Gateforum announced that they will release solutions by 7:30PM on Feb15th, but they didn't. None other coaching institute did. All of them released answer keys only and slept, keeping the students in confusion.

So, I decided, I will post the solutions to selected questions.

If you are just preparing for GATE, you should solve more and more questions. And for that I would recommend GKP Publisher's Question Bank. Although there are some mistakes in the book for solutions to some problems, but I recommend this book solely for the huge collection of problems it has.

Disclaimer: These are my developed solutions. I am a student and I can't guarantee the correctness of the solutions. And yeah, don't think that I solved them all correct. I am not a computer scince wizard either.

Note: Click on the images to enlarge.

First, get the question paper here or here(Thanks Sapan for the second version!)

Now, lets start!

Q-1: I fiddled with it and got the correct answer but didn't mark it cuz of fear of getting it wrong!

Q-2, 3,4,5 I skipped because I had skipped large amount of discrete maths portion and I'm not comfortable with serious topics.

Q-6:

The rightmost thing appearing in this image is P. Read it as PR'. I stripped the R' by mistake while cropping, and overwrote the image.

Q-7:

I hit B as explained below.

The refresh operation in a DRAM chip cycles through each row reading the contents of each cell in the row and writing them back.

For the given problem our RAM module will have 32 chips. Pretty easy calculation i.e. (4M*8bits)/(1M*8bits). One refresh cycle refreshes 1 row of each chip being used to construct the module. Thus we require 1K refresh cycles to refresh the wole module by refreshing 1 row of each chip in the module at once.

Hence time required will be 100*(2^10)ns

Q-8:

F 8 7 B is

1111 1000 0111 1011

Multiplying by 8 means multiplying by 2, three times successively. That accounts for 3, 1-bit left shifts i.e. 3-bit left shift.

On performing 3-bit left shift on the number we get

1100 0011 1101 1000

Thats C 3 D 8

Q-9:

Using the equation of multiplexer we have

f=RP'Q'+R'P'Q+R'PQ'+RPQ

=P'(Q'R+R'Q)+P(R'Q'+RQ)

=P'(Q XOR R)+P(Q XNOR R)........(since, A'B+AB' is A XOR B and AB+A'B' is A XNOR B)

=P'(Q XOR R)+P(Q XOR R)'

=P XOR (Q XOR R)

=P XOR Q XOR R..................(since, XOR is associative)

Q-10: I fiddled with it for quite some time and hit C. I believe I was wrong.

Q-11:

I marked B due to some misunderstanding and being in a hurry(dunno why?)!

Proceed to the function call. We have the address of i in p and address of j in q. Now the operation p=q; will transfer j's address in p. The second operation *p=2; will put 2 at j's address. i's address will remain untouched. Hence after returning from the function we get 0 2 as output from the printf function.

Q-12:

I hit D cuz of a calculation mistake.

Q-13B 14C 15D

Q-16: Seems to be an ambiguous question. A and D are the contenders for being the answer. I hit A.

Q-17:

Q-18:

Each node except root in a B+ Tree can hold between T-1 and 2T-1 keys where T is the order of the tree.

In the problem, we have

2T-1=5

=>2T=6

=>T=3

So the min. i.e. T-1 will be

=>T-1=2 hence option B

Q-19 C. Thats pretty obvious

Q-20: I hit A. But thats wrong. 2PL doesn't guarantee freedom from deadlocks. Timestamp ordering does. So B is the right answer.

Q-21: I hit D. I don't have any explanation for it. I had skipped this portion while studying Software Engineering.

Q-22: B is pretty obvious

Q-23: I skipped. I am not comfortable with this thing.

Q-24: 196 is the obvious answer. No need to explain.

Q-25: I hit B but thats wrong. D will be correct. Read Galvin's book.

Q-26: I skipped. I'm not comfortable with probability.

Q-27: Same as above.

Q-28: I drew some example graphs and guessed the answer to be D. And I got it correct!!!!

Q-29: D

Q-30: B

For all x, there exists y, there exists t, not F(x,y,t)

For every person x out there, there is a person y, and a time t such that x can't fool y at time t.

Option B i.e. No one can fool everyone all the time expresses the meaning best.

Q-31:

A is correct. This is a simple problem involving De Morgan's law. No need to explain.

