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Computer Science and Information Technology

GENERAL APTITUDE

One Mark Questions(Q01 One Mark Questions(Q01

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05)05)

01.

01. The The cost cost of 7 of 7 penpens, 8 s, 8 pencpencils ils and and 33 sharpeners is Rs 20. The cost of 3 pencils, sharpeners is Rs 20. The cost of 3 pencils, 4 sharpeners and 5 erasers is Rs 21. The 4 sharpeners and 5 erasers is Rs 21. The cost of 4 pens, 4 sharpeners and 6 erasers is cost of 4 pens, 4 sharpeners and 6 erasers is Rs 25. The cost of 1 pen, 1 pencil, 1 Rs 25. The cost of 1 pen, 1 pencil, 1 sharpener and 1 eraser is ________ (Rs) sharpener and 1 eraser is ________ (Rs)

Ans: 6 Ans: 6 Sol:

Sol: Let the costs of pens, pencil, eraser andLet the costs of pens, pencil, eraser and sharpener be p

sharpener be pnn, p, ppp, e and s respectively, e and s respectively

Given Given 7p 7pnn+ 8p+ 8ppp+ 3s = 20+ 3s = 20 3p 3ppp + 4s + 5e = 21 + 4s + 5e = 21 4p 4pnn+ 4s + 6e = 25+ 4s + 6e = 25

Adding all three equations Adding all three equations 11p

11pnn + 11p + 11ppp+ 11s + 11e = 66+ 11s + 11e = 66



1p 1pnn + 1 p + 1 ppp+ 1s + 1e = 6+ 1s + 1e = 6

02.

02. Sentence Sentence Completion:Completion:

Although some think the terms "bug" and Although some think the terms "bug" and "insec

"insect" aret" are ---, the form---, the former term actuer term actuallyally re

referfers tos to --- grgrououp of insp of insecects.ts. (A)

(A) paraparallellell - an - an idenidenticticalal (B

(B)) preprecicisese - an e- an exacxactt (C)

(C) intintercerchanghangeabeablele - parti- particulcularar (D)

(D) exclexclusiusiveve - a - a sepaseparaterate..

Ans: (C) Ans: (C) Sol:

Sol: The word "although" indicates that the twoThe word "although" indicates that the two parts of the sentence contrast with each parts of the sentence contrast with each

other: although most people think about the other: although most people think about the terms "bug" and "insect" one way, terms "bug" and "insect" one way, something else is actually true about the something else is actually true about the terms.

terms. ChoiceChoice (C) log(C) logically ically complecompletes thetes the sentence, indicating that while most people sentence, indicating that while most people think the terms are "interchangeable," the think the terms are "interchangeable," the term "bug" actually refers to a "particular" term "bug" actually refers to a "particular" group of insects.

group of insects. 03.

03. Sentence Sentence improvement:improvement:

Underestimating its value, breakfast is a Underestimating its value, breakfast is a

Hyderabad

Hyderabad DelhiDelhi BhopalBhopal PunePune BhubaneswarBhubaneswar BengaluruBengaluru | |LucknowLucknow PatnaPatna ChennaiChennai | |VijayawadaVijayawada VisakhapatnamVisakhapatnam TirupatiTirupati

CE

CE EngineEngineeriering ng cadecademymy

meal many people skip. meal many people skip. (A)

(A) Underestimating its value, breakfast isUnderestimating its value, breakfast is a meal many people skip

a meal many people skip (B)

(B) BreakfaBreakfast is st is skippeskipped by d by many pmany peopleeople bec

becausause of theie of theirr undeunderestrestimaimatinting itsg its value

value (C)

(C) Many Many peoppeople, le, undeunderestrestimaimatinting tg thehe value of breakfast, and skipping it. value of breakfast, and skipping it. (D) Many people skip breakfast because (D) Many people skip breakfast because

they underestimate its value. they underestimate its value.

Ans: (D) Ans: (D) Sol:

Sol: The problem with this sentence is that theThe problem with this sentence is that the opening phrase "underestimating its value" opening phrase "underestimating its value" modifies "breakfast," not "people." The modifies "breakfast," not "people." The order of the words in the sentence in choice order of the words in the sentence in choice (D) does not have this problem of a (D) does not have this problem of a mis

misplaplaced modced modifyifying phring phrasease. Choic. Choicee (D)(D) also clarifies the causal relationship also clarifies the causal relationship between the two clauses in the sentence. between the two clauses in the sentence. None of the other choices convey the None of the other choices convey the information presented in the sentence as information presented in the sentence as effecti

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: : 22:: CCSSIITT 04.

04. Spot the Spot the error, error, if if any:any:

If I were her / I would accept / his offer If I were her / I would accept / his offer ((AA)) IIf f I I wweerre e hheerr ((BB)) I I wwoouulld d aacccceepptt ((CC) h) hiis s ooffffeerr ((DD) N) No o eerrrroorr

Ans: (A) Ans: (A) Sol:

Sol: Rule we should use Subjective case of Rule we should use Subjective case of pronoun

pronoun after after BE BE forms…am, forms…am, isis, are was, are was were.,, has been, have been, had been. were.,, has been, have been, had been.

Her is

Her is an objean objective ctive casecase --- ---If I we

If I werere she. is she. is correctcorrect 05.

05. KisKishenhenkant kant walwalks 10 kks 10 kiloilometmetres tres towaowardsrds North. From there, he walks 6 kilometres North. From there, he walks 6 kilometres towards south. Then, he walks 3 kilometres towards south. Then, he walks 3 kilometres towards east. How far and in which towards east. How far and in which direction is he with reference to his starting direction is he with reference to his starting point?

point?

(A) 5 kilometres, West Direction (A) 5 kilometres, West Direction (B) 5 kilometres, North-East Direction (B) 5 kilometres, North-East Direction (C) 7 kilometres, East Direction

(C) 7 kilometres, East Direction (D) 7 kilometres, West Direction (D) 7 kilometres, West Direction

Ans: (B) Ans: (B) Sol:

Sol: The movements of Kishenkant are as The movements of Kishenkant are as shown in figure shown in figure A A to to B, B, B B to to C C and and C C to to DD AC = (AB AC = (AB – – BC) = (10 BC) = (10 – – 6) km = 4 km 6) km = 4 km Clearly, D is to the North-East of A Clearly, D is to the North-East of A



Kishenkant’s distance Kishenkant’s distance from starting point from starting point A A AD = AD = ACAC22



CDCD22







₍₍₄₄₎₎22 ₍₍₃₃₎₎22 25 25 = 5 km= 5 km

Two Marks Questions(Q06 Two Marks Questions(Q06

_–

10)10)

06

06.. ThThe ie infnfininitite se sum um 1+1+ 7 7 4 4 + + ₂₂ 7 7 9 9 + + ₃₃ 7 7 16 16 + + ₄₄ 7 7 25 25 + + --- - e- equaqualsls Ans: 1.8 to 2 Ans: 1.8 to 2 Sol:

Sol: We have to find the sum of the seriesWe have to find the sum of the series 1+ 1+ 7 7 4 4 + + ₂₂ 7 7 9 9 + + ₃₃ 7 7 16 16 + + ₄₄ 7 7 25 25 + + -Putting x = Putting x = 7 7 1 1 we get we get 1 + 2 1 + 222x + 3x + 322xx22 + 4 + 422xx33+ 5+ 522xx44 + - - - - - + -s s = = 1 1 + + 4x 4x + + 9x9x22 + 16x + 16x33 + 25x + 25x44 s.x = x + 4x s.x = x + 4x22 + 9x + 9x33 + 16x + 16x44 + - - - - - + -ss – – sx = 1 + 3x + 5x sx = 1 + 3x + 5x22 + 7x + 7x33 + 9x + 9x44 + - - - - - + -x(s x(s – – sx) = x + 3x sx) = x + 3x22 + 5x + 5x33 + 7x + 7x44 + - - - - + - (s (s – – sx) sx) – – x(sx(s – – sx) = 1 + 2x + 2x sx) = 1 + 2x + 2x22 + 2x + 2x33++ ---- ---- ---- ---- - - + + ttoo



(1 (1 – – x) x)22 s = 1+ s = 1+ x x 1 1 x x 2 2



; since; since xx



11 s = s = ₃₃ )) x x 1 1 (( x x 1 1





We may use it as direct formula for solving We may use it as direct formula for solving this type of problem

this type of problem Substituting x = Substituting x = 7 7 1 1 we get we get s = s = 3 3 7 7 1 1 1 1 7 7 1 1 1 1             = = 27 27 49 49 216 216 7 7 343 343 8 8





0 077.. IIff 55 aa 2 2 cc 3 3 zz cc 2 2 b b 3 3 y y b b 2 2 aa 3 3 x x













and a, b and a, b

and c are in continued proportion and b, c, and c are in continued proportion and b, c, aa are are in in cocontntininueued prd propoportortioion, n, ththenen

cc 3 3 zz b b 2 2 y y aa x x



is is ______________ (( a, b and c are a, b and c are

in continued proportion means b

in continued proportion means b22 = ac) = ac) 1 1 11 55 10 km 10 km 6 km 6 km 3 km3 km D D A A 4 km 4 km B B C C

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Cancel Anytime.

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:

: 3 3 :: PPrre e GGAATTE E QQuueessttiioon n PPaappeerr Also b, c, a are in continued proportion

Also b, c, a are in continued proportion



c c22 = ab = ab --- (--- (2)2) From (1) and (2) From (1) and (2) b b22cc22== aa22bcbc



a a22 = bc = bc --- --- (3)(3)

Conditions (1), (2) and (3) can only be Conditions (1), (2) and (3) can only be satisfied

satisfied when when a a = = b b = = c c = = k k (say)(say)



2525 k k zz k k y y k k x x 5 5 k k 5 5 zz k k 5 5 y y k k 5 5 x x









k k zz 3 3 1 1 k k y y 2 2 1 1 k k x x cc 3 3 zz b b 2 2 y y aa x x







= = 3 3 25 25 2 2 25 25 25 25



= = 6 6 5 5 45 45 6 6 275 275 6 6 11 11 25 25





08.

08. RaspRasputiutin wan was bors born in 3n in 3233 B233 B.C. T.C. The yhe yearear of birth of Nicholas when successively of birth of Nicholas when successively divided by 25, 21 and 23 leaves remainder divided by 25, 21 and 23 leaves remainder of 2, 3 and 6 respectively. If the ages of of 2, 3 and 6 respectively. If the ages of Nicholas, Vladimir and Rasputin are in Nicholas, Vladimir and Rasputin are in arithmetic progression, when was Vladimir arithmetic progression, when was Vladimir born? born? ((AA) ) 3322227 7 BB..CC ((BB) ) 3322229 9 BB..CC ((CC) 3) 322330 0 BB..CC ((DD) ) 3322331 1 BB..CC Ans: (C) Ans: (C) Solution:

Solution: The year of birth of NicholasThe year of birth of Nicholas 25 25 21 21 2323



 



2 2 3 3 66



3227 3227

The ages of Nicholas, Vladimir and The ages of Nicholas, Vladimir and Rasputin are in A.P

Rasputin are in A.P The

The ages ages of of Nicholas Vladimir Nicholas Vladimir RasputinRasputin 3

322227 7 ? ? 33223333



Vladimir age = Vladimir age =

2 2 Rasputin Rasputin Nicholas Nicholas



= = 2 2 3233 3233 3227 3227



= 3230 B.C = 3230 B.C 09.

09. RecRecent ent stustudiedies has have hive highlghlighighted ted thethe harmful effects of additives in food (colors, harmful effects of additives in food (colors, preservatives, flavor enhancers etc.). There preservatives, flavor enhancers etc.). There are no synthetic substances in the foods we are no synthetic substances in the foods we

Foo

Foods.ds. WhiWhich of thch of the folle followiowing, ing, if truef true,, would most weaken the contention of would most weaken the contention of Munchon Foods?

Munchon Foods? (A)

(A) Some Some synsynthetthetic suic substabstancences are s are notnot harmful

harmful

(B) Some natural substances found in foods (B) Some natural substances found in foods

can be harmful can be harmful

(C) Food without additives is unlikely to (C) Food without additives is unlikely to

taste good taste good (D)

(D) MunMunchochon n FFoodoods s prproducoduces es oonlynly breakfast cereals

breakfast cereals

Sol: Munchon’s contention is that buying their Munchon’s contention is that buying their products safeguards health. To weaken that products safeguards health. To weaken that argument we can show that, for some argument we can show that, for some reason, their foods might not be healthy. reason, their foods might not be healthy. So think about an alternative cause So think about an alternative cause 10.

10. To opTo open a en a loclock, a kk, a key iey is taks taken ouen out of t of aa collection of n keys at random. If the lock collection of n keys at random. If the lock is not opened with this key, it is put back is not opened with this key, it is put back into the collection and another key is tried. into the collection and another key is tried. The process is repeated again and again. It The process is repeated again and again. It is given that with only one key in the is given that with only one key in the collection, the lock can be opened. The collection, the lock can be opened. The probability

probability that that the the lock lock will will open open in in ‘n‘nthth’’ trail is _____. trail is _____. (A) (A) n n n n 1 1













_(B)_(B) nn n n 1 1 n n













 

(C) 1 (C) 1 – – n n n n 1 1 n n













 

_{(D) 1}_{(D) 1 –}_– nn n n 1 1













Sol: Probability that the lock is opened in a trailProbability that the lock is opened in a trail is is n n 1 1

(since there is exactly one key, which (since there is exactly one key, which opens the lock)

opens the lock)



The chance that the lock is not opened in The chance that the lock is not opened in a particular trail = 1 a particular trail = 1 – – n n 1 1

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Cancel Anytime.

