90000
Ratio of Read emails to Unread emails received by the organization
Q. What is the ratio of the number of emails read in January to those unread in the month of April in the organization?
Number of read emails in the month of January = (90000*17/100*8/15)
Number of unread emails in the month of April = (90000* 8/100 * 5/12)
Ratio = 68:25
answer which can fill both the blanks of both the sentences.
I. Most of our employees get _____ abroad at some stage
II. The aircraft and its crew were ______ missing.
Post means to send somebody to a place for a period of time as part of their job and to announce something publicly or officially.
Hence, option D is correct.
In the following question, out of the given four alternatives, select the one which is opposite in the meaning of the given word.
Scurrilous
The meanings of the given words are as follows:
Scurrilous— abusive
Coarse— rough or harsh in texture
Sophisticated— developed to a high degree of complexity.
Insolent— showing a rude and arrogant lack of respect
Complimentary— expressing a compliment; praising or approving
Clearly, option D is the antonym of the given word.
Statements:
All pillows are beds.
No fruit is pillow.
Some foods are fruits.
Conclusions:
I. At least some foods are pillows.
II. Some beds are definitely fruits.
Consider the following least possible Venn diagram,
Conclusions:
I. At least some foods are pillows → it’s possible but not definite, hence false.
II. Some beds are definitely fruits → it’s possible but not definite, hence false.
Direction: Study the following information carefully and answer the questions given below:
P, Q, R, S, T, U, V and F are sitting around a circle facing the centre. U is third to the right of Q, who is third to the right of F. P is third to the left of F. R is fourth to the left of P. T is third to the right of S. S is not a neighbour of P.Four of the following five are similar in a certain way based on their positions in the seating arrangement and so form agroup. Which of the following does not belong to that group?
Option D does not belong the group.
Following bar graphs shows time taken by different pipes to fill that particular percentage of tank which is described in tabular data. For Ex Pipe 1 fills the 10 % of tank in 6 mins.
Now, read the following data and answer the question carefully:
A Tank has three pipes; two of them are used to fil the tank while another one used to empty the tank namely: pipe 3, pipe 5 to fill the tank and a third pipe for making the tank empty. When all three pipes are open, 7/18^{th} part of the tank is filled in 1 hours. How much time will the third pipe take to empty the completely filled tank 10 times the bigger than the previous one.
part of work in one hour=
(P3+P5+Third pipE. X 1 = 7/18^{th} part of work
⇒ 7/18^{th} part of work = 1 hr
⇒ Total work done = 18/7 hr by all three pipes
Pipe 3 does 50% of work in 90 mins
100 % of work = 90/50 * 100 = 180 mins = 3 hr
Pipe 5 does 50% of work in 60 mins
100 % of work = 60/50 * 100 = 120 mins = 2 hr
Time Tw Efficiency
Pipe 3 = 3 hr Total work/3 = 6
Pipe5 = 2 hr Total work/2 = 9
All three = 18/7 hr Total work/(18/7)= 7
For Total work = LCM of (3,2,18/7) = 18
Efficiency of pipe 3 + Efficiency of pipe 5 + Efficiency of third pipe = total efficiency
6+9+X= 7
X = 715 = 8 (Negative sign shows the negative work herE.
Efficiency X Time = Total work
8 X Time = 18 * 10 ( as tank is 10 times bigger than the previous onE.
Time = 180/8 = 22.5 hrs
In a school, 12th class consists of 30% male students of which 30% male students failed in the class. Total 82% students passed in 12th examination out of 900 students. Calculate the total number of female passed students?
Short Trick:
Total students = 900
Boys : Girls = 30 % : 70% = 3 : 7
So, exactly Boys : Girls = 270 : 630
Total failed students = 18% of 900 = 162
Failed boys = 30% of 270 = 81
Failed girls = 162  81 = 81
Passed female students = 630  81 =549
Basic Method:
⇒ Let us consider total number of students be 100x
⇒ ∴ Males students = 30x , female students = 70x
⇒ According to condition given in the problem, of this 30x male students, 70 % male students pass the exam.
ie. Passed males =
⇒ Total passed students =
⇒ ∴ female students passed = 82x – 21x = 61x
⇒ ∴ Girls passed = 61% of total
∴ Answer is
Directions: Rearrange the following six sentences (A), (B), (C), (D), (E) and (F) in the proper sequence to form a meaningful paragraph: then answer the questions given below them.