Q-32: I skipped. No idea. I ain't an electronics wizard.

Q-33: B

Q-34:

I took an example sequence and hit A after applying some unexplainable heuristics. :P

Q-35:

Just follow the recursive call, you'll get 12+(7-(13-(4+(11-(6))))) which is 15.

Q-36:

D is the obvious answer. Moving the last element to the first means setting the pointer of second last element to null and setting head to point to the value of last element(now first) and setting its next pointer point to the to the previous first node.

Q-37:

d=a+b(requires 3 registers. 2 for operands, 1 for result)

e=c+d(a is not used as an operand anywhere ahead so we can overwrite it as e. so, still 3 registers)

f=c+e(we don't need d as an operand, so overwrite it as f. still 3 registers)

b=c+e(we don't need c as operand, so overwrite it as b. still 3 registers)

e=b+f(we don't need b as operand, so overwrite it as e. still 3 of them)

d=5+e(we can overwrite e with the new d. still 3)

return d+f

We're done with 3 registers. But I marked 4 registers. Because in the middle of this whole evaluation I toppled on a variable and needed an extra one.

Q-38: I am uncomfortable with compiler design. I skipped.

Q-39: This one seems to be ambiguous. Consider the string 1010011. It has even number of ones. But none of the given regular expressions can generate it. I didn't mark anything.

Q-40: A CFG can keep the count of 1 thing. D seems to be the obvious answer. I hit C. Even I don't know why!!!!

Q-41: I hit B because I thought, to accept a substring we need equal or lesser number of states. Dunno whether I am write or wrong, but everyone says I am wrong. Rest in peace.

Q-42:

I hit C cuz of a confusion. *damn*

Q-43:

Q-44:

Go through the test conditions T1 through T4.

You'll deduce that:

T1->S1

T2->S3

T3->S1

T4->S2,S4

so T1,T2,T4->{S1.S2,S3,S4)

Q-45 and 46 I skipped because I am scared of these things.

Q-47: D is the obvious answer. No need to explain. Just bitwise AND 255.255.255.224 to 10.105.1.113 and 10.105.1.91 and you will get different network addresses.

Q-48 and 49:

Q-50 and 51:

Label the rows and columns of the matrix as 0,1,2,3,4 and draw the graph. Apply Kruskal's algorithm on Q-50 skipping the second 1 weight edge.

Edit: For Q-51 the required path is 1-0-4-2 and that is 8 units.. I made the same mistake(rather blunder) in exam hall as well which will set me back by -2.66 marks andaround 150-250 rank positions.

Q-52: Pretty easy. Ans. C

Q-53: I made some heuristics and guessed the ans. to be C 30. And that seems to be right! Al the coaching institutes converge at that answer!

Q-54 and 55: I think I got both of them wrong. I hit B and D respectively.

So, I decided, I will post the solutions to selected questions.

If you are just preparing for GATE, you should solve more and more questions. And for that I would recommend GKP Publisher's Question Bank. Although there are some mistakes in the book for solutions to some problems, but I recommend this book solely for the huge collection of problems it has.

Note: Click on the images to enlarge.

First, get the question paper here or here(Thanks Sapan for the second version!)

Now, lets start!

Q-1: I fiddled with it and got the correct answer but didn't mark it cuz of fear of getting it wrong!

Q-2, 3,4,5 I skipped because I had skipped large amount of discrete maths portion and I'm not comfortable with serious topics.

Q-6:

The rightmost thing appearing in this image is P. Read it as PR'. I stripped the R' by mistake while cropping, and overwrote the image.

Q-7:

I hit B as explained below.

The refresh operation in a DRAM chip cycles through each row reading the contents of each cell in the row and writing them back.

For the given problem our RAM module will have 32 chips. Pretty easy calculation i.e. (4M*8bits)/(1M*8bits). One refresh cycle refreshes 1 row of each chip being used to construct the module. Thus we require 1K refresh cycles to refresh the wole module by refreshing 1 row of each chip in the module at once.

Hence time required will be 100*(2^10)ns

Q-8:

F 8 7 B is

1111 1000 0111 1011

Multiplying by 8 means multiplying by 2, three times successively. That accounts for 3, 1-bit left shifts i.e. 3-bit left shift.

