Firstranker's choice
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1.Ans. C
Sol. 'Breaks down' is a transitive phrasal verb which means to divide something such as a total amount into separate parts.
Option (c) is most suitable.
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2. Ans. A
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Sol. The search engine business model revolves around the fulcrum of trust.
Fulcrum is any thing that plays a central or essential role in an activity, event, or situation.
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3.Ans. B
Sol. Speed of car A = 50 km/hr
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Speed of car B = 60 km/hr
Since, both cars A and B are moving in same direction, the relative speed = 60 - 50 = 10 km/hr
Distance required between them = 20 km
Time = Distance / Speed = 20 / 10 = 2 hrs
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4.Ans. C
Sol. Let share of each student = x
Total cost of gift = 10 x x
x = 8(x + 150)
x = 600
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Total cost = 10 × 600 = 6000
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5.Ans. D
Sol. A 'court' is for a 'judge' as a 'school' is for a 'teacher'.
Court is a place where a judge works.
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Similarly, school is a place where a teacher works.
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6.Ans. B
Sol. Case I:
Criminals P Q R S
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Assumption F T F F
Result QNC SC Rc Sc
S and R are criminal in the result is impossible because only one person committed the crime.
Case II:
Criminals P Q R S
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Assumption T F F F
Result QC SNC Rc Sc
Q and R are criminal in the result is impossible because only one person committed the crime.
Case III:
Criminals P Q R S
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Assumption F F T F
Result QNC SNC RNC SC
SNC and Sc in the result which is contradiction. [S committed crime and same time not committed crime which is contradiction]
Case IV:
Criminals P Q R S
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Assumption F F F T
Result QNC SNC RC SNC
R is criminal in the result.
Hence this case satisfies only one person committed the crime.
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7.Ans. C
Sol. Percentage of Administrators > Administrators
Total × 100 = 50 / 160 x 100 = 31.25
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8.Ans. B
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Sol. The passage states that the underlying disease behind begging is the failure of the state to protect citizens who fall through the social security net.
option b can be concluded from the above.
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9.Ans. C
Sol.
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Dance 60
Drama 80
Maths 30
Other 38
Both Dance & Drama 5
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Both Drama & Maths 10
All 2
Total number of students = 60 + 80 + 30+38 + 5 + 10 + 2 = 225
25% = 225 / 2
100% = (225 / 25) x 100 = 900
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10.Ans. D
Sol. According to conditions mentioned in the question 'D' is the best suited option.
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11.Ans. D
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Sol. Cache memory size = 16 kB
Block size = 16 B
Main memory address = 32 bit
The number of possible ordered pairs (x, X) where x ? X is k. "Ck for a given value of k from 1 to n.
So total number of ordered pairs in A
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k.
n
-IAI-K.CK()
So II is correct.
(Note that k = 0 is excluded since empty set has no elements and cannot form an order pair such as (x, X)).
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But since by the combinational identity
16K = 214
= 210
Number of lines (N) = 16 = 24
Fully associative cache memory (N-way)
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N = 210 / 24 =1
So, number of sets (S)
P-way
Sk.
n
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2-1
So I is also correct.
So both I and II are correct.
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12.Ans. A
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Sol.
0 A15 A14 A13 A12 A11 A10 A9 A8
1 1 0 0 1 0 0 0
Chip select (CS) = [C800 to CFFF]
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13.Ans. D
Sol. LR parser is a bottom up parser.
Hence it uses right most derivation in reverse order.
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14.Ans. C
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Sol.+28 ? 0000 0000 0001 1100
-28 ? 1111 1111 1110 0100 (2's complement form)
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15.Ans. C
Sol.A = {(x, X), x ? X and X = U}
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The number of k element subsets of a set U with n elements
(R) = Ck
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16.Ans.B
Sol.(a) Address format:
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32 bit
TAG WO
(b) 32 bit
TAG = 28 bit
log-16=4b
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TAG = 28 bit
Index = 0 bit (No address)
x + y = (xy + x'y ')'
= (xy)'
= xy, it is valid.
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(x+y) z = (x+y).z + (x + y)z
= xyz + xz + yz
x(y+z) = x(y+z)+x(y+z)
=xy+xz+yz
So option (b) is invalid.
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(c)
(z ? x) ? x = z ? (x ? x)
Associativity is true on Ex-OR operator so it valid.
(d)
x+y=(x+y)(x+y)
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= (x+y)xy
= (x + y)0
= (x + y), so it is valid.
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17.Ans. B
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Sol. If L is regular, L LR is also regular by closure property.
