Analysis

GATE CSE 2020 Admission Responses

March 24, 2021 by soujanya Leave a Comment

Source Add your Response Rank Predictor for GATECSE 2021 4,174 total views, 24 views today

4,174 total views, 24 views today

GATE CSE 2021 Result Responses

March 22, 2021 by soujanya Leave a Comment

Source Add your Response 7,350 total views, 32 views today

7,350 total views, 32 views today

TIFR Previous Year Papers with Solution

August 1, 2020 by soujanya Leave a Comment

Computer Scitifrence 5,862 total views, 2 views today

5,862 total views, 2 views today

Previous Year Information Technology GATE Papers with Solution

July 18, 2020 by soujanya Leave a Comment

3,524 total views, 10 views today

3,524 total views, 10 views today

GATE CSE 2020 Result Responses

March 14, 2020 by Arjun Suresh Leave a Comment

Source Add your Response 27,247 total views, 4 views today

27,247 total views, 4 views today

GATE Overflow
Aptitude Overflow

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: GO 2021 Verbal Ability 1: 2

Had I not informed you on time, you would have missed the opportunity. [...]

Answered: GO 2021 Verbal Ability 1: 1

confine – keep or restrict someone or something within certain limits of (space, scope, or time). [...]

Answered: NTA NET NOV 2020

This question is solved: https://gateoverflow.in/349662/ugcnet-oct2020-ii-11 [...]

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… [...]

%d bloggers like this: