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TIFR Mathematics Previous Year Papers with Solution

October 18, 2020 by soujanya Leave a Comment

10,573 total views, 14 views today

10,573 total views, 14 views today

Spatial Aptitude

October 5, 2020 by soujanya Leave a Comment

Syllabus Transformation of shapes: translation, rotation, scaling, mirroring, assembling, and grouping Paper folding, cutting, and patterns in 2 and 3 dimensions 7,038 total views, 20 views today

7,038 total views, 20 views today

Analytical Aptitude

October 5, 2020 by soujanya Leave a Comment

Syllabus Logic: deduction and induction Analogy Numerical relations and reasoning 5,084 total views, 14 views today

5,084 total views, 14 views today

GATE 2021 Syllabus Including Changes

August 24, 2020 by Arjun Suresh Leave a Comment

GATE 2021 syllabus is changed after 5 years. Like in 2016 there are more removals than additions. More importantly there are better clarifications regarding topics this time — some topics implicitly present have been added explicitly. This can also mean that it is very highly unlikely that non mentioned topics — dynamic pipelining for example […]

14,691 total views, 64 views today

TIFR Previous Year Papers with Solution

August 1, 2020 by soujanya Leave a Comment

Computer Scitifrence 5,882 total views, 22 views today

5,882 total views, 22 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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