On performing 3-bit left shift on the number we get

1100 0011 1101 1000

Thats C 3 D 8

Q-9:

Using the equation of multiplexer we have

f=RP'Q'+R'P'Q+R'PQ'+RPQ

=P'(Q'R+R'Q)+P(R'Q'+RQ)

=P'(Q XOR R)+P(Q XNOR R)........(since, A'B+AB' is A XOR B and AB+A'B' is A XNOR B)

=P'(Q XOR R)+P(Q XOR R)'

=P XOR (Q XOR R)

=P XOR Q XOR R..................(since, XOR is associative)

Q-10: I fiddled with it for quite some time and hit C. I believe I was wrong.

Q-11:

I marked B due to some misunderstanding and being in a hurry(dunno why?)!

Proceed to the function call. We have the address of i in p and address of j in q. Now the operation p=q; will transfer j's address in p. The second operation *p=2; will put 2 at j's address. i's address will remain untouched. Hence after returning from the function we get 0 2 as output from the printf function.

Q-12:

I hit D cuz of a calculation mistake.

Q-13B 14C 15D

Q-16: Seems to be an ambiguous question. A and D are the contenders for being the answer. I hit A.

Q-17:

Q-18:

Each node except root in a B+ Tree can hold between T-1 and 2T-1 keys where T is the order of the tree.

In the problem, we have

2T-1=5

=>2T=6

=>T=3

So the min. i.e. T-1 will be

=>T-1=2 hence option B

Q-19 C. Thats pretty obvious

Q-20: I hit A. But thats wrong. 2PL doesn't guarantee freedom from deadlocks. Timestamp ordering does. So B is the right answer.

Q-21: I hit D. I don't have any explanation for it. I had skipped this portion while studying Software Engineering.

Q-22: B is pretty obvious

Q-23: I skipped. I am not comfortable with this thing.

Q-24: 196 is the obvious answer. No need to explain.

Q-25: I hit B but thats wrong. D will be correct. Read Galvin's book.

Q-26: I skipped. I'm not comfortable with probability.

Q-27: Same as above.

Q-28: I drew some example graphs and guessed the answer to be D. And I got it correct!!!!

Q-29: D

Q-30: B

For all x, there exists y, there exists t, not F(x,y,t)

For every person x out there, there is a person y, and a time t such that x can't fool y at time t.

Option B i.e. No one can fool everyone all the time expresses the meaning best.

Q-31:

A is correct. This is a simple problem involving De Morgan's law. No need to explain.

Q-32: I skipped. No idea. I ain't an electronics wizard.

Q-33: B

Q-34:

I took an example sequence and hit A after applying some unexplainable heuristics. :P

Q-35:

Just follow the recursive call, you'll get 12+(7-(13-(4+(11-(6))))) which is 15.

Q-36:

D is the obvious answer. Moving the last element to the first means setting the pointer of second last element to null and setting head to point to the value of last element(now first) and setting its next pointer point to the to the previous first node.

Q-37:

d=a+b(requires 3 registers. 2 for operands, 1 for result)

e=c+d(a is not used as an operand anywhere ahead so we can overwrite it as e. so, still 3 registers)

f=c+e(we don't need d as an operand, so overwrite it as f. still 3 registers)

b=c+e(we don't need c as operand, so overwrite it as b. still 3 registers)

e=b+f(we don't need b as operand, so overwrite it as e. still 3 of them)

d=5+e(we can overwrite e with the new d. still 3)

return d+f

We're done with 3 registers. But I marked 4 registers. Because in the middle of this whole evaluation I toppled on a variable and needed an extra one.

Q-38: I am uncomfortable with compiler design. I skipped.

Q-39: This one seems to be ambiguous. Consider the string 1010011. It has even number of ones. But none of the given regular expressions can generate it. I didn't mark anything.

Q-40: A CFG can keep the count of 1 thing. D seems to be the obvious answer. I hit C. Even I don't know why!!!!

Q-41: I hit B because I thought, to accept a substring we need equal or lesser number of states. Dunno whether I am write or wrong, but everyone says I am wrong. Rest in peace.

Q-42:

I hit C cuz of a confusion. *damn*

Q-43:

Q-44:

Go through the test conditions T1 through T4.

You'll deduce that:

T1->S1

T2->S3

T3->S1

T4->S2,S4

so T1,T2,T4->{S1.S2,S3,S4)

Q-45 and 46 I skipped because I am scared of these things.

Q-47: D is the obvious answer. No need to explain. Just bitwise AND 255.255.255.224 to 10.105.1.113 and 10.105.1.91 and you will get different network addresses.

Q-48 and 49:

Q-50 and 51:

Label the rows and columns of the matrix as 0,1,2,3,4 and draw the graph. Apply Kruskal's algorithm on Q-50 skipping the second 1 weight edge.

Edit: For Q-51 the required path is 1-0-4-2 and that is 8 units.. I made the same mistake(rather blunder) in exam hall as well which will set me back by -2.66 marks andaround 150-250 rank positions.

Q-52: Pretty easy. Ans. C

Q-53: I made some heuristics and guessed the ans. to be C 30. And that seems to be right! Al the coaching institutes converge at that answer!

Q-54 and 55: I think I got both of them wrong. I hit B and D respectively.

## 72 comments:

for the Q no. 48 and 49....if we transfer a block from L2 cache to L1 cache then it needs 4 cycles. that means 4 times to read L2 and 4times to write in L1.

Based on this the answer of 48 will be 22*4=88 and answer of 49 will be 200*4+20*4+88=968 ms

:cheers

@Raja Thats a disputed question. Check out the related post on gatementor. 902 and 968 are the obvious answers. I'm in the team that believes 902 is the answer!

:grin@REY Mysterio

In question no.51 you have clearly missed the right answer:

1-0-4-2 gives you 8

Yeah, gani u r right! I realise my blunder now! So, another minus 2.66 for me. :damn I'm ruined!

For Q51) Weight(1-0-4-2) = 1+4+3 = 8

i guess for 54=>c and for 55=>b

hey prabhakar...r u sure for

18=>b

30=>b

i m in a big fix.....

plz help me out!!!

54 and 55 k ans bhi bata do !!

http://forum.gatecounsellor.com/viewtopic.php?f=108&t=62no

@sujit Yeah, u r right.

@freaky_ss Yeah, I am sure about 18-B and 30-B. For 54 and 55, the correct aswer is C and B respectively.

yuppy !!!! :)

hey !!@prab...hw can u say that made easy ans r worst...when most of yurs and its ans match??

Well, thats because they have given wrong answers to some very simplest of the questions in the aptitude section like 65, 61, 56.

Later I saw, they have given most correct answers in the CS section i.e. 1-55. But I realized that after I posted criticism for them. LOL

By the way, around how much are you getting according to them?

i m getting 44.7 according to MADE EASY....i hav nt attempted 65th question ..bt i think made easy is correct...

65=>b

61=>c(this is 100 %)

56=>b(i ticked A...made easy is banking on C...bs yahi ans galat hai unka sayad)......

63 ka option kya hai according to u???

i ticked D

k...now i got u...i guess u hav older keys of made easy....it has put recheckd keys on its site...

according to me GATE FORUM ans r d worst...

Oh god! I didn't know they put an updated key!!! lol I sould recheck now with the latest key.

A/C to me

65:B

61:C

56:B

Old keys of made easy say 65:D, 61:A

Yeah, u r right with 56:B I hit A and I know I am wrong.

I ticked 63:A

I'm getting 42.66. Lets see what happens on 15th March. Sach ka saamna usee din hoga!!! I'm damn scared!

Yup, I agree. Gateforum keys are the worst.

What did u hit in 48 and 49??? What do you think should the answer be? I marked 22 and 902. Made Easy says 88 and 968. There is a big debate about it on gatementor in a post. Some advocating the 22-902 duo and others the 88-968.

i didnt attempt q48 n 49...

oh ok...

@prab if possible....q 54 n 55 k ans bata do along with solns

wht is yur result in GATE2010?

I got AIR1504, Marks 42.33, GATE Score 614.

My roll no.: 3095251

I got AIR 1649, Marks 41.67, GATE Score 605.

My roll no.: 3099429

@prab do we have any chance??? agar kucch pata chale to kindly discuss.....

We can try for IIT Roorkee. Also, M.S. in IITD. And M.Tech in IITB under RA category.

hey...u r solutions were almost right can u tell me that which college u get..

and can u also tell me that how to prepare for gate imean the books,shedule etc...

Hi Arunesh! You seem to be a GATE 2011 aspirant. All the best! I'll post in my blog all kind of help that I can for the latest GATE aspirants. All the links to resources and articles on GATECS preparation will be posted soon. But Now a days I am busy with admission helter-skelter. Lets see which college I get. I'm trying for 4 IITs but the chances are very lean. As a backup I have chosen IIIT Bangalore.

yeah i'm a 2011 aspirant but i scora only 12 no.s in 2010..as i'm not prepared well...

best of luck to u.......

hi prabhakar

i m in cs 3rd year....i appeared in gate 2010...my gate score is 605 with a rank of 1649....wich is close 2 urs....so pls tell d cutoffs in d various IITs nd where u r getting admission...also wat shud b a safe rank for IITB....i plan to appear again next year....so any help regarding preparation...

@living_mystery Getting a rank within 110 will guarantee you a seat in IIT Bombay. I've applied in 4 IITs and 2 IIITs. The interview season hasn't yet started. I'll post in the blog as I appear in the interviews.

hey thnx prabhakar 4 ur rply....shud i join any coaching institute like gateforum or shud i prepare on my own....also hw shud i prepare....any tips

Joining classroom coaching is not much needed, but join the Gateforum test series. Although the questions they ask aren't upto the mark, but it will be very beneficial for you since it will expose those areas from the syllabus you aren't good at so u can polish the areas :)

wat about joining correspondence course??

I don't think that the quality of material they provide in correspondence course is good.

Use your B.Tech text books for the prearation. And get the previous year GATE questions

thnx dude....thn i'll join the test series only....nd prepare on my own...for my b.tech. i m following these books :

OS - galvin

DBMS - Korth+Navathe

Algorithms - Sahani

Network - Forouzan

S/w Engg - Rajib Mall

Automata - Mishra & Chandrasekharan

Organization - M.Mano

C & C++ - Schildt

DS - Sahni

Digital - Salivahanan

cn u suggest some better books?

DBMS-Only Navathe is enough. No need of Korth.

Networks-Tanebaum is preferred by GATE question setters

S/W Engg.-Pressman is used for setting questions

Automata-opcroft and Ullman is the standard book

Comp. Org.-Morris Mano is Ok. But for numerical problems related to cache memory, you also have to follow Patterson. For pipelining and caches and some other important topics, also use the resources available on IISc site here http://hpc.serc.iisc.ernet.in/~govind/hpc/

C++ is not in syllabus. For C, Let Us C from Yashvant Kanitkar is perfect.

Data Structures and Discrete Mathematics- Seymour Lipschutz

Digital Logic- Morris Mano

These are all the most standard books. Most of the GATE questions are set from these books only. The tiniest points are captured by setters and then asked.

thnx dude...wat bout maths??....

For Discrete Maths u have Seymour Lipschutz.

I skipped large parts of Maths syllabus during preparation, that was because I never actually practiced maths in past 4 years!!! Hehehe... But I would say, you don't do the same mistake as me. Maths is very crucial.

hehe....same here....anyways thnx

hey prab..wassup...

tell me...can i convert 12 marks into 12*4...or s'thing like that..

i 'm desperately want to do m.tech...

suggest me some books...or anything if u want to ...

plz..help me i'm much in need..

Yeah.... Easily. You can convert into 12*6.

12*4 won't guarantee IIT.

Just keep patience while studies for one year and u can convert it into 12*6 without any problems. Practice is the key to success.

hey,it's very disguisting abt the iit-b...;i read all u write....

then what u should now....

any other college...

and ur right that there is hardly any student without a back...in b.tech..

do u apply for iii-t allahabad.. it's a good college either,, and near to my college.having good infrastructure...

The most unethical thing IIIT-Bians did was that they didn't mention this fact in the eligibility criteria at the time of applying. I hate them for this.

Yeah, I've applied in IIIT-A. I've my test on 1st of June there. It is the best IIIT out there. But seats are very less in number. Lets see what happens.

I'm also applying to NIT Trichy headed central counselling in which 6 NITs are participating.

hey...wass up ...got some college...

