Pipe A can fill a tank three times as fast as pipe B. If together two pipes can fill the tank in 48 min, the slower pipe alone will be able to fill the tank in:
A = 3B
Ratio of efficiency, A : B = 3 : 1
Ratio of times, A : B = 1 : 3 Total capacity = Total efficiency × Total time = 4 × 48 = 192 unit Time taken by slower pipe
B = Total Capacity / Efficiency of B = 192 / 1 = 192 min
In the following question, out of the four alternatives, select the word opposite in meaning to the given word.
Gratuitous
Gratis means without charge; free.
Find the area bounded between parabola and the line y^{2 }= x2y = 2
⇒ x = 2^{2} = 4
: Parabola and line intersect at the point (4,2) ∴ Area =
⇒ A = = 8/3 sq.units
Direction: The bar graph shows the number of employees working under the six different Departments (A, B, C, D, E, F) of a certain company. Study the diagram and answer the following questions.
If departments F and D are merged to create a new department G, then which department will have the least number of employees?
Employees in department A = 25
Employees in department B = 6
Employees in department C = 10
Employees in department E = 15
Employees in department G = 8
∴ Department B has the least number of employees.
In the following question, a sentence is given with a blank to be filled in with an appropriate word. Select the correct alternative out of the four and indicate it by selecting the appropriate option.
Confusion prevails in madrasas in Uttar Pradesh over the distribution of free NCERT textbooks at the academic session ____________ from August.
A sum of Rs. 400 amounts to Rs. 480 in 4 years. What will it amount to if the rate of interest is increased by 2 % for the same time?
We know that, S.I = P x R x T / 100
A = S.I + P
480 = 5.1+400
⇒S.I = 480 − 400 = 80
⇒S.I = P × R × T100
⇒80 = 400 × R × 4100
⇒R = 5%
Now rate is increased by 2% So, new rate is 7% New S.I=400 × 7 × 4 / 100 = Rs.112
New Amount = S.I + P = 112 + 400 = Rs.512
The question below consists of a set of labelled sentences. These sentences, when properly sequenced form a coherent paragraph. Select the most logical order of sentences from among the options.
P: The Information and Broadcasting Ministry plans to conduct an independent study to gauge the
impact of government advertisements on people.
Q: The advertisements are carried on various platforms, including print and visual media.
R: The Directorate of Advertising and Visual Publicity (DAVP) is the nodal agency of the government
for advertising on behalf of the various ministries.
S: The initiative comes ahead of the Lok Sabha election in 2019 for which the government is expected
to reach out to the people and highlight the works done by it in the past 4 years.
The paragraph talks about the plans and advertisements of The Information and Broadcasting Ministry, which is given in sentence P. P is the first statement. The word ‘initiative’ given in the sentence S is talking about the plans. Hence, S must follow P. Now, the introduction of the advertising agency is given in the sentence R, which must be the next statement. Thus, the sequence after rearrangement is PSRQ and option B is the correct answer.
In the following question, some parts of the sentence may have errors. Find out which part of the sentence has an error and select the appropriate option. If the sentence is free from error, select 'No error'.
The gold foil used liberal (1)/ in Thanjavur paintings serves (2)/ many objectives that makes the painting more attractive. (3)/ No error
The error is in part (1) of the sentence. Change ‘liberal’ to ‘liberally’ because in this sentence it is in adjective form while the proper usage of liberal is in its adverb form i.e. ‘liberally’ as it qualifies the gold foil here.
A,B and C can do a job in 6 days, 12 days and 15 days respectively. C works till 1/8 of the work is completed and then leaves. Rest of the work is done by A and B together. Time taken to finish the remaining work by A \& B together is how much?
Remaining work = 1  1/8 = 7 / 8
(A+B)′s1 day′s work =
∴ Time taken in doing 7/8 part of work = ⅞ x 4 = 7/2
= days
A invests 1/3 part of the capital for 1/6 of the time, B invests 1/4 part of the capital for 1/2 of the time and C invests the rest of the capital for the rest of the time. Out of a profit of Rs. 23000, B’s share is?
Ratio of their investment,
=
=
=
= 4 : 9 : 10
B's share = 23000 x 9/23 = Rs. 9000
Identify the correct translation into logical notation of the following assertion.
“You can not ride the roller coaster if you are under 4 feet tall unless you are older than 16 years old.”
Where q = “You can ride the roller coaster”
r = “You are under 4 feet tall”
s = “You are older than 16 years old”
The third option seems the most appropriate as ~Q is on the righthand side.
(r∧= s)→q . If you try to interpret this mathematical statement, you will get that this is the most appropriate.
The traditional computer system stores the data in the form of binary. Arjun is a professor who teaches binary systems to the students. He observes that if ternary (radix 3) is used instead of binary, space can be utilised more efficiently. If n be the no of bits in string when data is stored in the form of binary, then ternary system will need
represented = 2^{n}
No of bits required to stare it intaternary format= log_{3}2^{n }= nlog_{3}2
Consider a processor that includes a base register with indexing addressing mode. Suppose an instruction is encountered that employs this addressing mode and specifies a displacement of 7B2, in hexadecimal. Currently the base and index register contain the decimal numbers 48022 and 8, respectively. What is the address of the operand (In Decimal)?
= 48022 + 8+ 1970
= 50000
A nonpipelined processor has a clock rate of 2.5 GHz and an average CPI (cycles per instruction) of 4. An upgrade to the processor introduces a fivestage pipeline. However, due to internal pipeline delays, such as latch delay, the clock rate of the new processor has to be reduced to 2 GHz.
What is the speedup achieved for executing 100 instructions and MIPS of the upgraded processor?
Time for execution= no of cycle per instruction × cycle time* no of instruction
= 5 × 100 × 0.4 nsec
= 200 nsec
For a pipelined processor,
Time for execution= [no of stages + (no of instruction 1)]× cycle time
= [5 + 99] × 0.5 nsec
= 104 × 0.5 nsec
= 52 nsec
Speed Up= 200/ 52
= 3.84
For upgraded processor, clock frequency= 2GHz
Cycle time= 1/2 nsec = 0.5 nsec
In pipelined processor, the average time of executing an instruction is 1 cycle
Therefore no of instructions in 1 sec= 1/0.5 × 109 instructions
=2000 MIPS
Consider a relation R(A B C) with attribute size of A as 8 bytes. Disk block size is 512 bytes and block pointer is 8 bytes. The best choice for degree (maximum value) for B+ tree, if B+ tree was used for creating indexing on R(A B C) is _________.
(p  1) keys + p Block pointers should fit in a block ie (p1) keys + p Block pointers size ⇐ 512
(2p  1) × 8 ⇐ 512
p ⇐ 65/2
p = 32
If you take p = 33 node size becomes 520 bytes so not possible to fit in a block hence the correct answer is 32.
Find the number of states in minimal FA for the following language :
L = {(a+b) ×  number of a's = (0 mod 4) and (0 mod 8)}
Let set A = number of a's = 0 mod 4
and set B = number of a's = 0 mod 8
clearly , B is a subset of A
Hence A ∩ B = B
So the number of states will be determined only by set B
Hence , the number of a's = 0 mod 8 requires 8 states.
Hence answer = 8
Find the number of MN equations for the language, L = [a^{n}b^{m}  m,n 0]
Number of MN equations = number of states in DFA
The DFA for given language is :
Hence MN equations = 4
Suppose a processor uses a round robin scheduling algorithm to schedule the process. Previously, it was using a time slice of 2 units while scheduling the process. An update is made to the system and the time slice is now changed to 4 units. Then the turnaround time of the process will
Which of the following is/are true?
1. Option A is false, because semaphores are the solution used to avoid busy waiting.
2. Option B is false, Since Priority inversion causes the problem of LiveLock and priority Inheritance is the solution for Live Lock.