Suffix (L) and Prefix (L) are also regular by closure property.
However option (b) {wwR |w ? L} need not be regular since if L is an infinite regular language, then {wwR |w ? L} will not only be infinite, but also non-regular. Since it involves string matching and we can increase in length indefinitely and then finite automata FA will run out of memory.
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18.Ans. C
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Sol. For example:
Let, X = +6, n = 4
Y = -5, n = 4
4(X-Y) = +11
Hence,
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Z = 11 which required 5 bits which is (n + 1) bits
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19.Ans. D
Sol. Both I and II are equivalent statements.
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20.Ans. B
Sol.R1: va, b ? G, a R1 b if and only if ag ? G such that a = g-1bg
Reflexive: a = g-1ag can be satisfied by putting g = e, identity "e" always exists in a group.
So reflexive
Symmetric: aRb ? a = g-1bg for some g
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? b = gag-1 = (g-1)-1 ag-1
g-1 always exists for every ge G.
So symmetric
Transitive: aRb and bRc = a = g1-1bg1 and b = g2-1 cg2 for some g1g2 ? G.
Now a = g1-1( g2-1cg2 )g1 = (g2g1)-1cg2g1 = (g2g1)-1cg2g1
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g1 ? G and g2 ? 2 ? G ? g2g1 e G since group is closed so aRb and aRbaRc hence transitive
Clearly R1 is equivalence relation.
R2 is not equivalence it need not even be reflexive, since aR2 a ? a = a-1 va which not be true in a group.
R1 is equivalence relation is the correct answer
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21.Ans. C
Sol.I. Strict 2PL guaranteed conflict serializable because of 2PL condition and also strict recoverable.
II. Thomas Write timestamp ordering ensures serializable. Thomas write rule timestamp ordering allowed to execute schedule which is view equal serial schedule based on timestamp ordering.
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22.Ans. D
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Sol. In a complete graph we can traverse the n vertices in any order and return to the starting vertex and form a Hamiltonian cycle. The number of such cycles will be n!
However, since circular rotations will have to ignored. Since for example K4 with vertices {1, 2, 3, 4}, the cycle 1-2-3-4 is same as 2-3-4-1 is same as 3-4-1-2 etc. we now get only (n - 1)! distinct Hamiltonian cycles. Further, the cycle 1-2-3-4 and 1-4-3-2 are also same (clockwise and anticlockwise).
So ignoring this orientation also we finally get (n-1)! / 2 distinct Hamiltonian cycles.
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23.Ans. C
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Sol. limx->1 (x3 - 2x + 5) / (x3 - 4x - 5) 0/0 form.
So apply L'H rule
lim 4x3 / x3 - 4x - 5 = 108 / 7
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24.Ans. B
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Sol.B+ tree non leaf node have pointer to data records is false statement.
B+ tree non leaf node consists of only keys and tree pointers (node pointers).
Below is the structure of B+ tree non leaf node
B1 K1 B2 K2 B3
Tree Tree Tree
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pointer pointer pointer
K1 and K2 are keys
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25.Ans. D
Sol. L = {a2 + 3k or b10 + 12k} for k = 0
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= a2 (a3)* or b10 (b12)*
= {a2, a5, a8, ..., b10, b22, b34 .....}
The pumping length is p, than for any string w e L with |w| = p must have a repetition i.e. such a string must be breakable into w = xyz such that y = 0 and y can be pumped indefinitely, which is same as saying xyz ? L ? xy*z e L.
The minimum pumping length in this language is clearly 11, since b10 is a string which has no repetition number, so upto 10 no number can serve as a pumping length.
Minimum pumping length is 11. Any number at or above minimum pumping length can serve as a pumping length. The only number at or above 11, in the choice given is 24.
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26.Ans. B
Sol.SMTP is push protocol and to send email and POP3 is pull protocol i.e. to retrieve email.
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27.Ans. (31)
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28.Ans. (26)
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29.Ans. (31)
Sol. S -> Aa
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A -> BD
B -> b|e
D -> d|
Follow (B) = {d, a}
Hence their index in descending order is 31.
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30.Ans. (0.08)
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31.Ans. (2)
Sol. By Fermat's theorem
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3(5-1) mod 5 = 1
34 mod 5 = 1
351 mod 5 = (34)12. 33 mod 5 = 2
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32.Ans. (0.502 to 0.504)
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33.Ans. (80)
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34.Ans. (6)
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35.Ans. (29)
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36.Ans. D
It will not print anything and will not terminate
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37.Ans. B
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38.Ans. C
Sol.