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: : 44:: CCSSIITT T T11 (e , f, X (e , f, X,, a c)a c) Fk Fk T T22 (e g) (e g) RC RC (e (e f g f g XX aa c)c)

Techinical Questions

One Mark Questions (Q11 One Mark Questions (Q11

_–

35)35)

11.

11. ThThe e resresulult t evaluaevaluatiting ng tthe he popoststfifixx expression expression 8 2 4 * 8 2 4 * / 2 3 * / 2 3 * + 5 1 *+ 5 1 * – – is __________. is __________. Ans: 2 Ans: 2 Sol: Sol: 12

12.. FoFor wr whihich ch vavalulue oe of f



the following system the following system of equations is inconsistent? of equations is inconsistent? 3x 3x + + 2y 2y + + z z = = 1010 2x 2x + + 3y 3y + + 2z 2z = = 1010 x x + + 2y 2y ++



z = 10z = 10 Ans: 1.4 Ans: 1.4 Sol : Sol : 00 2 2 1 1 2 2 3 3 2 2 1 1 2 2 3 3







3(3 3(3



– – 4) 4) – – 2(2 2(2



– – 2) + (42) + (4 – – 3) = 03) = 0 13.

13. ConConsidesider tr the he folfollowlowing ing E-R E-R diadiagramgram

Total number of attributes in each of the relation Total number of attributes in each of the relation schemes in relational model is ______

schemes in relational model is ______

Ans: 13 Ans: 13 Sol: Sol:

The

The relation relation schemes schemes in in relational relational modelmodel

B

B (c (c dd11 d d22))

Total number

Total number of attributes of attributes = 13= 13 14.

14. The The SmaSmallellest Nst Negategative ive intintegeeger thr that cat can bean be represented in signed 2’s complement represented in signed 2’s complement notation with 10 bit is _______.

notation with 10 bit is _______.

Ans:

Ans: – – 512512 Sol:

Sol: Range for n bit is Range for n bit is – – (2(2nn – – 11) to +(2) to +(2nn – – 11 – – 1)1) Given n = 10, Given n = 10,



– – (2(21010 – – 11) to + (2) to + (21010 – – 11 – – 1)1)



– – (2(299) to + (2) to + (299 – – 1)1) 8 8 88 8 8 2 2 4 4 1 1 ₁₁ 2 2 3 3 2 * 4 = 8 2 * 4 = 8 8 / 8 = 18 / 8 = 1 6 6 1 1 2 * 3 = 6 2 * 3 = 6 7 7 6 +1 = 7 6 +1 = 7 7 7 5 5 1 1 5 5 7 7 5 * 1 = 5 5 * 1 = 5 2 2 7 7 – – 5 = 2 5 = 2 C C R R XX S S A A b b B B S S d d d d11 d d22 aa f f gg ee AS AS (a(a c bc b)) Fk Fk

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:

: 5 5 :: PPrre e GGAATTE E QQuueessttiioon n PPaappeerr

II11 3 3 II22 6 6 II44 11 11 II66 21 21 II88 41 41 II99 75 75 II77 34 34 II55 19 19 II33 2 2 1 1 33 44 99 55 55 10 1010 10 1515 2020 15.

15. The The maximaximum mum numbnumber er of of artarticuiculatlationion points of any binary tree of 100 nodes is points of any binary tree of 100 nodes is ______. ______. Ans: 98 Ans: 98 Sol: Sol:

Maximum number of articulation points Maximum number of articulation points are possible if the binary tree is left skewed are possible if the binary tree is left skewed (or)

(or) rigright skeweht skewed.d. In

In left left skewed skewed (or) (or) right right skewed skewed binary binary treetree with ‘n’ nodes contains maximum (n– with ‘n’ nodes contains maximum (n– 2)2) articulation points.

articulation points. Since

Since we we have have 100 100 nodes, nodes, so so we we have 98have 98 articulation points.

articulation points. 16.

16. A A grgrououp p of of N N ststatatioions ns shashare re a a 64 64 KbKbpsps pure ALOHA channel. Each station pure ALOHA channel. Each station outputs a 2000 bits frame on an average of outputs a 2000 bits frame on an average of once every 100 sec, even if the previous once every 100 sec, even if the previous one has not yet been send. What is the one has not yet been send. What is the maximum value of N ? maximum value of N ? Ans: 576 Ans: 576 Sol: Sol:

Pure ALOHA throughput = 18 % Pure ALOHA throughput = 18 % Utilization bandwidth = 0.18

Utilization bandwidth = 0.18



64 Kbps 64 Kbps = 11.52 Kbps = 11.52 Kbps This has

This has to be shareto be shared by N statiod by N station.n. Transmission rate = 2000/100 = 20 bps Transmission rate = 2000/100 = 20 bps N = 11520/20 = 576

N = 11520/20 = 576 1

177.. LLeet t FF11, F, F22, ... F, ... Fnn be files with length L be files with length L11,,

L

L22... L... Lnn we would like to merge all of the we would like to merge all of the

files together to make a single file. The files together to make a single file. The

5, 3, 10, 20, 15, 10, 5, 1, 2, 4 is 5, 3, 10, 20, 15, 10, 5, 1, 2, 4 is _______. _______. Ans : 219 Ans : 219 Sol: Sol:

First arrange all files in the increasing First arrange all files in the increasing order of length and then in each iteration, order of length and then in each iteration, select two files which are having least select two files which are having least number of records and merge them into number of records and merge them into single file.

single file.

Continue this process until all files are Continue this process until all files are completed.

completed.

Minimum cost of merging all files Minimum cost of merging all files

= 3 + 6 + 9 + 11 + 19 + 21 + 34 + 41 + 75 = 3 + 6 + 9 + 11 + 19 + 21 + 34 + 41 + 75 =

= 219219 18

18.. inint ft f(l(lonong ig int nt n)n) {{ int sum; int sum; if (n = = 9) if (n = = 9) return 1; return 1; if if (n (n < < 9 9 )) return return 0;0; sum sum = = 0;0; while while (n (n > > 0)0) {{ sum sum + + = = n%10;n%10;

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Cancel Anytime.