3. Option C is true.
To realise a 128x1 MUX , how many 4x1 MUX are required _____________
128/4 = 32
32/4 = 8
8/4 = 2
2/4 = 0.5 ≅ 1
Total = 32 + 8 + 2 + 1 = 43
Consider the following Relational Schema:
Sailors (sid: integer, sname: string, rating: integer, age: real)
Boats (bid: integer, bname: string, color: string)
Reserves (sid: integer, bid: integer, day: date)
Consider the Following Statements:
S1:SELECT S.sname
FROM Sailors S
WHERE NOT EXISTS (SELECT *
FROM Sailors S2
WHERE S2.age < 21 AND S.rating <= S2.rating )
S2: SELECT S.sname
FROM Sailors S
WHERE S.rating > ANY (SELECT S2.rating
FROM Sailors S2
WHERE S2.age < 21 )
Which of the following is true regarding S1 and S2?
S1: it will generate the name of sailors whose rating is more than the rating of every sailor whose age is less than 21.
S2: it will generate the name whose rating is more than the rating of the same sailor with age less than 21.
Consider a binary tree with the following more conditions :
Condition 1: The elements present in the left subtree of a specific node are smaller than that node and elements present in the right subtree of a specific node are greater than that node.
Condition 2: The tree is always balanced i.e. the height of the left subtree and right subtree is approximately the same. (maximum differ by 1)
What is the worstcase time complexity to run a membership test for an element on this tree?
Consider a lower triangular matrix. When this lower triangular matrix is stored in array format then only the elements a[i][j] with i≥j are stored in array i.e only the elements present in lower triangular matrix are stored. Hence less size is consumed to store the array. Consider a lower triangular matrix as [25100 , 25100] with base address as 1000 and size of each element in matrix is 10, If the array is stored in column major order then find the address of the element a [80][45] stored in array ________________.
In general , if the array is a [lb1ub1 , lb2ub2]
Base address = BA
Size of element = c
Number of rows = nr = ub1 lb1 + 1
Number of columns = nc = ub2  lb2 + 1
and lower triangular matrix stored in column major order then
=
So,
Suppose there are certain number of peoples who’s usage are ALOHA dependent and they are producing 100 requests/sec. Consider time to be slotted in 50msec units. After considering all of the above cases find out the chances of win on the 1^{st} attempt?
Number of slots per second = 1/50msec = 20
Channel load = = 100/20 = 5
We know, Poisson Distribution
or the success on the first attempt
K=0
P_{0} = e^{G} = e^{5}
= 6.73 x 10^{3}
You have a class B network 172. 16. 0. 0. You use 11 bits for subnetting. Which of the following is a correct range of IP addresses that belong to the same network?
If we use 11 bits for subnetting, we have 5 networking bits and 11 node bits
From the subnetting formulas above:
M = 5, N=11
The first range = 255.255.X.1 to 255.255.Y.254 where X = 2^{(N8)} + 1 and Y = 2^{(N81)} – 2
For the next ranges, just add 2(N8) at each end of the range
The first range = 255.255.9.1 to 255.255.14.254 = 2^{(N8)}= 23=8
The second range = 255.255.17.1 to 255.255.22.254 so on and so forth.
Consider the following relations, SQL query and given instances of relations: (where keys are underlined)
Student (snum, sname)
Enroll (snum, cname)
SELECT S.name FROM Student S WHERE
S.snum NOT IN (SELECT E.snum FROM Enroll E)
Number of tuples returned by the SQL query is ________.
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.
Consider two languages, L_{1}and L_{2}:
L1_{=} {a^{n} n > = 0} and L_{2} = {b^{n}  n > = 0}
Which of the following correctly represents L_{1}⋅ L_{2}, where ‘⋅’ is the concatenation operation?
Consider the following regular expressions over the alphabet {0, 1}.
I. 1* 0(0 + 1)*
II. (0 + 1)* 01*
III. 1* 0(1 + 0)* + (0 + 1)* 01*
Which of the above regular expressions are equivalent?
Hence all 3 regular expressions are equivalent.