A). Indian Government remain concerned about potential spillover effects from the unconventional monetary policies of the advanced economies, which could cause disruptive volatility of exchange rates, asset prices and capital flows.
B). The challenges are related to high public debt and unemployment, poverty and inequality, lower investment and trade, negative real interest rates along with signs of prolonged low inflation in advanced economies.
C). In this context, emerging markets and developing countries (EMDCs) continue to be major drivers of global growth.
D). The global recovery continues, although growth remains fragile, with considerable divergences across countries and regions.
E). It is important to strengthen the framework of international financial cooperation, including through instruments such as swaplines, to mitigate the negative impacts of monetary policy divergence in reserve currency issuing countries.
F). Structural reforms, domestic adjustment and promotion of innovation are important for sustainable growth and provide a strong and sustainable contribution to the world economy.
Which of the following will be the Second sentence?
Refer to the last question of the series.
17 25 34 98 121 339
The pattern followed here is,
⇒ 17 + 2^{3} = 17 + 8 = 25
⇒ 25 + 3^{2} = 25 + 9 = 34
⇒ 34 + 4^{3} = 34 + 64 = 98
⇒ 98 + 5^{2} = 98 + 25 = 123
⇒ 123 + 6^{3} = 123 + 216 = 339
There are 3 red balls and 5 green balls in a bag. If two balls are picked at random from the bag then find the probability of getting a red ball?
Required probability
=^{3}C_{1}*^{5}C_{1}/^{8}C_{2}
= 3*5/28
= 15/28
The size of a proper subgroup of a finite group G is 31. The size of the smallest possible group G is ____.
The order of a subgroup divides the order of a group. Here the order of the subgroup is 31 therefore the smallest possible size of group is 62 which is evenly divisible by 31.
Tags: 1 Mark
Find the number of seven digit integers with sum of the digits equal to 11 and formed by using the digits 1, 2 and 3 only.
Total possibility with sum = 11 and 7 digits
3,3,1,1,1,1,1
3,2,2,1,1,1,1
2,2,2,2,1,1,1
The number of 7 digit integers = 21+105+35 = 161
Consider the following statements:
i. Selection sort performs minimum number of swaps.
ii. Insertion sort performs worst in case of sorted array.
iii. Floyd Warshall uses dynamic programming to calculate all pairs of shortest paths
Q. Which of the statements are true?
Selection sort uses minimum number of swaps. In each pass, there can be at most 1 swap. So there can be at most n swaps in worst case.
In case of insertion sort, if every element is less than every element to its left, the running time of insertion sort is Θ(n^{2}). Thus insertion sort is worst for reverse sorted data. Floyd Warshall algorithm uses dynamic programming for all pairs of shortest path computation.
Consider the following recursive function find.
int find (int A[], int n)
{
int sum=0;
if(n==0) return0;
sum = find(A, n1)
if (A[n1]<0) sum=sum+1;
return sum;
}
Q. What is the worst case running time above function find (A[], n) when array A has 0 to n1 elements?
Recurrence relation for the function find:
F(n)=0; if n=0
=F(n1)+1 ;n>0
Time complexity of F(n)=O(n)
Let A be a collection of objects. An efficient method for converting it into a set is first to sort the objects of A. Then, walk through the sorted sequence and remove all duplicates. What is the running time of this method?
This takes O(nlogn) time to sort and n time to remove the duplicates. Overall, therefore, this is an O(nlogn) time method.