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: : 66:: CCSSIITT What is the return value of f(49698589) is

What is the return value of f(49698589) is ________. ________. Ans: 0 Ans: 0 Sol: Sol:

A number is divisible by 9 iff the sum of A number is divisible by 9 iff the sum of the digits of the number is divisible by 9. the digits of the number is divisible by 9. f(49698589) f(49698589)



4 + 9 +6 + 9 + 8 + 5 + 8 +9 4 + 9 +6 + 9 + 8 + 5 + 8 +9 = 58 = 58 f(58) f(58)



5 + 8 = 13 5 + 8 = 13 f(13) f(13)



1 + 3 = 4 1 + 3 = 4 f(4) f(4)



return ‘0’ return ‘0’ 19.

19. MiMininimumum num numbmber oer of 6 × 6f 6 × 64 De4 Decocodederr needed to implement 8× 256 Decoder with needed to implement 8× 256 Decoder with enable inputs (without using extra enable inputs (without using extra Hardware) is ______. Hardware) is ______. Ans: 5 Ans: 5 Sol: Sol: 1

1 master master + + 4 4 slavesslaves

Total 5 decoders are minimum needed. Total 5 decoders are minimum needed. 20.

20. In wIn which hich of tof the mhe modeodes of s of operoperatiation ton thehe transmission error does not affect the transmission error does not affect the future bits.

future bits.

(A) Cipher block chaining (A) Cipher block chaining (B) Cipher feedback (B) Cipher feedback (C)

(C) OutOutput feedbput feedback ack (D) B and C

(D) B and C

21.

21. AssAssume ume intintermermediediate ate routrouter rer receieceived ved aa IP

IPVV6 datagram. If the datagram is too large6 datagram. If the datagram is too large

to be forwarded over the outgoing link, the to be forwarded over the outgoing link, the router is _______.

router is _______. (A

(A)) DrDropopss (B) Fragments (B) Fragments (C) Forwa

(C) Forwardrd withowithout fragmenut fragmentingting (D) None

(D) None

Sol: Drops and sends “Packet Too Big” ICMPDrops and sends “Packet Too Big” ICMP error message

error message 22.

22. ConConsidesider tr the fhe follollowiowing eng enumnumeraeratiotions &ns & choose the false answer:

choose the false answer:



1:Effective enumeration of all cfls that is1:Effective enumeration of all cfls that is countably infinite.

countably infinite.



22:A :A coucounnttaablbly y iinnffiinniittee eeffffececttiivvee enumeration of all pairs of real enumeration of all pairs of real numbers.

numbers.



3:A 3:A countably countably infinite infinite effectiveeffective enumeration of all TMs

enumeration of all TMs



4:An effective enumeration of all4:An effective enumeration of all recursive sets that is countably infinite recursive sets that is countably infinite (A)

(A)



1 ×1 ×



2 does not exist2 does not exist (B)

(B)



2 ×2 ×



3 does not exist3 does not exist (C)

(C)



3 ×3 ×



4 does not exist4 does not exist (D)

(D)



1 ×1 ×



3 does not exist3 does not exist

Sol: Since the descriptions of TMs are finite &Since the descriptions of TMs are finite & all cfls are finite we can enumerate all cfls are finite we can enumerate effectively all the descriptions in effectively all the descriptions in increasing order of size. So

increasing order of size. So



1 1 ××



3 3 isis

E E E E 6 × 64 6 × 64 6 × 64 6 × 64 6 × 64 6 × 64 6 × 64 6 × 64 E E E E 6 I/P’s 6 I/P’s 6 I/P’s 6 I/P’s 6 I/P’s 6 I/P’s 6 I/P’s 6 I/P’s 6464 64 64 64 64 64 64 0 0 1 1 2 2 3 3 63 63 6 × 64 6 × 64 Mas Mas 2 I/P’s 2 I/P’s 4 I/P’s 4 I/P’s EN EN 1 1 .... .. .. 256 256 Slaves Slaves

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Cancel Anytime.

:

: 7 7 :: PPrre e GGAATTE E QQuueessttiioon n PPaappeerr 23.

23. WhWhat iat is ths the Tie Time Cme Comomplplexiexity ty of tof thehe following piece of code?

following piece of code? void A( ) void A( ) { { int int nn



2222KK ;; for(i

for(i = = 1; 1; i i < < n; n; i++)i++) {{ j j = = 2;2; while(j while(j < < = = n)n) {{ j = j j = j22;; printf(“ACE”); printf(“ACE”); }} }} } }

((AA) ) OO((n n lloog g nn)) ((BB) ) OO((n n lloogg22 lloogg ))nn₂₂

((CC) ) OO((nn)) ((DD) ) OO((lloog g nn))

Sol: For each value of i, the while loop will beFor each value of i, the while loop will be executed K + 1 Times.

executed K + 1 Times. So,

So, the the Total Total number number of Times of Times while while looploop executed = n (K + 1) executed = n (K + 1) but but nn



2222KK



2 2k k = = loglognn₂₂



k = log k = log22 loglognn₂₂



Total Time complexity Total Time complexity = n(log

= n(log22 loglog + 1)nn₂₂ + 1)

(A) Every key of R contains A (A) Every key of R contains A (B) No

(B) No key of key of R contains R contains AA

(C) Some key of R contains A while some (C) Some key of R contains A while some

other

other key key does notdoes not (D) None of the above (D) None of the above

Sol:Based on given dependencies candidate keyBased on given dependencies candidate key is AD (or) A $.

is AD (or) A $.

If $ is A or B or C then candidate key is A. If $ is A or B or C then candidate key is A. If

If $ $ is is E E then then candidate candidate key key is is AD AD and and AE.AE. 25.

25. ConConsidesider thr the mie minimnimum sum stattate dfe dfa M fa M for tor thehe set denoted by the complement of the set denoted by the complement of the intersection of the sets denoted by (rs + r) intersection of the sets denoted by (rs + r)**ss & r(sr + r)

& r(sr + r)**

Choose the correct answer: Choose the correct answer: (A)

(A)M has one staM has one statete (B)

(B) M has twM has two stateo statess (C)

(C) M is not unique becM is not unique because of ause of



(D)M has more than two but a finite (D)M has more than two but a finite

number of states number of states

Sol: The given regular expressions haveThe given regular expressions have nothing in common. The complement of nothing in common. The complement of the intersection is the set (r + s)

the intersection is the set (r + s)**. This. This needs one state.

needs one state. 26

26.. CoConsinsider der ththe e fofollllowowiningg (S

(S11) ((~p) ((~p



q) q)



~q) ~q)



p p

(S

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: : 88:: CCSSIITT

S

S22:: whwhen p en p hahas trs trututh vah valulue fae falslse, e, q haq hass

truth value false and r has truth value truth value false and r has truth value true, we have S

true, we have S22 is false. is false.