A 4bit preset table UP counter has preset input 0111. The preset operation takes place as soon as the counter becomes maximum, i.e. 1111. The modulus of this counter is
Preset → 8 clock pulses, it is repeating
So mod 8.
Let R be a recursive language. Consider the following operations on languages.
I. Union
II. Intersection
III. Complement
Let X be the number of operations under which R is closed. Then the value of X is ________.
Consider the following synchronization construct used by two process P0 and P1 which need to access a critical section:
P_{0} :
while (true) {
turn_0 = true;
while (turn_l = = true)
turn_0 = false;
CRITICAL SECTION
}
/ * Remainder section*/
P_{1} :
while (true) {
turn_1 = true;
while (turn_0 = = true)
turn_1 = false;
CRITICAL SECTION
}
/*Remainder section*/
Here, turn_0 and turn_1 are shared variables, which are initialized to false. Which of the following statements is true about the above construct?
Consider the following sentence:
“There is exactly one God”
Let G(x) : x is a God
Now consider the following predicate logic statements.
I. ∃ × God(x)
II. ∃ × God(x) ∧∀y (God(y) ⇒ x = y))
III. ∃ x ∀y(God(x) ∧ (∼ God(y) ∨ x =y))
The number of statements which are equivalent to the above sentence is ________.
* I is not correct because it says "at least one God" instead of "exactly one God”.
* II is clearly the correct statement.
* III is same as II and can be obtained using,
p ⇒ q ≡ ∼ p ∨ q
\)God\y) \Rightarrow(x = y \equiv(\sim \operatorname{God}(y) \vee x = y\)
A 3×3 matrix has 3 eigenvalues 1,0.5,3 . What will be the eigenvalues of P^{2} +2P + I?
P^{2} + 2P + I can be written as,
= P^{2}+ 2P.I + I
= (P + I)^{2}
Eigen Values of I are (1, 1, 1)
Eigen Values of P are (1, 0.5, 3)
Eigen Values of (P + I) will be (1 + 1, 0.5 + 1, 3 + 1) = (0, 1.5, 4)
Eigen values of (P + I)2 will be (0^{2}, 1.5^{2}, 4^{2}) = (0, 2.25, 16) =
(0,9/4,16)Hence (A) is the most appropriate choice.
Let (G, *) be a group such that O(G) = 8, where O(G) denotes the order of the group G. Which of the following is true?
Therefore (A) is false, as O(A) = 6 and 6 doesn’t divide 8.
However the converse of the above theorem need not be true, therefore (B) is false.
(C) says there are at least two elements in G with order 1, however this violates the fact that every group can have only one identity element. So (C) is also wrong. So the appropriate choice is (d).
The number of oneone functions possible from a set having n elements to a set having m elements is
Hence option (d) is correct.
Consider a computer system with 32bit virtual addressing and 44bit physical addressing and page size is 4 KB. Each page table entry contains 2 valid bits, 3 protection bits and 2 permission bits. Size of the page table when virtual memory uses single level paging ________ (MB).
Number of pages = = 2^{32} / 2^{12} = 2^{20}
Number of frame = 2^{44} / 2^{12} = 2^{32}
Number of bits in page table entry =32+2 valid bit +3 protection bit +2 permission bit =39 bit = 5B
Page table size = 2^{20} x 5B = 5MB
Consider a 2way set associative cache with total 6 cache blocks and the following sequence of memory block requests arrived:
21, 7, 20, 32, 21, 16, 27, 22, 7, 16, 22
If LRU replacement policy is used then the hit ratio will be ________. (Upto 2 decimal places.)
21 mod 3 = 0 → miss
7 mod 3 = 1 → miss
20 mod 3 = 2 → miss
32 mod 3 = 2 → miss
21 mod 3 = 0 → hit
16 mod 3 = 1 → miss
27 mod 3 = 1 → miss
22 mod 3 = 1 → miss
7 mod 3 = 0 → miss
16 mod 3 = 1 → hit
22 mod 3 = 1 → hit
Hit Ratio = Total Hit / Total Reference = 3 / 11 = 0.27
Consider the following relation R(A, B, C, D, E, F, G) and set of functional dependencies.
