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GATE CSE 2019 Admission Responses

February 14, 2020 by Arjun Suresh Leave a Comment

Source Add your Response Rank Predictor for GATE CSE 2020 Result Responses: GATE CSE 2019 Result Responses Admission Responses: GATE CSE 2021 Admission Responses GATE CSE 2020 Admission Responses GATE […]

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GATE CSE 2019 Result Responses

February 14, 2020 by Arjun Suresh Leave a Comment

Source Add your Response Rank Predictor for GATECSE 2020 Admission Response: GATE CSE 2019 Admission Responses Result Responses: GATE CSE 2021 Result Responses GATE CSE 2020 Result Responses GATE CSE […]

7,550 total views, 2 views today

2018 Year Placement Statistics for CSE

June 10, 2018 by Arjun Suresh Leave a Comment

Placement Statistics for Year 2018 for Computer Science & Engineering

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GATE CSE College Reviews for Masters

June 10, 2018 by Arjun Suresh Leave a Comment

Reviews for Computer Science & Engineering Course for Masters by GATE Overflow members.

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GATE CSE 2018 Admission Responses

June 7, 2018 by Arjun Suresh Leave a Comment

Admission Responses for GATE 2018 Computer Science and Engineering branch given by GATE Overflow users. You can know what rank people got into which all Institutes.

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GATE Overflow
Aptitude Overflow

Answered: made easy test series

$T_{0}=0\ ,\ T_{1}=1\ given\ in\ question $ Let's there is $T{_{2}}=T{_{1}}+T {_{0}}$ $T{_{1}}\ will\ call\ first\ and\ return\ 1\ and\ make\ space\ for \ T{_{0}}$ $it\ means\ T{_{1}}\ and\ T{_{0}}\ will\ share\ same\ space$ $say\ n=4$ $T_{1}$ $T_{2}$ $T_{3}$ $T_{4}$ $T_{1}\ and\ T_{0}\ values\ are\ store\ in\ table\ so \ no\ need\ to\ evaluate\ them.$ $in\ this\ question\ maximum\ number\ of\ function\ call\ before\ stack\ overflow\ is\ $ $n*4=48$ n=12 [...]

Answered: GATE CSE 2014 Set 1 | Question: 39

compare in pairs like- compare arr[0],arr[1] and arr[1],arr[2] ….. total 50 comparisions for each comparison store bigger one in A list and smaller one in B list maximum of arr will be in A and min will be in B, finding max in A and min in B each will take 49 comparisions total 50+49+49=148 [...]

Answered: GATE CSE 2017 Set 2 | Question: 31

Answered: GATE CSE 2017 Set 2 | Question: 31

This is the toughest question of 2017 GATE exam but you can easily understand the answer like this….….. [...]

What is the difference between frame & fragmented packet? [...]

The DMA mechanism can be configured in a variety of ways, which are: Single-bus detached DMA. Single-bus integrated DMA-I/O I/O bus In Question – Single bus detached DMA is mentioned and in this two “system bus” cycles are needed. @Chaitanya Kale, When the data is about to be transferred, the DMA controller gets access to the system bus from the CPU. But if we consider this then bus is busy for data preparation as well as data transfer but this shouldn't happen as bus should be busy only during data transfer time. When the device has a lot of data to be transferred, we will surely… [...]

Answered: Simplified Boolean expression for A'BC+AB'C'+A'B'C'+AB'C+ABC

option d correct [...]

Answered: CAT 1996 | Question: 48

DACB [...]

Answered: NIELIT 2022 Feb Scientist C - Section D: 26

(p+1)^3 = p^3 + 1 + 3p(p+1) =[p^3+3p^2+3p] + 1 =7+1=8 therefore p+1=2 => p=1 so answer is 3 [...]

Answered: NIELIT 2022 Feb Scientist D - Section A: 30

When two values given a,b then Number of numbers between a,b (including extreme values) = a-b+1 Number of numbers between a,b (excluding extreme values) = a-b-1 From question, 1,2,3…….7position………23positon……..Last position[L] We can write as (23-7)-1=(L-23)-1 [ becoz the person exactly between me(L) and my friend(7position) is on the 23rd position ] 15=L-24 L=39(Answer) [...]

Answered: NIELIT 2022 Feb Scientist D - Section C: 1

let x = amount for he bought + amount spent on repair x--->[x+(20%of x)=1.2x]--->[1.2x-(10% of 1.2x)=1.08x]--->[1.08x+(10%of1.08x)] Given in question 1.08x+(10% of 1.08x)=1188 from this x=1000 you get Required answer = 1000 – 110 = 890. Therefore answer is option D [...]

Answered: NIELIT 2017 OCT Scientific Assistant A - Section A: 33

Part filled in $1 $ minute: $\frac{1}{10}+\frac{1}{30}=\frac{2}{15}$ Part filled in $5 $ minute: $\frac{2}{15}\times 5=\frac{2}{3}$ Unfilled part= $1-\frac{2}{3}=\frac{1}{3}$ so this is filled by tape first by alone,hence will be filled in $30\times \frac{1}{3}=10$ minute. So option (C) is correct. [...]

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