The following function test takes argument a queue and uses stack to perform some function:
void test(Queue *A)
{
Stack T;
while (!isEmpty(A))
push(&T, deQueue(A));
while (!isEmpty(&T))
enQueue(A, pop(&T));
}
Q. What does the function test do?
The first while loop copies the data of the queue into the stack and the next while loop copies into the queue but stack follows LIFO. Hence, the last element will be enqueued first and then so. The order of the queue is reversed.
Consider the following code
int DO(char *gate)
{
char *gate1 = gate;
char *gate2 = gate + strlen (gate) – 1;
while (gate1 < =gate2)
{
if (*gate1 ++! = *gate2 )
Return 0;
}
return 1;
}
Q. What is the functionality of above function Do ()?
Here two pointers are used gate 1 and gate 2. One pointer points to beginning and other to the end. Loop is a set up that compares the characters pointed by these two pointers. If the characters do not match then it’s not a palindrome. It returns 1 for both even and odd palindrome.
{
printf(“%d%d%d%d%d”, ‘ \n ‘ ,printf(“\0”),printf(“\n”),’ \0 ‘ , ‘ \b ‘ );
}
What will be printed by this above code?
ASCII value of ‘ \n ‘ is 10.that is printed.
printf(“\0”) // it prints nothing ,only returns zero. This returned zero value is printed.
printf(“\n” // it prints nothing ,only returns one ,This returned 1 is printed
‘ \0 ‘. //ASCII value of this null character is zero.that is printed .
‘ \b ‘ // ASCII value of this backspace escape sequence is 8,that is printed.
Given T_{s} : Transfer time
B : Number of bytes to be transferred
N : Number of bytes on a track
R : Rotational speed.
Q. Which of the following expression gives the total aveage access time?
Average access time is
Cache with 4 Blocks of 4 bytes. Give the number of cache hits and cache miss respectively
Block address =Byte Address/ Block size
Cache index=Block address% No. of Cache blocks
Tag=block address/No. of cache blocks
We have three stations P, Q, R connected in serial manner. P is connected to Q through a 3Gbps fibre optic link and length is 500Km. Q is connected to R through 60Mbps link and length is 15Km.All the links are full duplex in nature. A file is sent from station A to C. Packet size is 1KB.We use sliding window protocol such that SWS=RWS. Find the optimal SWS packets.
First we have to calculate RTT.
RTT= Transmission period + 2*propagation delay
= (8192/3*10^{9}) +(8192/60 *10^{6}) + 2*(5*10^{5}/2*10^{8})(15*10^{3}/2*10^{8})
= 5.29 ms.
Number of packets to be sent =5.29 *60 Mbps =31.74≈32 packets.
Consider a 3bit number A and 2 bit number B are given to a multiplier. The output of multipier is realized using AND gate and one bit full adders. If minimum number of AND gates required are X and one bit full adders required are Y, then X + Y =
Number of AND gates required X=6
Number of one bit full adders required Y=3
X+Y=6+3=9
The output F of the 4to1 MUX shown in figure is
F = xy(1) + xy^{'}(1) + x^{'}y(0) + x^{'}y^{'}(0)
=xy + xy^{'}
= x
Let L= {M accepts some string} where M is a Turing machine. Find the language L?
Simulate M on all strings of length atmost n for n steps and keep increasing n. We accept if the computation of M accepts some string. Language generated by a grammar is recursively enumerable hence it is turing recognizable. So it is also recursively enumerabl language.
Let A, B, C and D are problems. Consider the following polynomial reductions to known about the problem B.
(i) A≤B (A is reducible to B)
(ii) C≤ D
(iii) B≤ D
Find the correct statement from the following.
A≤B
C≤ D
B≤ D
A. A is decidable ⇒ B need not be decidable.
B. D is undecidable ⇒ B need not be undecidable.
C. C is decidable ⇒ D need not be decidable.
D. A is undecidable ⇒ B is undecidable.
Consider the relation scheme of the relation SCHEDULE shown below. What is the highest Normal form of this relation?