F = {BCD → A, BC → E, A → F, F → G, C → D, A → G}
Which of the following is the minimal cover of F?
(1) RHS contain single attribute
A → F …..(1)
F → G …..(2)
A → G …….(3)
Hence FD' (3) is redundant.
Now consider
BCD → A
(BC)+ = BCEDAFG
It contains attribute D, D is redundant.
Minimal cover is {BC → A, BC → E, A → F, F → G, C → D}
So option (c) is correct.
Assume that the time complexity of the most efficient algorithm to find the median in a list of n characters is θ(n logn). It is experimentally found that, when this algorithm is made to run on a list of 16 elements, it takes 256 units of time respectively. Then the time taken by this algorithm when the input list contains 64 elements is _______.
Given, T(n) = θ(n logn) = c.nlog n
Now, T(16) = 256 units
Therefore, c.16.log (16) = 256
⇒ c = 4
Now we have to find T(64).
T(64) = 64.c.log (64)
Substitute the value of c and solve the equation to get
T(64) = 1536 units
Consider the following graphs:
The number of graphs which are Euler ________.
For a graph to be Euler, it must obeys the following conditions:
1. Every vertex of G must have an even degree.
2. G should be connected.
G_{1} is Euler, as it is connected and all vertices in G_{1} have even degree.
G_{2} has 2 vertices with odd degree ⇒ G_{2} is not Euler.
G_{3 }is not connected ⇒ G_{3} is not Euler.
The cube root of a natural number n is defined as the largest natural number m such that m3 ≤ n. The complexity of computing the cube root of n (n is represented in binary notation) is:
You need to find upper and perhaps lower bounds on the complexity of finding an integer cube root m of n. At least one upper bound is trivial, and rules out answers A and B: m can be found in O(log n) time using binary search.
Also note that the input size is O(log n) because the minimum number of bits needed to represent an arbitrary n in binary notation is proportional to log n. Because all bits of the number must be processed to solve the problem, θ(log n) is a lower bound on the time to solve the problem, and therefore the problem cannot be solved in time O((log log n)^{w}) [where w is some constant > 0] because that isn't O(log n).
Thus, answer C applies.
For a complex number z, find the value of
=
Applying L’ Hospital rule
=
=
=
= 2i
Identify the language accepted by the following deterministic finite automata over the input alphabet
Is equivalent to
The given DFA accepts the language of all strings where every string ends with a.
The maximum number of edges in a bipartite graph on 12 vertices is_______.
The number of edges in a bipartite graph on nvertices is atmost n^{2}/4
The maximum number of edges in a bipartite graph on 12 –vertices is n^{2}/4 = 12 x 12 / 4 = 36
Which of the following helps in conversion of deterministic finite automata into a regular expression?
Given a DFA, Kleene’s algo helps to convert it into regular expression. FloydWarshall algorithm is out of context because it is used in graphs and Thompson’s Construction Algorithm transforms a regular expression into an equivalent NFA.
In a 8 bit SISO register, if 16bit data 1011 0101 0010 1010 is applied at the input. The minimum number of clock pulses required to transfer 16bit data at the output is.
Given that 8 bit SISO register n = 8, input data = N = 6 bits.
After application of 16th clock pulse out of 16bit data 8 will already transferred and remaining 8 will be inside register, so to transfer these 8 we need 7 more clock pulse, so total = 16 + 7 = 23
Minimum number of clock pulse=N + (n 1) = 16 + 8  1 = 23
A transaction state changes from active to after the transaction has been rolled back and the database restored to its state prior to the start of the transaction
A transaction state changes from active to aborted, after the transaction has been rolled back and the database restored to its state prior to the start of the transaction.
An instruction is stored at location 300 with its address field at location 301. The address field has the value 400. A processor register R1 contains the number 200. What will be the addressing mode of the effective address of the instruction is 702?
Relative: 302 + 400 = 702
In this mode the content of the program counter is added to the address part of the instruction in order to obtain the effective address.
is__________
=
=
Using L' Hospital rule
⇒ = √2 = 1.414
The difference between local maximum and local minimum value of f(x)=2x^{3 } 24x + 107is
f(x)=2x^{3 }− 24x + 107
⇒ f′(x) = 6x^{2 }− 24=0
⇒ x = ±2 are stationary points
f′′(x) = 12x
⇒ f′′(2) = 24>0, 0,therefore,local minima
f′′(−2) = −24>0, 0,therefore,local minima
f(−2) = 2(−2)^{3 }+ 48 + 107 = 139
f(2) = 2(2)3 − 48 + 107 = 75
∴ Difference = f(−2)  f(2) = 64
Regular languages are
According to Chomsky hierarchy, regular languages come under type 3 i.e. level 3 language class.
Which of the following is correct output for the program code given below?
main( )
{
void pr( );
pr ( );
pr ( );
pr ( );
void pr ( )
{
static int i = 1;
printf (“%c”, (65+ i ++));
}
}
The correct output is “BCD” when the function pr ( ) is first called the value of i is initialized to 1.
After the pr ( ) completes its execution i = 2 is retained for it’s next call as “i” is a static variable.
∴ 65 + 1 = 66 = B
65 + 2 = 67 = C
65 + 3 = 68 = D
∴ BCD is the correct output.
A survey of 150 college students reveals that 83 own cars, 97 own bikes, 28 own motorcycles, 53 own a car and a bike, 14 own car and a motorcycle, 7 own a bike and a motorcycle and 2 own all three. How many students own a bike and nothing else?
A→ The set of students owning cars B→ The set of students owning bikes C→ The set of students owning motorcycles. A = 83, B = 97, C=28, A ∩ B = 53
A∪C = 14, B∩C = 7, A ∩ B ∩ C = 2
A ∪ B ∪ C=150
(B + ABC − AB − BC) denotes the set of students who owns bike and nothing else
If s is a string over (0+1)∗ then let n0(s) denote the number of 0 's in sand n1(s) the number of 1 's in s. Which one of the following languages is not regular?
a) 3 digit primes are finite, hence regular
b) Only finite combinations, hence regular
c) L={s∈(0+1)∗n0(s)−n1(s)≤4} is not regular because to keep track of
a number of 0 stacks is required.
d) regular, it is a standard mod machine
Let X be a random variable following normal distribution with mean +1 and variance 4. Let Y be another normal variable with mean –1 and variance unknown. If P(X ≤ 1) = P(Y ≥ 2), the standard deviation of Y is
X(,4),4(−1,σ^{2})
P(X ≤ −1)
Z = −1 − 1 / 2 = −1
\( = P(Z \geq1) = P(Z \geq 1) = 0.5  P(0 = 0.5 − 0.3413 = 0.1587
If σ = 3 then
P(Z ≥ 2),Z = 2 + 13 = 1
f(x,y)=tan(y^{3}/x^{3})y^{2}x^{2} +xy^{3}+y^{4}
What is the value of f_{x}+2f_{y} at point(3,6)
Given the function f(x,y) is a homogeneous function,so can be applied Euler’s theorem.
Applying Euler’s theorem on f(x,y),we get
xf_{x} + yf_{y} = nf(x, y) [n =4 here]
Substituting (x,y)=(3,6) we have
3f_{x} + 6f_{y} = 4f(3,6) =4(tan(216/27)36 × 9 + 3 × 216 + 216 × 6)
f_{x}+2f_{y}=4(108tan(216/27)+648)
=432 tan (8) + 2592.
A disk has 200 tracks(from 0 to 199). At time t = 0, the pending requests for read are for the following disks in the order:
46, 110, 32, 52, 14, 120, 36, 96.
Current head position is at 50 and moving towards 0.
The total head movement(in number of tracks) incurred if LOOK is applied is
If LOOK is applied, the requests are served in the order: 46, 36, 32, 14, 52, 96, 110, 120
But it moves from 14 to 52 without moving upto 0 and then to 52.