S S22 is not valid is not valid

S

S33 : : whwhen p ien p is fas false lse anand q is td q is trurue; we; we hae have,ve,

S

S33 is false. is false.



S S33 is not valid is not valid

S S44:: ((~~((pp



q)q)



p) p)



~q ~q



((p ((p



~q)~q)



p)p)



~q ~q



((~q ((~q



p)p)



p)p)



~q ~q By By commutativitycommutativity



(~q(~q



(p(p



p))p))



~q By associativity ~q By associativity



(~q (~q



p) p)



~q ~q

By idempotent law By idempotent law



T T ((SSiimmpplliiffiiccaattiioonn)) 27.

27. The The belbelow dow diagiagram ram actacts as s as 2 bi2 bit __t ______________ when mode switch is connected to

when mode switch is connected to ‘1’.‘1’.

(A) 2 bit up counter (A) 2 bit up counter (B) 2 bit Down counter (B) 2 bit Down counter

How much time can be saved using design How much time can be saved using design D

D22 over D over D11, , for for executing executing ‘100’‘100’

instructions? instructions? ((AA) ) 999 9 nnss ((BB) ) 2200991 1 nnss (C) (C) ‘0’ ns ‘0’ ns (D) 309 ns(D) 309 ns Ans: (D) Ans: (D) Sol: Sol: DD11 DD22 K K ==66 KK== 88 nnss ttpp = = 6 6 nnss, , n n == 110000 ttpp = 3 ns; = 3 ns; (K+n (K+n – – 1)× t1)× tpp= = 11005 5 × × 66 n n = = 110000 = 630 ns = 630 ns



t tpp = 107 × 3 = 107 × 3 = 321 = 321 Time Time



saved = 630 saved = 630 – – 321 = 309 321 = 309 29.

29. The The ConcConcateatenatination of on of two ltwo linkinked led listists is ts is too be performed on O(1) time. Which of the be performed on O(1) time. Which of the following implementation should be used? following implementation should be used? (A) Circular Linked List

(A) Circular Linked List (B) Singly Linked List (B) Singly Linked List (C) Doubly Linked List (C) Doubly Linked List (D)

(D) CirculCircular doubly lar doubly linked lisinked listt

Sol: The startThe starting node holing node holdsds the addrethe address of thess of the last node

last node 30.

30. An oAn orgarganizanizatiotion han has a Cs a Class lass C neC netwotwork ark andnd want to form subnets for four departments want to form subnets for four departments with hosts as follows.

with hosts as follows. A

A – – 35, B35, B – – 20, C 20, C – – 54, D 54, D – – 4040

What is the possible arrangements of What is the possible arrangements of

Clk Clk T T Mode switch Mode switch 1 1 TT QQ Q Q QQ 1 1

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:

: 9 9 :: PPrre e GGAATTE E QQuueessttiioon n PPaappeerr 31.

31. ConConsisideder a r a sysyststem em wiwith th nn – CPU’s – CPU’s (n > (n > 1)1) and ‘z’ processes. (z < n). The maximum and ‘z’ processes. (z < n). The maximum number of processes that can be in the number of processes that can be in the running & block states are respectively. running & block states are respectively. (A)

(A) n n & & zz – – n n ((BB) z &) z & zz (C) (C) n n & & zz (D (D) ) n n & & nn - - zz Ans: (B) Ans: (B) Sol:

Sol: Since the number of processes are less thanSince the number of processes are less than the number of CPU’s hence all can be the number of CPU’s hence all can be running state. In a situation sometimes all running state. In a situation sometimes all can be blocked.

can be blocked. 32.

32. ConConsisideder a r a SySyststem em wiwith th nn – – processesprocesses arriving at time zero. Scheduling overhead arriving at time zero. Scheduling overhead is ‘s’ seconds. Using Round Robin CPU is ‘s’ seconds. Using Round Robin CPU Scheduling what must be the value of time Scheduling what must be the value of time quantum ‘q’ such that each process is quantum ‘q’ such that each process is guaranteed to gets its turn on the CPU in guaranteed to gets its turn on the CPU in its subsequ

its subsequent run exactly twice within ‘t’ent run exactly twice within ‘t’ sec (inclusive)? sec (inclusive)? (A) (A) n n 2 2 tt (B) (B) n n ns ns tt



(C) (C) 1 1 n n ns ns tt



(D) (D) 2 2 n n 2 2 ns ns tt



33

33.. If If ththe de deteterermiminanant nt of nof n×× n matrix is zero, n matrix is zero,

then then (A) rank(A) (A) rank(A)



n n – – 2 2 (B) trace of A is zero (B) trace of A is zero

(C) zero is an eigen value of A (C) zero is an eigen value of A (D) X = 0 i

(D) X = 0 is the ons the onlyly SoluSolutiotionn of AX = 0of AX = 0

Sol: If det(A If det(Ann××nn) = 0 then zero is an eigen value) = 0 then zero is an eigen value

of A. Since det(A) = product of eigen of A. Since det(A) = product of eigen values.

values. 34.

34. Let Let B = B = {2, {2, 3, 63, 6, 9, , 9, 12, 12, 18, 18, 24} 24} and and letlet A

A = = BB ×× B. Define the following relation B. Define the following relation

on

on A: A: (a(a, b, b) R) R(c(c, d, d) i) if af and nd ononly ly if if {a | c and b

{a | c and b



d} Then R is _____. d} Then R is _____. S

S11) reflexive and symmetric) reflexive and symmetric

S

S22) antisymmetric and transitive) antisymmetric and transitive

S

S33) an equivalence relation) an equivalence relation

S

S44) a partial order relation) a partial order relation

Which of the following is true? Which of the following is true? (A) S

(A) S11 and S and S33 are true are true

(B) S

(B) S22 and S and S44 are true are true

(C) S

(C) S11, S, S33 and S and S44 are true are true

(D) Only S

(D) Only S44 is true is true

Ans: (B) Ans: (B)

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:

: 1100:: CCSSIITT



(a, b) R (e, f) (a, b) R (e, f)



R is transitive R is transitive

Hence R is a partial order Hence R is a partial order 35.

