arjun

Master College Reviews for GATE CSE

June 5, 2017 by arjun Leave a Comment

Here, you can find reviews about various Colleges for doing Masters in Computer Science & Engineering in India. If you are studying Masters, it would be great if you can also give a review using this form. For admission and cutoff details you can see gatecselastrank.com Source Send Form Loading… 16,652 total views, 10 views today

16,652 total views, 10 views today

The truth behind the RBR Controversy- lets hear from the fake legend himself

May 2, 2017 by arjun Leave a Comment

Please see: http://drive.google.com/drive/folders/0B8_aYGBndW4HOGR4SVY2T2tUUkU 4,153 total views, 2 views today

4,153 total views, 2 views today

GATE CSE 2017 Result Responses

April 30, 2017 by arjun Leave a Comment

Source Send Form Loading… 29,391 total views, 10 views today

29,391 total views, 10 views today

GATE CSE 2017 Admission Responses

April 30, 2017 by arjun Leave a Comment

Source Add your Response Loading… 36,766 total views, 12 views today

36,766 total views, 12 views today

Updates of Rank Predictor

February 15, 2017 by arjun 1 Comment

The rank predictor for GATE 2017 is now live at http://gateoverflow.in/mymarks thanks to Pragy. You can know your marks, normalized marks and even expected ranks all of which are not mere prediction but a realistic estimate. You should not get much deviation from this but Pragy is trying to make an even better app. Please […]

14,191 total views, 8 views today

GATE Overflow
Aptitude Overflow

Answered: GATE CSE 2014 Set 3 | Question: 39

Answered: GATE CSE 2014 Set 3 | Question: 39

I have found an alternative solution, with ref to Coreman textbook exercise, Q.12.2-8. If you can prove this (solution: https://stackoverflow.com/questions/40561235/time-complexity-of-finding-k-successors-in-bst) Then we know that since there are m nodes having value between L and H. We can also say that ‘m’ inorder successors of node with value ‘L’. So searching for L : O(log n) ‘m’ inorder successors : O(m+h) = O(m + log n) So overall time complexity : O(m + log n). a=1,b=1.c=1,d=0 Answer : 110. [...]

Answered: GATE CSE 2020 | Question: 10

https://cs.stackexchange.com/q/138479/118790 Check this out and the concept said in Peter Linz. I guess it is much harder to say that whether a LANGUAGE is $LL(k)$ or not while the question is much easier to answer if the grammar is given. See the $Example$ $7.11$ in Peter Linz, the language from the arguments given by other answers to the GATE question would make it look that the language of the grammar in Example 7.11 is not LL(k) for any $k$ but then suddenly the author comes up with a bit difficult grammar and proves that the problem with the grammar in… [...]

Answered: GATE CSE 2021 Set 2 | GA Question: 7

In this type of question, we join the missing piece, then a rectangle should be formed. In the question figure, the $L$ shape part of the shape doesn’t fit in options B and D. In option C, the size of the $U$ shape part is not the same as the question figure. So, the correct answer is $(A).$ [...]

Answered: GATE CSE 2019 | Question: 48

[youtube https://www.youtube.com/watch?v=B8ElmECvmWw]\ GATE CS 2019,Q48: Let Σ be the set of all bijections from {1, …, 5} to {1, …, 5}, where id denotes - YouTube [...]

Answered: MadeEasy Subject Test 2019: Compiler Design - Parsing

S1: False Refer this: https://stackoverflow.com/questions/34707467/is-every-ll1-grammar-also-an-lr0-grammar S2: False Refer this : https://gateoverflow.in/1251/gate-cse-2007-question-53 S3: TRUE Refer this: https://www.csd.uwo.ca/~mmorenom/CS447/Lectures/IntermediateCode.html/node3.html [...]

Answered: NIELIT 2016 DEC Scientist B (IT) - Section B: 43

D. None of the above As we have only two basic operation which are union and set difference [...]

Answered: NIELIT 2019 Feb Scientist D - Section D: 12

$\frac{1}{2^{2}-1}+\frac{1}{4^{2}-1}+\frac{1}{6^{2}-1}+\cdots+\frac{1}{20^{2}-1}$ $= \frac{1}{(2+1)(2-1)} + \frac{1}{(4+1)(4-1)} +\frac{1}{(6+1)(6-1)} +\cdots +\frac{1}{(20+1)(20-1)}$ $= \frac{2}{2}\left [\frac{1}{(2+1)(2-1)} + \frac{1}{(4+1)(4-1)}+\frac{1}{(6+1)(6-1)} +\cdots +\frac{1}{(20+1)(20-1)}\right]$ $= \frac{1}{2}\left [\frac{2}{(2+1)(2-1)} + \frac{2}{(4+1)(4-1)} +\frac{2}{(6+1)(6-1)} +\cdots +\frac{2}{(20+1)(20-1)}\right]$ $= \frac{1}{2}\left [\frac{(2+1)-(2-1)}{(2+1)(2-1)} + \frac{(4+1)-(4-1)}{(4+1)(4-1)} + \frac{(6+1)-(6-1)}{(6+1)(6-1)}+\cdots +\frac{(20+1)-(20-1)}{(20+1)(20-1)}\right]$ $= \frac{1}{2}\left [\frac{1}{(2-1)} - \frac{1}{(2+1)} + \frac{1}{(4-1)} - \frac{1}{(4+1)} + \frac{1}{(6-1)} - \frac{1}{(6+1)}+\frac{1}{(8-1)} - \frac{1}{(8+1)}+\cdots +\frac{1}{(20-1)} - \frac{1}{(20+1)}\right]$ $= \frac{1}{2}\left [\frac{1}{1} - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7}+\frac{1}{7} - \frac{1}{9}+\cdots +\frac{1}{19} - \frac{1}{21}\right]$ $= \frac{1}{2}\left[\frac{1}{1} - \frac{1}{21}\right]$ $= \frac{1}{2}\left[ \frac{20}{21}\right]$ $= \frac{10}{21}$ Similar question: https://gateoverflow.in/358957/tifr2021-a-11 [...]