Total track movements = 10+4+18+(52−14)+96−52+110−96+10 = 138.
Consider the following arguments about Bankers algorithm.
i. Each process must have apriori claim of its maximum requirement
ii. There are multiple instances of each resource type
iii. It is a deadlock prevention algorithm
iv. It is a resource allocation algorithm
Which of the following are true?
Bankers algorithm is a deadlock avoidance algorithm. Requires that the system has some additional a priori information available.
Simplest and most useful model requires that each process declares the maximum number of resources of each type that it may need. The deadlockavoidance algorithm dynamically examines the resourceallocation state to ensure that there can never be a circularwait condition.
Resourceallocation state is defined by the number of available and allocated resources, and the maximum demands of the processes. So except statement iii, all are true.
A disk has 200 tracks(from 0 to 199). At time t = 0, the pending requests for read are for the following disks in the order:
46, 110, 32, 52, 14, 120, 36, 96.
Current head position is at 50 and moving towards 0.
If the SSTF algorithm and SCAN algorithm are used, difference in total movement of tracks is
In case of SSTF algorithm, the requests are served in this order
52, 46, 36, 32, 14, 96, 110, 120⇒Total track movements = 146
In case of SCAN, the requests are served in the order: 46, 36, 32, 14, 52, 96, 110, 120
Total track movements = (4636)+(3632)+(3214)+(140)+(520)+(9652)+(11096)+(120110) = 170.
Hence extra movement =20.
A system uses FIFO policy for page replacement. It has 4 page frames with no pages loaded to begin with. The system first accesses 50 distinct pages in some order and then accesses the same 50 pages in reverse order. How many page faults will occur?
50 distinct pages(forward)  50 page replacements
Reverse order first 4 pages will be already loaded then remaining 46 page replacements Total  96.
Let L_{1 }be a regular language, L_{2 }be a deterministic contextfree language and L_{3} a recursively enumerable but not recursive language.
Which one of the following statements is false?
A) L_{1}∩L_{2} is a deterministic CFL, because Regular ∩ DCFL is
always DCFL
B) L_{3}∩L_{1} is need not be recursive, But always REL
C) L_{1.}L_{2} is context free, because Regular. DCFL is always DCFL
D) L_{1}∩L_{2}∩L_{3} is recursively enumerable, because all are REL
Let f be the fraction of a computation (in terms of time) that is parallelizable, P the number of processors in the system, and sp the speed up achievable in comparison with sequential execution – then the sp can be calculated using the relation:
If 'f' be the fraction of a computation that can be parallelized than speedup = is the relation that can be used to calculate speedup.
Consider a problem Z, in it
“Suppose there is a Turing Machine L over the input alphabet ∑ any state of L and a word s € ∑, does the computation of L on s visit the state q?
Choose the correct statement about Z?
This problem is a State Entry Problem. State entry problem can be reduced to a halting problem. We construct a turing machine L with final state ‘q’. We run a turning machine E (for state entry problem) with inputs : L, q, s. We give ‘s’ as input to L. If L halts in the final state ‘q’ then E accepts the input. So, the given problem is partially decidable. If L goes in an infinite loop then L can not output anything. So, E rejects the input. So, the given problem becomes undecidable.
Suppose Suresh and Aditya are shown combinely P_{1} is called the 3 SAT problem. Which of the following is correct?
P1 is 3 SAT problem and also NPCproblem. P1 is solvable in polynomial time.
Now, the 3SAT problem is a Pproblem.
⇒ P = NPC
⇒ P = NP = NPC
∴ Option B is correct.
Which one of the following is not involved in Preventing the dead lock
To prevent deadlock, at least one of the following conditions cannot occur.
• Mutual exclusion.
• Hold and wait
• No preemption
• Circular wait.
No “mutual exclusion “, No '' Hold and wait”, preemption or no circular wait condition occurs to prevent deadlock.
“Nopreemption“not occur to prevent deadlock.
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