SCHEDULE (Stud_ID, Class, Stud_Name, Stud_Major, Class_Time, Building_Room, Instructor)
Assume the following functional dependencies
Stud_ID →Stud_Name
Stud_ID →Stud_Major
Class →Class_Time
Class →Building_Room
Class →Instructor
As per the given data there are some patial dependencies on the key,
For example : Class →Building_Room or Stud_ID → Stud_Major
Therefore, the highest form of the relation is 1NF
In a database system, unique time stamps are assigned to each transaction using Lamport’s logical clock. Let TS(T_{1}) and TS(T_{2}) be the timestamps of transactions T_{1} and T_{2} respectively. Besides, T_{1} holds a lock on the resource R, and T_{2} has requested a conflicting lock on the same resource R. The following algorithm is used to prevent deadlocks in the database system assuming that a killed transaction is restarted with the same timestamp.
if TS(T_{2}) < TS(T_{1}) then
T_{1} is killed
else T_{2} waits.
Assume any transactions that is not killed terminates eventually. Which of the following is TRUE about the database system that uses the above algorithm to prevent deadlocks?
Given,
if TS(T2) <TS(T1) then
T1 is killed
else T2 waits.
According to algo, TS(T2) <TS(T1) then T1 is killed else T2 will wait. So in both cases neither deadlock will happen nor starvation.
Therefore, option A is correct
Consider the following statements
(i) SQL’s aggregation can’t be expressed in relational algebra
(ii) SQL’s grouping construct can be expressed in relation algebra.
Which of the following is correct?
Only (i) and (ii) is correct. Grouping construct can’t be expressed in relational algebra.
Which of the following SQL query find all the tuples having temperature greater than that of Paris.
Inner query returns the temperature of the city Paris and this will be compared with temperature of all tuples in outer query.
Consider the following assembly code
MOV R_{1},b ; R_{1}b
MOV R_{2},c ;
MUL R_{1},R_{2} ;
MOV t_{1},R_{1} ;
MOV R_{1},a ;
MOV R_{2},t_{1} ;
ADD R_{1},R_{2} ;
MOV t_{2},R_{1} ;
MOV R_{1},t_{2} ;
MOV d,R_{1} ;
Find the correct expression which is equivalent to the above code.
Consider the following grammar with corresponding synthesized attributes.
F.L {F. val=L. val}
LLB {L.len= L_{1} .len+1, L. val= L_{1}. val + 2^{L.len}B.val}
LB {L. len=1, L. val=B. val/2}
B0 {B. val =0}
B1 {B. val =1}
If “F.val” gives the value of the binary fraction generated by F in the above grammar then that will be the value of F.val on input .101?
Given grammar
F.L
LLB
LB
B0
B1
Input string .101
The dining Philosphers problem with N philospheres has the solution for synchronization without deadlock. Which of the following is correct solution
Both (a) and (b) are correct solutions for the dining Philospher problem without deadlock for the synchronization.
Cansider a system has 4 processes with arrival times and burst times as shown in the following table
Compute the average waiting time of processes using the FCFS scheduling?
Consider the following three concurrent processes.
Initially A = 3 , B = 0
Q. What is the minimum number of 1’s printed by the execution of the above processes?
I. A = 3, B = 0
First process – 1 execute three times and process – 1 is blocked (A=0, B=3).
II. Process – 3 executes next three times and process – 3 is blocked (A=0,B=0).
III. Process – 2 executes last then it will be blocked by executing P(B) first time.
No. ‘1’ is printed by the execution of three processes and all are blocked
Number of 1’s printed = 0.
Calculate the value of
Assume that ‘e’ is the number of edges and n is the number of vertices. The number of nonisomorphic graphs possible with nvertices such that graph is 3regular graph and e = 2n – 3 are .
For 3 regular: 3n=2.e
Given e=2n3, substitute in above equation.
3n= 2.(2n3)
3n = 4n6
n =6
The number of vertices =6.
n=6 e=2n3=9
Number of 3 regular graphs with 6 vertices and 9 edges are 2.
Both graphs are non –isomorphic but 3 –regular graphs.
Two graphs are possible.