35. MeMessassageges s arare e trtranansmismitttted ed oovever r aa communication channel using two signals. communication channel using two signals. The transmission of one signal requires 1 The transmission of one signal requires 1 microsecond and the transmission of the microsecond and the transmission of the other signal requires two microseconds. other signal requires two microseconds. The recurrence relation for the number of The recurrence relation for the number of different messages consisting of sequences different messages consisting of sequences of these two signals (where each signal is of these two signals (where each signal is immediate followed by the next signal) immediate followed by the next signal) that can be send in

that can be send in n n microseconds (n microseconds (n



2) is2) is _____. _____. (A) a (A) ann = a = ann – – 11 + a + ann – – 22 (B) a (B) ann = a = ann – – 11 + 2a + 2ann – – 22 (C) a (C) ann = 2a = 2ann – – 11 + a + ann – – 22 (D) a (D) ann = a = ann – – 11 + a + ann – – 33 Ans: (A) Ans: (A) Sol: Sol: Let a

Let ann = number of different messages that = number of different messages that

can be sent in n microseconds. can be sent in n microseconds.

Case(i):

Case(i): If the first signal required one If the first signal required one microsecond, then the remaining part of microsecond, then the remaining part of the message can be sent in a

the message can be sent in ann – – 11 ways. ways.

Two Marks Question (Q36 Two Marks Question (Q36

_–

65) 65)

3

366.. AA k-array k-array heap is like a binary heap, but heap is like a binary heap, but instead of two children, nodes have

instead of two children, nodes have k-children. A

k-children. A k-array k-array heap can be heap can be represented in a one-dimensional array as represented in a one-dimensional array as follows. The root is kept in A[1], its ‘k’ follows. The root is kept in A[1], its ‘k’ children are kept inorder in A[2] through children are kept inorder in A[2] through A[k+1], their children are kept inorder in A[k+1], their children are kept inorder in A[k+2] through A[k

A[k+2] through A[k 22+k+1] and so on.+k+1] and so on. In

In 4-array 4-array heap, to what index does the 2 heap, to what index does the 2ndnd child of 3

child of 3rdrd index node map _______. index node map _______.

Ans: 11 Ans: 11 Sol: Sol:

From the above 4-array tree, we can see From the above 4-array tree, we can see index of 2

index of 2ndnd child of 3 child of 3rdrd index node is 11. index node is 11. 37.

37. ConConsidesider thr the foe follollowinwin ionion

1 1 2 2 3 3 4 4 ₅₅ 21 21 20 20 19 19 18 18 17 17 16 16 15 15 14 14 13 13 12 12 11 11 10 10 9 9 8 8 7 7 6 6

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: : 111 1 :: PPrre e GGAATTE E QQuueessttiioon n PPaappeerr The operation which are underlined (which

The operation which are underlined (which are not conflict operations) can be are not conflict operations) can be interleaved in interleaved in ((22 22))!! 2 2!! 22!!





= 6 = 6

For serial schedule T

For serial schedule T22



T T11

R

R22 (A) W(A) W22 ((AA)) RR22 (B) W(B) W22 (B) R(B) R11 ((A)A)

W

W11 (A) R(A) R11 (B) W(B) W11 (B)(B)

The operations which are underlined The operations which are underlined (which are not conflict operations) can be (which are not conflict operations) can be interleaved in interleaved in ((22 22))!! 2 2!! 22!!





= 6 = 6 Total 6 + 6 = 12 Total 6 + 6 = 12 38

38.. CoConsnsidider a er a sysyststem uem usisingng dedemamand pnd pagaginingg with a page size of 1000 words. Disk with a page size of 1000 words. Disk access time for a page is 100 ms. Memory access time for a page is 100 ms. Memory cycle time is 10

cycle time is 10



s. It requires 1s. It requires 1



s s toto access current page and an additional 3 access current page and an additional 3



ss to access other than the current one of the to access other than the current one of the instru

instructions ctions accessaccessed, 10% ed, 10% accesseaccessedd otherother than the current one of the instructions than the current one of the instructions accessed other than the current one 20% accessed other than the current one 20% were not in memory.The effective memory were not in memory.The effective memory access time (approximately) is access time (approximately) is



s.s. j = j = 2 2 1 1 n n



;; for(num

for(num = 1; num = 1; num < =< = n * n ; numn * n ; num++)++) {{

a[i][j]

a[i][j] = = num;num;

i++; i++; jj – – – – ;; if if (num%n (num%n = = = = 0)0) {{ ii – – = 2; = 2; j++; j++; }} else else if if (i (i = = = = n)n) i i = = 0;0; else else if if (j (j = = == – – 1) 1) j = n j = n – – 1; 1; }} }}

If the above program if we give input value If the above program if we give input value n = 3, then what is sum of elements of n = 3, then what is sum of elements of every row of matrix is _______.

every row of matrix is _______.

Ans: 15 Ans: 15 Sol:

Sol: If we give input n = 3 then we can fill theIf we give input n = 3 then we can fill the matrix as follows matrix as follows 2 2 99 44 7 7 55 33 6 6 11 88 Sum

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:

: 1122:: CCSSIITT

41.

41. An 8 An 8 way way set set AssoAssociaciativtive cae cache che memmemoryory unit with a capacity of 32 kB is built using unit with a capacity of 32 kB is built using a block size of 8 words. The word length is a block size of 8 words. The word length is 16 bits. The size of the physical Address 16 bits. The size of the physical Address space is 16 GB. The number of bits in tag space is 16 GB. The number of bits in tag field is _________.

field is _________.

Ans: 22 Ans: 22 Sol:

Sol: Associativity = 8Associativity = 8 Size

Size of of cache cache memory memory = = 221515BB

Size of a Block = 8 × 2 B = 16 Bytes Size of a Block = 8 × 2 B = 16 Bytes

= 2

= 244 Bytes Bytes Number o

Number of blocks in caf blocks in cache memoryche memory = 2= 21111 Size

Size of of main main memory memory = = 223434BB Number

Number of of blocks blocks in in main main memory memory = = 223030 Number

Number of of sets sets in in cache cache memory memory = = 221111 /8 /8 = 2 = 288 Number

Number of of Tag Tag bits bits == 88

30 30 2 2 2 2 2 2 lloog g == lloog g = 22₂₂2222 = 2222 42.

42. If a sIf a simimplple gre grapaph G hah G has 2 cos 2 connnnecectetedd components with 6 vertices and 8 vertices components with 6 vertices and 8 vertices

How big would the window size (in How big would the window size (in packets) have to be for the channel packets) have to be for the channel utilization to be greater than 80% is utilization to be greater than 80% is _______.

_______.

Ans: 1706

_–

1707 1707 Sol:

Sol:Bandwidth-delay productBandwidth-delay product = 512 = 512



10 1066



40 40



10 10 – – 33 = 2048 = 2048



10 1044 bits. bits. 80% utilization, 80% utilization, it is 2048 it is 2048



10 1044



0.8 = 16384 0.8 = 16384



10 1033.. N

Nuummbbeer r oof f ppaacckkeettss 11770077 8 8 1200 1200 10 10 16384 16384 33







44.