100.10.5.2
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M 255.255.255.252
104.56.10.0
100.10.5.5
N 255.255.255.252
194.50.10.4
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2 00000010
252 11111100
0 00000000
5 00000101
252 11111100
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4 00000100
P
100.10.5.6
255.255.255.252
194.56.10.4
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6 00000110
252 11111100
00000100
N and P belongs to same subnet.
Hence, C is correct answer.
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39.Ans. C
X sends an ARP request packet with broadcast MAC address in its local subnet
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40.Ans. A
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Sol.f1. f2 = S(2, 8, 14)
f1 = f3 (f1f2)
= S(7, 8, 11)
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41.Ans. C
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Sol. (a) {wwR |w ? {a,b}* } is a CFL
(b) {wan bn wR |w ? {a, b}*, n = 0} is a CFL, since we can first push w, then a's, b's pop with a's and wr pops with the w.
So PDA can accept the language.
(c) {wanwR bl |w ? {a, b}*, n = 0} is a not CFL because after pushing w, we need to push a's into stack which will stop the w from being matched with wR. If we don't push a's after w, than later we cannot match with bn. So this language is not acceptable by a PDA and hence not a CFL.
(d) {ambl | l ? {n, 3n, 5n}, n = 0}
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= anbn u anb3n U anb5n is CFL since each of the three parts is a CFL and closure under union guarantees that result also is a CFL.
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42.Ans. C
Sol.X(PQRS) {QR ?S, R?P, S ? Q} decomposed into
Y(PR) {R ? P}
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Z(QRS) {QR ? S, S ? Q}
Candidate key: R Candidate key : QR, RS
Relation Y in BCNF Relation Z in 3NF but not BCNF
Common attribute between Y and Z relations is R which is key for relation Y.
So that given decomposition is lossless join decomposition.
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R ? P in Y
QR ? S
S?Q are in Z
and dependency preserving decomposition.
Hence, C is the correct answer.
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43.Ans. B
Sol. 1 word = 4 bytes
Page size = 8 kB = 213 B
Number of words in 1 page = 213 / 22 = 211
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TLB can hold 128 valid entries so, at most 128 x 211 memory address can be addressed without TLB miss.
128 x 211 = 256 x 210
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44.Ans. B
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45.Ans. C
Sol.?x[vzx ? ((z = x) v (z = 1)) ? ?w (w > x) ^ (?z zw ? ((w = z) v (z = 1)))]
The predicate & simply says that if z is a prime number in the set then there exists another prime number is the set which is larger.
Clearly $ is true in S2 and S3 since in set of all integers as well as all positive integers, there is a prime number greater than any given prime number.
However, in S1: {1, 2, 3, .....100} f is false since for prime number 97 ? S1 there exists no prime number in the set which is greater.
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So correct answer is C.
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46.Ans. A
Sol.SDT for inserting type information in the symbol table
D ? TL {L.idtype = T.stype}
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T? int {T.stype = int}
T? float {T.stype = float}
L ? L1, id {L1.itype = L.itype}
addtype(id.entry, L.itype)
L? id addtype(id.entry, L.itype)
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47. Ans. C
Sol.S1: The set Lre is known to be countably infinite since it corresponds with set turing machines.
S2: Since syntactically valid C programs surely run on Turing machines, this set is also a subset of set of Turing machines, which is countable.
S3: Set of all languages 2S* which is known to be uncountable. S* countably infinite ? 2S* is uncountable.
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S4: Set of all non-regular languages includes set LNOT RE which is uncountable infinite and hence is uncountable.
So, S3 and S4 are uncountable.
Hence, B is the correct answer.
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48.Ans. C
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Sol. If no two edges of G have same weight surely G will have unique spanning tree is true.
So I is true
Also if, for every cut of G, there is a unique minimum weight edge crossing the cut then G will have unique spanning tree is also true. So II is true
[Note: The converse of II is not true, but that is not relevant to this question]
So both I and II are true.
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Option (d) is correct.
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49.Ans. A
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50.Ans. A
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51.Ans. (2)
Let's assume,
Z = 2
P1 P2 P3 P4
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0 1 2 3
Arrival time CPU time Completion time Waiting time
P1 0 3 4 1
P2 1 1 2 0
P3 3 3 9 3
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P4 4 6 10 0
Average waiting time (WT)
= 1+0+3+0 / 4 = 1 ms
Hence,
Z = 2
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52.Ans. (4.0 to 4.1)
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53.Ans. 5
Sol.
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S -> .S
S -> .L>
L -> .LS
L -> .S
S -> .4
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Sid .
Total number of items in the set GOTO (I0, () is 5.
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54.Ans. (12)
Sol. Product of eigenvalues is same as the determinant of a matrix.
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| 1 3-2 3-22 33-23 |
| 1 2 22 22 |
| 1 1 1 |
1 3 9 15
1 2 4 19
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1 1 1 0
| 1 4-2 42-22 43-23 |
| 1 5-2 52-22 53-23 |
(3-2)(4-2)(5-2)
= 1.2.3.2= 12
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55.Ans. (160)
Sol. Total time to transfer a cache block = 1 + 3 + 8 = 12 cycles
12 cycles = 32B
8 * 106 / sec
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= 160 x 106 bytes/sec
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56.Ans. 4.25
No. of pairs with path length 0=8.0=8.
No. of pairs with path length 1=0.1=0.
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No. of pairs with path length 2=8.2=8.
No. of pairs with path length 3=0.3=0.
No. of pairs with path length 4=16.4=16.
No. of pairs with path length 5=0.5=0.
No. of pairs with path length 6=32.6=32.
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Total number of possible pairs =8×8=64=8×8=64
So, expected path length,E(x),
=0*8/64+2*8/64+4*16/64+6*32/64=272/64=4.25
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57.Ans. (0.8)
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Sol. It is given that, Polynomial 3x2 + 6xY + 3Y + 6 has only real roots
b2 - 4ax = 0
(6Y)2-4(3) (3Y+ 6) = 0
Y2 - Y + 2 = 0
? (-8, -1] n [2, 8)
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? ? ? [2, 6)
Since y is uniformly distributed in (1, 6)
Probability distributed function,
F(Y) = 1/5 12 cycles
[Y]--0.8
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58.Ans. (120)
Sol. The DFA for accepting L will have 5! = 120 states, since we need one state for every possible permutation function on 5 elements. The starting state will be "id" state, named as
| 12345 |
| 12345 |
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and from there n! arrows will go the n! states each named with a distinct permutation of the set {1, 2, 3, 4, 5}. Since composition of permutation function is closed every arrow has to go to some permutation and hence some state.
Since the language only has those strings where ?(?) = id only the starting state ("id" state) will be the final state.
Sample machine with only 2 states is shown below
| 12345 |
| 12345 |
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| 12345 |
| 12345 |
| 12345 |
| 13245 |
(| 12345 |)
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| 12345 |
| 12345 |
| 12345 |
| 12345 |
| 13245 |
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(| 12345 |)
| 13245 |
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59.Ans. (3)
Sol.3 switches of ethernet are required to connect 15 computers.
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Hence, 3 is correct answer.
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60.Ans. (3)
Sol.f = Sm(0, 2, 5, 7, 8, 10, 13, 15)
f = ?(1, 3, 4, 6, 9, 11, 12, 14)
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C. D
A, B
00 01 11 10
00 0 0
01 1 1 (B+D)
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11 0 0
10 1 1
f=(B+D)(B+D)
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61.Ans. (5)
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Sol.
Student
Performance
Roll no. Student name
Roll no.
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Student code
Marks
1 Amit T
A
80
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2 Priya B
05
3 Vinit X
C
00
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4 Rohan Z
A
80
5 Smita Z
C
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02
C
80
Total 5 different student names all 5 group records in result. (In where condition no condition over Roll_no so query produces all groups.)
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62.Ans. (5)
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63.Ans. (10)
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64.Ans. (97)
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Sol. n = px q = 3007
f(n) = (p – 1) (q - 1) = 2880
By RSA algorithm, n = 31 × 97 in which 97 is prime factor which greater than 50.
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65.Ans. (1)
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Sol.
C. D
A, B
00 01 11 10
00 X Y Z R
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Y V X1 Y1
01 Z1 Y1 V1 X1 Y2
5 Y3 V2 X1 Y1
6 Y2 V3 X3 Y3
11 1 22 Y3 V2 1 Y2
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2 24 Y2 V2
10 Y4 24
?. (s xPY-R.Y.R.V. V2) (Px))x2
Q X Y T
R Y V
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X2 Y1 2
Y1 V1
X1 Y2 5
Y3 V2
X1 Y1 6
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Y2 V3
II (QY-R.Y-QT-2 (Q×R)) = X1
I-II ?
f X
X2 one record in result.
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This download link is referred from the post: GATE Previous Last 10 Years 2010-2020 Question Papers With Solutions And Answer Keys
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