Find an eigen vector corresponding to largest eigen value of given matrix
and
If the probability of a bad reaction from a certain injection is 0.001, and then the probability that out of 2000 individuals more than two will get a bad reaction will be
Probability of occurrence is very small so it will follow poisson’s distribution Mean (m)
=np
= 2000*0.001 = 2 Probability that more than 2 will get a bad reaction
=1 [p(0)p(1)p(2)]
=1[
=1[
=1[
= 1 
A certain problem is having an algorithm with the following recurrence relation.
Q. How much time would the algorithm take to solve the problem?
Here Master’s theorem is not applicable directly.
Consider the following bubble sort function.
bubblesort (A)
{
for i ← 1 to length [A]
do for j ← length [A] down to i + 1
do if (X) then exchange A[j] A[j – 1]
}
Q. What is the missing statement at X, if the bubble sort function implemented correctly to sort the
elements in the increasing order?
Sol.If A[j] is smaller than A[j – 1] then exchanges A[j] and A[j – 1] to get the smaller first.
Consider the following structure for creating nodes of a double linked list.
Struct node
{
int data;
struct node *x_{1};/*left pointer*/
struct node *y_{1};/*right pointer*/
};
Let us assume P be the node in the existing linked list. Which of the following is true about the following code C_{1}?
Code C_{1} :
Assume memory is made free automatically by the compiler after deletion.
Consider the following Binary search tree with the following traversals
Inorder: A, M, N, O, P, Q, R, S, T, X, Z
Preorder: Q, P, A, M, N, O, X, S, R, T, Z
Find the number of elements in the 4^{th}, 5^{th} and 6^{th} level respectively. Assume root is at level 1.
The resultant BST obtained is
∴ The number of elements in the 4^{th}, 5^{th} and 6^{th} level are 3, 1, 1.
∴ Option C is correct.
Consider a system employing an interrupt driven I/O for a articular device that transfers data at an average of 8 KB/s on a continuous basis. Assume that interrupt processing takes about 100 micro seconds (i.e. jump to the interrupt service routine (ISR); execute it and return to the main program). Determine what fraction of processor time is consumed by this I/O device when it is interrupted for every byte.
The device generates bytes/sec
i.e. 1 second 8192 bytes
Given that each interrupt consumes 100 us.
∴ Fraction of proce ssor time consumed in (for every byte)=0.82
Find how much disk space (in surfaces) will be required to store 300,000 120byte logical records, if the disk is fixed sector with 512 bytes/sector, with 96 sectors/track, 110 tracks per surface, and 8 usable surfaces. Ignore any file header records and track indices and assume that records cannot span two sectors.
Each sector can hold 4 logical records.
The required number of sectors is 303,150/4 = 75,788 sectors
= 75,788/96 tracks = 790 track
= 790/110 = 8 surfaces.
A 5 stage pipelined processor has IF, ID, EXE, MEM and WB. WB stage operation is divided into two parts. In the first part register write operation and in the second part register read operation is performed. The latencies of all those stages are 300, 400, 500, 500 and 300 (in nano second) respectively. Consider the following code is executed on this processor
Find minimum number of nop instructions (no operation) to eliminate hazards without using operans forwarding. (Assume each instruction takes one cycle to complete its operation in every stage)
Since I_{1} and I_{2} are dependent. So I_{2} will enter into execute stage when I_{1}completes WB stage.It creates 2 nop. Similarly I_{3} and I_{4} are dependent. They create 2 nop.
Total 4 nop instructions are required to eliminate hazards.
In a leaky bucket system if the output rate is 5 KB/sec and input burst of 50 KB/sec for 10 sec and 10 KB/sec for 50 sec then bucket size in KB is ..............
Input burst = 50KB/sec for 10 sec and 10KB/sec for 50sec.
Output rate = 5KB/sec
Total time = 60sec ,Total Length = 500KB + 500KB = 1000KB.
Now output rate = 5KB  1 sec
?  60sec
In 60sec we can transmit 300KB.
We should have a Bucket of size 1000KB  300KB = 700KB
Assume 10 Mbps Ethernet and two stations A and B on it’s same segment. The RTT between two nodes is 650 bit times. A and B start transmitting frame and collision occurs and both sends 30bit jam signal. Find the time at which both nodes A and B sense an idle channel (in)
Propagation delay=
Both nodes detect collision
at t=325, jam signal takes 30bit times.
Both A and B finish transmitting JAM signal at t=325+30=355
The last bit of jam signal of A, reaches B after 1 propagation delay i.e. after 355 +325=680
(Similarly, from B→A at 680)
At=68 they both will sense idle channel.
Four channels are multiplexed using TDM. If each channel sends 100 bytes/second and we multiplex 1 byte per channel, then the bit rate for the link is ……………bps
Firstly we understand the working of Multiplexer using TDM.
Each frame contain 1 byte from each channel; { 1 Byte = 8 Bit}
we have 4 channels
Then ,the size of each frame is
1x4= 4 bytes = =~ 32 bits.
Each channel is sending 100 bytes/second
A frame carries 1 byte from each channel,
The frame rate must be 100 frames per second. { its given}
Then the total rate is = 32*100 = 3200 bits/second
Consider CSMA/CD protocol, after 4^{th} collision, what is the probability that node choose k=4.
First collision, i = min(10, 1)
K={0,……,2^{1}1} = {0,1}
Second Collision, i = min(10, 2)
K= {0,…..,2^{2}1} ={ 0,1 ,2 ,3}
Third Collision i= min(10, 3)
K= {0, ……, 2^{3}1} = {0, 1,2 ,3 ,4 ,5 ,6 ,7}
Fourth collision i=min(10, 4)
K= {0,…….., 2^{4}1} ={0, 1, ….., 15}
Probability to choose k=4 after fourth collision 1/16 = .0625
We have a network 210.50.60.0, we have to divide it into 6 equal subnets, give network and broadcast address of each subnet.
As it is a class C network, it’s subnet mask will be 255.255.255.0è11111111.11111111.11111111.00000000 to have 6 sub networks we have to convert 3 leftmost bits in the host Id so new subnet mask will be 11111111.11111111.11111111.11100000 ie. 255.255.255.224
Network addressè
Ip address: 11010010.00110010.00111100.00000000
Subnet mask: 11111111.11111111.11111111.11100000
ANDING
11010010.00110010.00111100.00000000
Ie. 210.50.60.0
Broadcast Addressè
Ip address: 11010010.00110010.00111100.00000000
Inverted Subnet mask: 00000000.00000000.00000000.00011111
XORING
11010010.00110010.00111100.00011111
Ie. 210.50.60.31
The number of 4 line to16 line decoders required to make an 8lineto256 line decoder is
8line to256line decoder using 4line to16line decoders is shown below:
Number of 4 x 16 MUX required
Consider the following synchronous counter made up of JK, D and T flipflops.
Find the modulus value of the counter.
Consider characteristics equation of JK flipflop:
Consider characteristic equation of D FlipFlop :
Consider characteristics equation of TFlipflop:
_{Using equations (i), (ii) and (iii)}
The number of used states = 5
Modulus value of the counter = 5.
The following transition diagram is for:
(i) >
(ii) >=
The given transition diagram is for the pattern > = and >, Its start state is sate 0. In state 0 we read the next input character. The edge labeled > from state 0 is to be followed to state 6 if this input character is >. Otherwise, we have failed to recognize either > or <=.
On reaching state 6, we read the next input character. The edge labeled = from state 6 is to be followed to state 7 if the input character is an = otherwise, the edge labeled other indicates that go to state 8. The double circle on state 7 indicates that it is accepting state, a state in which the token >= has been found.
Consider the following DFA
Q. Which of the following two states are equivalent?
Using partition method:
Step1: {TOC, CD, OS, Prog, DS, Algo, CO, CN} {SWT}
Step2: {TOC, CD, OS, Prog, DS Algo} {CO} {CN} {SWT}
Step3: {TOC, CD, OS, Prog} {DS} {Algo} {CO} {CN} {SWT}
Step4: {TOC, CD, OS} {Prog} {DS} {Algo} {CO} {CN} {SWT}
Step5: {TOC, OS} {CD} {Prog} {DS} {Algo} {CO} {CN} {SWT}
∴ OS & TOC are equivalent
So option (b) is correct.
Consider the following grammar:
S > aBA
A >aAE  E
B >bBC  b
C > cCD  D
D >aDC
E >ED  eE  e
F > SeF
After removal of useless symbols, how many productions violate the rules of
greibach normal form?
C, D and F are the useless symbols in given grammar.
After removal of useless symbol the grammar will be
S > aBA
A > aAE  E
B > b
E > eE  e
Then AÃ E is the only production which is not according to GNF.
What the following relational calculus represents?
{t∃s ∈ supp(t[s_name]=s[s_name]^
∃u∈ supp(u[s_address]=s[s_address]^
u[s_name]=”tarun”))}
The above query lists the name of all the suppliers that have the same address as the address of the supplier name tarun.
AB→C
C→D
D→EA
E→F
F→B
Find the number candidate keys of R.
Considering the following grammar
S → iEtS iEtSeSa
E → b
Then after left factoring we get _________
Given grammar is,
S →iEtS iEtSeSa
E → b
After applying algorithm for left factoring a grammar we get
S → iEtSS’a
S’ → eS€
E → b
Suppose two processes P and Q are as follows where m1, m2 and m3 are binary semaphores having values m1=1, m2=0, m3=1.
P: Q:
While (1) While (1)
{ {
A C
P (m3); P (m3);
Print (0); Print (1);
V (m3); V (m3);
B D
} }
For output “010101”, what will be the correct option for A, B, C, D?
For string “010101”, P has to execute 1st then Q and then same thing alternatively for 2 more times. For P executing first, P(m1) is must for process P as m1=1, m2=0.
A disk drive has 100 cylinders, numbered 0 to 99. Disk requests come to the disk driver for cylinders 12, 26, 24, 4, 42, 8 and 50 in that order. The driver is currently serving a request at a cylinder 24. A seek takes 6 msec per cylinder moved. How much seek time is needed for shortest seek time first (SSTF) algorithm?
Present position of the cylinder is 24, the nearest to it is 26. So it will go to 26 first. From 26 the nearest is 12 so it will move to 12. Likewise it will traverse in the order 24,26,12,8,4,42,50.
The total number of traversed will be ((2426) + (2612) + (128) + (84) + (442) + (4250)
(2+14+4+4+38+8)=70 cylinders. And time taken to traverse one cylinders is 6 ms.
So 70*6= 420s/1000=.42sec
Consider a virtual system with 2 level paging 32 bit virtual address is used to support 32 bit physical address, Page size is 4096 byte pages and page table has 4 bytes of page table entry. How many memory accesses required to read or write a single word.
∴ 3 memory accesses are required.
Consider the following predicates for the domain of real number
P(x, y): x>y
Q(x, y): xy
R(x): x7 = 2
S(x):x>9
This of the following proposition predicates gives the FALSE as the truth value.
A.
For x = 9. R(x) evaluated to True
x7 = 2 ⇒ 97 = 2
∴True V evaluated to True
B.
evaluated to True.
There exist y for every x in the real domain.
C.
is True.
For every x and y, x>y or x≤y always satisfies.
∴ All statements are evaluated to True.
Consider the below given matrix :
Let the Eigen values of this matrix be a , b and c. Then find the value of ab + bc + ac __________.
The following is a scheme for floating point number representation using 12 bits
Let ‘S’, ‘E’ and ‘M’ be sign, exponent and mantissa field respectively Then the floating point number is represented as .
Q. What is the percentage error if stored in above format?
This contains 7 bits while the mantissa part of the given number representation can store only 6 bits.
If p and q are two statements, then statement (p>q) V ¬q is
(p>q) V ¬q
= ¬p V q V ¬q
= True
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