Answered: NIELIT 2019 Feb Scientist D - Section D: 13

$1-6+2-7+3-8+\dots\dots$ to $100$ terms $=\underbrace{1-6}_{-5}+\underbrace{2-7}_{-5}+\underbrace{3-8}_{-5}+\dots\dots \underbrace{50-55}_{-5}$ $= (1+2+\dots\dots 50) - (6+7+\dots\dots 55)$ $= (1+2+3+4+5)+\underbrace{(6+7+\dots\dots 50) - (6+7+\dots\dots 50)}_0-(51+52+53+54+55)$ $= (1+2+3+4+5)-(51+52+53+54+55) $ $=\frac{5}{2}(2(1)+(5-1)(1)) - \frac{5}{2}(2(51)+(5-1)(1))$ $(\because$ Sum of $n$ terms of A.P. sequence is $ \frac{n}{2}(2a+(n-1)d)$ $=\frac{5}{2}(6) - \frac{5}{2}(106)$ $=5(3)-5(53)$ $=15-265$ $=-250$ [...]

Answered: NIELIT 2017 OCT Scientific Assistant A - Section A: 52

Let Brian time be Y months and given Adrian invested for whole 12 months. partnership ratio 85000 ×12 : 42500 × y we need to make this as 3:1 in terms of profit so simply dividing by 85000 ×4 on both sides of ratio we'll now have 3 : y/8 that means Brian invested for 8 months [...]

Answered: NIELIT 2019 Feb Scientist C - Section D: 1

Ans is option (D) The minute covers $360^{\circ}$ in $60$ minutes. Hence, it will cover $\frac{360^{\circ}\times 35}{60}=210^{\circ}$ in $35$ minutes. Now, area of a sector with radius $r$ units and $\theta$ angle subtended at the centre ,is : $\frac{1}{2}\times r^{2}\times \theta$ units ($\theta$ in radian) . So, $210^{\circ}=210^{\circ}\times \frac{\pi}{180^{\circ}}$ radians $=\frac{11}{3}$ radians $\therefore$ Area of the sector: $\frac{1}{2}\times 100\times \frac{11}{3}=183.33$ cm$^{2}$ [...]

Answered: NIELIT 2019 Feb Scientist C - Section D: 28

Ans is option (D) Let the original price be $x$ Rs. Price after $p\%$ increase: $x(1+\frac{p}{100})$ Rs. Price after $p\%$ decrease: $x(1+\frac{p}{100})-[\frac{p}{100}\times x(1+\frac{p}{100})]=1$ (given in question) $\therefore$ $x(1+\frac{p}{100})(1-\frac{p}{100})=1$ $\Rightarrow$ $x=\frac{10000}{10000-p^{2}}$ Rs. [...]

Answered: CAT2019-2: 86

$\frac{n^{2}+7n+12}{n^{2}-n-12} = \frac{n^{2}+3n+4n+12}{n^{2}+3n-4n-12}$ $= \frac{n(n+3)+4(n+3)}{n(n+3)-4(n+3)}$ $= \frac{(n+3)(n+4)}{(n+3)(n-4)} = \frac{(n+4)}{(n-4)}$ $\therefore \frac{n^{2}+7n+12}{n^{2}-n-12} = \frac{(n+4)}{(n-4)}$ Now let’s substitute $n$ values given in options $ n = 8 $ $\frac{(n+4)}{(n-4)} = \frac{(8+4)}{(8-4)} =\frac{(12)}{(4)} = 3$ (Positive Integer) $ n = 12 $ $\frac{(n+4)}{(n-4)} = \frac{(12+4)}{(12-4)} =\frac{(16)}{(8)} = 2$ (Positive Integer) $ n = 16 $ $\frac{(n+4)}{(n-4)} = \frac{(16+4)}{(16-4)} =\frac{(20)}{(12)} \sim 1.67$ (Not an integer) $ n = 6 $ $\frac{(n+4)}{(n-4)} = \frac{(6+4)}{(6-4)} =\frac{(10)}{(2)} = 5$ (Positive Integer) Options A, B, D have produced the result as a positive integer. In question, given that we need to find “largest positive integer $n$ such that… [...]

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