Yes! I got selected in IIIT-Allahabad! I'll make a blog post sharing my experience soon.

hey congrats...it's take almost 1 month to congrats u...

finally u made it so...how u feel there...

one of my seniors is also there pursuing ms name nerrej kumar..

i'm also live almost one kilometer for from the iiit campus....

all the best to u....

@Arunesh Thanks! I'll try to find your senior once classes start.

And we too may meet someday :)

hi bro, hw r u

finally u got selected.

congrates!!!!!

are the books that u mention r enough to crack gate??????

i want to appear in gate 2011.

hi..wass up... need u favour that.how can i manage the subjects apart from the 4th year subjects...is it possible to just read the subjects and do q's at last..or it should be done parallel with subjects..

Everything fine here. U say. How is life going?

Yes, u can manage 4th yr parallely. B.Tech papers can be cleared with a one night stand! :p Thats what I did in final yr.

GATE needs more attention. You should concentrate on the basic and fundamental concepts in the subjects as well as solve previous years papers.

By the way, I have a collection of online resources for GATE preparation which I mailed to a few of my acquaintences who are preparing for GATE. If you wish, give me your mail address, I'll forward that mail to you as well.

hi bro..your blog really shows me a pathway..im gate 2011 aspirant.. terribly want to get into IIT's..can you plzz mail me the resources which you used.. i wud be grateful..:)

Yeah sure! But I dont have your email address.

My gmail ID is prabhakar97

Just mail me.

hi bro..thanks a lot..ultimateresources..:):) and congrats for making it to iiit..:)good luck:)

You are most welcome :-)

hello prab..how r u?

i jus wanted some help..im studying theory of computations..but it becomes harder if we go deeper..so is it necessary to learn every theorem..? im stuck here.:(..

Well, if you feel TOC is tough then you can skip most parts of that.

But do study one topic thoroughly in TOC. And that topic is "Identifying the language and grammar". You can scan through all the previous GATE papers and notice that in every paper there is at least one question related to that. Most year papers have more than 2 questions from that topic. Do study that seriously.

cool..ok i will do that..thanks man:)

yeah fine..life is same as it were before..sorry for replying late...yeah i also think that last one month preparation is enough..thanks for ur kindness..my email.id is avi.cindrella@gmail.com.

i'm stuck in the file-indexing in dbms..feeling bored..

Hi Arunesh. File indexing isn't a big deal. You can master it by studying with Navathe's book and solving the numerical examples therein.

By the way, my article for GATE preparation was published at the URL below. You can access useful online resources from the links at the end of that article.

http://www.techvyom.com/2010/09/gate-faq-s-all-you-need-to-know-about.html

thanks..bro..

these are truly good resources.

ya i follow navathe..i understand it but it bores me...

trying to increase my potential.

hey wass up..

do me a favour i am confuse bt'n the galvin and william stallng of os.which one is the best and i have the galvin.

Galvn is good enough. Stick to it. It is good for you :-)

hi Prab, can you please explain me Q.7(the 1 related to dram). Since there are 32 chips, total no of refreshes reqd = 32 * 1000 right. So the total time should be 3200 * 2^10 I think(obviously i know its wrong since the option isn't there. But I would like to knwo where I'm goin wrong). Also, please mail me some resources to sachindraac@gmail.com

none of the question paper links are working prab.. please if u can repair them...

Hey Prab! How come the logic of Q17.

L1-L3 is always recursive which implies L1-L3 is re

The question asked which one is not necessarily true

I think L2 U L3 is not necessarily re always

What do you say about this?

hello sir ,myself nikhil i am going through with ur post in ur blog and u have written excellent post regarding gate ...as i am a gate aspirant so i want some tips from u regarding gate preparation....if u tell me that how much time we have to give for gate and how to start preparing for it from scratch as i am a average student not that much brilliant so what i have to do so i can qualify gate with good marks

is it necessary to get coaching ....if not how can i start my preparation on my own as i cant able to decide that frm where to start.....

and plze provide some material for gate preparation if u have........

i will be very thankfull to u ......

An excellent job done by you. No doubt you are great and helpful for visitors those who are related to learning activities.

..................... Excellent

q33 shouldnt the OF operation be after the perform operation plz reply !!

Q33 shoudnt OP operation be after the PO operation because the ans from perform operations is used in the next instructions

Q33 shoudnt OP operation be after the PO operation because the ans from perform operations is used in the next instructions

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