44. CerCertaitain Cn CPU PU useuses exs expanpandinding Og Opcopcode. de. ItIt has 32 bit instructions with 8 bit has 32 bit instructions with 8 bit Addresses. It supports One Address and Addresses. It supports One Address and Two Address instructions only. If there are Two Address instructions only. If there are ‘n’ Two

‘n’ Two Address instructions; TheAddress instructions; The maximum number of one Address maximum number of one Address

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: : 113 3 :: PPrre e GGAATTE E QQuueessttiioon n PPaappeerr 45

45.. # i# incncluludede<s<stdtdioio.h.h>> int main( )

int main( ) {{

char

char arr[15];arr[15]; arr

arr = = “ACEHYDERABAD”;“ACEHYDERABAD”; printf(“%d”,

printf(“%d”, arr);arr); return

return 0;0; }}

What will be output of above program? What will be output of above program? (A) ACEHYDERABAD (A) ACEHYDERABAD (B) A (B) A (C) NULL (C) NULL (D) Compilation ERROR. (D) Compilation ERROR. Ans: (D) Ans: (D) Sol:

Sol: Compilation Compilation Error: Error: L L value value RequiredRequired Array name is constant pointer and we can Array name is constant pointer and we can not assign any value in constant datatype not assign any value in constant datatype after declaration.

after declaration.

47.

47. A faA fair cir coin oin is tois tossed ssed untuntil oil one of ne of the the twotwo sides occurs twice in a row. The sides occurs twice in a row. The probability that the number of tosses probability that the number of tosses required is even is ________. required is even is ________. (A) (A) 3 3 2 2 (B) (B) 3 3 1 1 (C) (C) 2 2 1 1 (D) None (D) None Ans: (A) Ans: (A) Sol: Sol: A = {HH, HTHH, HTHTHH, ...}A = {HH, HTHH, HTHTHH, ...} B B = = {TT, {TT, THTT, THTT, THTHTT, THTHTT, ...}...} P(A) P(A) == 3 3 1 1 & P(B) = & P(B) = 3 3 1 1 P(A P(A or or B) B) == 3 3 2 2 48.

48. The The intintermermediediate ate form form of a of a + b * c + b * c is gis giveivenn below in triples, quadruples, postfix, trees below in triples, quadruples, postfix, trees & DAGs choose the correct answer.

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:

: 1144:: CCSSIITT

49.

49. ConConsisideder r twtwo o foformarmal l lalangnguaguageses L & L

L & L11 == LL



{ {



}. It is given that L is}. It is given that L is



--free. Let L

free. Let L22=L=L11



LL22. Which of the. Which of the

following statements is true? following statements is true? (A)If L is regular then L

(A)If L is regular then L22 is always is always

recursive recursive (B)

(B) If L is a If L is a cfl thecfl then Ln L22 is always recursive is always recursive

(C)L

(C)L11, , LL22 can never be recursive but L can never be recursive but L

may be recursive may be recursive (D)L, L

(D)L, L11& L& L22 can possibly be all r.e. & can possibly be all r.e. &

recursive recursive

Sol: A recursive set cannot containA recursive set cannot contain



.. 50.

50. ConsConsideider r the the folfollowlowing ing concconcurreurrentnt processes ‘i’, ‘k’ and ‘l’.

processes ‘i’, ‘k’ and ‘l’. S,T

S,T and and Z Z are are Binary Binary SemaphoresSemaphores initialized as follows :

initialized as follows : B Sem S = 1, T = 0, Z = 0 B Sem S = 1, T = 0, Z = 0

51

51.. ThThe e aarea rea bouboundnded ed bby y tthe he curcurveve y

y = = xx44 – – 2x2x33 + + xx22 + 3, the x-axis and two + 3, the x-axis and two ordinates corresponding to the points of ordinates corresponding to the points of minimum of this function is;

minimum of this function is; (A) (A) 30 30 1 1 (B) (B) 30 30 91 91 (C) (C) 480 480 758 758



(D) (D) 480 480 728 728 Ans: (B) Ans: (B) Sol: Sol: yy



= 0 = 0



4x 4x33 – – 6 x 6 x22 + 2x = 0 + 2x = 0



x = 0, x = 0, 2 2 1 1

, 1 are stationary points , 1 are stationary points y

y



= 12x = 12x22 – – 12x + 2 12x + 2



y(x) has min at x = 0 & x = 1 y(x) has min at x = 0 & x = 1



Required Area Required Area = =

 



30 30 91 91 dx dx 3 3 x x x x 2 2 x x 1 1 0 0 2 2 3 3 4 4

_

_

_



52.

52. The The ChaiChained ned matmatrix rix proproducduct prt probloblem em isis defined as follows:

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: : 115 5 :: PPrre e GGAATTE E QQuueessttiioon n PPaappeerr i = 1, 2, 3, ...n

i = 1, 2, 3, ...n – – 11 m

mi i , , i+si+s = = expr expr , , if if 1 1 < < s s < < n n andand

i = 1, 2, 3, ...n i = 1, 2, 3, ...n – – s ;s ; Which one of the following options is Which one of the following options is

correct

correct_{for expr?}_{for expr?}

(A) (A) ss ii k k

iiminmin  (m(mik ik + m + mk+1k+1,, i + s i + s + d + dii – – 11××ddk k ×× d di+si+s))

(B) (B) _ik_ik _k_k₁₁_,,_ii _ss _ii ₁₁ _k_k₁₁ _ii _ss ss ii k k iiminmin  mm



mm  



dd 



dd 



dd  (C) (C)

 

_ik_ik _k_k_,,_ii _ss _ii ₁₁ _k_k _ii _ss



ss ii k k iiminmin  mm



mm 



dd 



dd



dd  (D) (D)

 

ik ik k k ,,ii ss ii 11 k k ii ss



ss ii k k iiminmin  mm



mm 



dd 



dd



dd  Ans : (A) Ans : (A) Sol: Sol:

When s > 1, the diagonal ‘s’ contains the When s > 1, the diagonal ‘s’ contains the elements m

elements mii,,i+si+s corresponding to the corresponding to the

products of the form M

products of the form Mii , , MMi+1i+1... M... Mi+si+s..

We can make the first cut in the product We can make the first cut in the product

Sol: only option(D) satisfies all conditions.only option(D) satisfies all conditions. tr(A)

tr(A) == – – 11 – – 2 + 3 = 0, 2 + 3 = 0, det(A) = (

det(A) = ( – – 1) (1) ( – – 2) (3) = 62) (3) = 6 The

The eigen eigen values values of of (A+I) (A+I) are are 4,4, – – 1, 01, 0 det(A

det(A + + I) I) = = 00 54.

