Analysis

GATE CSE 2015 Admissions Responses

October 23, 2015 by Arjun Suresh 3 Comments

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GATE CSE 2014 Admissions Responses

September 27, 2015 by Arjun Suresh Leave a Comment

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GATE CSE 2015 Result Responses

September 27, 2015 by Arjun Suresh 1 Comment

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Previous Year GATE Papers and Official Keys

September 27, 2015 by Arjun Suresh 10 Comments

Information Technology

Solution and Discussion

http://gateoverflow.in/previous-years

TIFR Papers

ISRO Question Papers for CSE

CBSE (UGC) NET Question Papers for CSE

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GATE Overflow
Aptitude Overflow

Answered: Galvin Edition 9 Exercise 9 Question 36 (Page No. 456)

I think working set should be for the entire process [...]

Answered: UGCNET-June-2019-II-17

Arbiters are electronic devices that allocate access to shared resources. Example: A bus arbiter is [...]

Answered: Grade up

it is may be O(n) [...]

Answered: GATE1993-7.9, UGCNET-Dec2012-III-41

Condition for deadlock free R >= Sum(p×(n-1))+1 Here R= total no. Of resources n= Max demand of [...]

Answered: Michael Sipser Edition 3 Exercise 2 Question 27 (Page No. 157)

for the string "if condition then if condition then a:=1 else a:=1" has two different pars [...]

Answered: minimum number of registers

Solution is right answer is 3 [...]

Answered: TCS DIGITAL APTI

I think answer shold be B(51,50) you can take first small size for understanding suppose i took (4,8 [...]

Answered: CAT2000-95

Answered: self doubts

so both A and B worked for 20 days so reaming work = 24-20 = 4 days so B completed 4 days work in 10 [...]

Answered: (number system) self doubts

so this question says N mod13=11 and Nmod17=9 we need to find that common N but i am also not able t [...]

Answered: arithmetic progression

sun of n terms of AP = N/2{2a+(n-1)*d} where a= first term .d=common difference ,n= number of terms [...]

Answered: time and distance (self doubts)

Average speed = total distance covered/total time taken total distance covered = 10+12 = 22 km total [...]

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