54. If tIf the phe probarobabilbility ity of hof hittitting ing a taa target rget isis 5 5 1 1 and if 10 shots are fired, what is the and if 10 shots are fired, what is the

conditional probability that the target being conditional probability that the target being hit atleast twice assuming that atleast one hit atleast twice assuming that atleast one hit is already scored?

hit is already scored?

((AA) ) 00..66999999 ((BB) ) 00..662244 ((CC) ) 00..889922 ((DD) ) 00..226688 Ans: (A) Ans: (A) Sol: Sol: P(x P(x



2| x 2| x



1) = 1) =

 



 

xx 11



P P 2 2 x x P P



= = _n_n 1 1 n n n n q q 1 1 npq npq q q 1 1



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(31)

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Cancel Anytime. : : 1166:: CCSSIITT We have (A We have (A



B) B)



C = A C = A



(B (B



C) C)



* is associative on P(S) * is associative on P(S) we have, A

we have, A

 

= A = A



A A



P(S) P(S)

 

is identity element in P(S) w.r.t. *. is identity element in P(S) w.r.t. *. We have, A

We have, A



A = A =

 

A A



P(S) P(S)



For each element of P(S), inverse For each element of P(S), inverse exists, because inverse of A=A

exists, because inverse of A=A



A A



P(S). P(S).



(P(S), *) is a group. (P(S), *) is a group. 56.

56. In In ththe e folfollolowiwing ng fifigugurere

Frames are generated at node A and sent to Frames are generated at node A and sent to node C through node B. Determine the node C through node B. Determine the minimum transmission rate required minimum transmission rate required between nodes B and C. So that buffers at between nodes B and C. So that buffers at node B are not flooded based on the node B are not flooded based on the

Transmission

Transmission Time Time (x) (x) = = 1000/R1000/R A

A → → B:B: 3 frames transmission takes 503 frames transmission takes 50 msec,

msec, B

B → C → C one frame taone frame takes 10 + kes 10 + x msecx msec



30 + 3x = 50 30 + 3x = 50



3x = 20 3x = 20



x = 6.66 msec x = 6.66 msec R R = = 1000/6.66 1000/6.66 = = 150 150 kbpskbps 57

57.. In a nIn a new new numumbeber syr systestem; x am; x and y nd y areare successive digits such that (xy)

successive digits such that (xy)rr = = 25251010 and and

(yx)

(yx)rr = = 31311010; x, y and r values respectively; x, y and r values respectively

are. are. ((AA) ) 44, , 33, , 77 ((BB) ) 33, , 44, , 77 ((CC) ) 77, , 44, , 33 ((DD) ) 77, , 33, , 44 Ans: (B) Ans: (B) Sol:

Sol: xr + y = 25, yr + x = 31 and y = x + 1 fromxr + y = 25, yr + x = 31 and y = x + 1 from these expressions; x = 3, y = 4 and r = 7. these expressions; x = 3, y = 4 and r = 7.

C C B B A A 4000 km4000 km 1000 km1000 km

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Cancel Anytime.

: : 117 7 :: PPrre e GGAATTE E QQuueessttiioon n PPaappeerr

Ans: (C) Ans: (C) Sol: Sol: I. I.TRUETRUE

 Suppose ‘f’ belongs to T*Suppose ‘f’ belongs to T*

 Deleting ‘f’ from T* disconnects T* andDeleting ‘f’ from T* disconnects T* and

Which of the following queries not returns Which of the following queries not returns the

the cuscustomtomerer namnames, wes, who pho purcurchase hase allall products?

products? (A)Select C (A)Select Cnamename

From customer From customer Product Product P Pidid PPnamename 9 988 DDeettooll 9 999 LLuuxx 1 10000 CCoollggaattee H H G G F F E E ee f f A A B B C C D D T* T* S S

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Cancel Anytime.

:

: 1188:: CCSSIITT

(A) S

(A) S11 is true and S is true and S22 is false is false

(B) S

(B) S11 is false and S is false and S22 is true is true

(C) Both S

(C) Both S11 and S and S22 are true are true

(D) Both S

(D) Both S11 and S and S22 are false are false

Ans: (D) Ans: (D) Sol: Sol:

S

S11 is false is false

Proof by counter example: Proof by counter example: For the lattice shown below For the lattice shown below

Sol:

Sol: If f(x) is probability density function, thenIf f(x) is probability density function, then

  



f f xx dxdx



11

  



f f xx dxdx == 00 22xxee dxdx 11 x x22

_



 



f(x) is a probability density function f(x) is a probability density function 62.

62. A biA binary nary min hmin heap ceap consonsistiisting ong of intf integeegers isrs is implemented using array A[1....n] in which implemented using array A[1....n] in which root node is stored at A[1] and locations root node is stored at A[1] and locations A[1] through A[n] store the ‘n’ integer A[1] through A[n] store the ‘n’ integer values in the heap. The minimum number values in the heap. The minimum number of comparisons required to find maximum of comparisons required to find maximum

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: : 119 9 :: PPrre e GGAATTE E QQuueessttiioon n PPaappeerr 64

64.. CoConsnsidider a ser a syyststem wem witith nh n – – processes and aprocesses and a ssingle resource ‘R’ each process Pingle resource ‘R’ each process Pii isis

assigned ‘x

assigned ‘xii’ copies of ‘R’ and further ’ copies of ‘R’ and further

requests

requests ‘‘yyii’ co’ copipies oes off ‘R’‘R’at at timtime e t, t, thetherere

are two processes a & b whose requests is are two processes a & b whose requests is zero. Further,

zero. Further, ‘K’ ‘K’ copies copies of of ‘R’ ‘R’ are are freefree available at time

available at time ‘‘tt’’. System is said to be. System is said to be not approaching deadlock if the minimum not approaching deadlock if the minimum need of process request is satisfiable. need of process request is satisfiable. Which condition indicates that system is Which condition indicates that system is not approaching deadlock?

not approaching deadlock? (A)

(A) maxmax

  

65.

65. ConConsidesider thr the fole followlowing ing setsets oves over a fir a finitnitee alphabet

alphabet

S1: Descriptions of all LR(0) grammars S1: Descriptions of all LR(0) grammars S2: Descriptions of all LR(1) grammars S2: Descriptions of all LR(1) grammars S3: Descriptions of all SLR(1) grammars S3: Descriptions of all SLR(1) grammars S4: Description of all LL(1) grammars S4: Description of all LL(1) grammars S5: Description of all operator grammars S5: Description of all operator grammars S6: Description of all LALR(1) grammars S6: Description of all LALR(1) grammars Choose the correct answers: