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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.

29,472 total views, 14 views today

GATE CSE 2020 Syllabus

October 24, 2015 by Arjun Suresh 4 Comments

Detailed syllabus for GATE CSE 2016 with changes and links to previous year questions

75,405 total views, 16 views today

GATE Overflow Book- Q/A (Only Previous GATE/TIFR/CMI/ISI)

October 23, 2015 by Arjun Suresh 32 Comments

GATE Overflow Question & Answer- subject wise – only previous GATE CS/IT papers.

251,305 total views, 94 views today

GATE Overflow
Aptitude Overflow

Answered: GATE1988-2-ii

This will clear all your doubts https://www.youtube.com/watch?time_continue=4&v=QRAL0v6yWzo&feature=emb_title [...]

Answered: Doubt Question on Transaction schedules

number of view equivalant schedules = 10 and conflict equivalant = 6 anyone can suggest how to rotate the image? [...]

Answered: Andrew S. Tanenbaum (OS) Edition 4 Exercise 3 Question 41 (Page No. 259)

i) For 4096 bytes 65536 / 4096 = 16 pages(total) text = 32768 / 4096 = 8 pages data = 16386 / 4096 = 5 pages stack = 15870 / 4096 = 4 pages 8+5+4 = 17, the answer is no. ii) For 512 bytes 65536 / 512 = 128 pages(total) text = 32768 / 512 = 64 pages data = 16386 / 512 = 33 pages stack = 15870 / 512 = 31 pages 64+33+31 = 128, the answer is yes. [...]

Answered: GATE2010-1

Draw two non isomorphic graphs like this You can clearly see that |s| = |t| and both trees are different to each other but they satisfy the property mentioned in the question ANSWER: C [...]

Answered: GATE2013-26

Everything you need to know about LINE GRAPHS You won’t find anything better than this anywhere as its from NPTEL IIT MADRAS. [...]

Answered: GATE2004-54

For access control Ethernet(802.3) uses CSMA/CD. In CSMS/CD collision is resolved by binary backoff algorithm which says if a collision occurs then the stations which involved in collision have to wait $wait = k * Time$ here K is a number chosen from ($0 - 2^n-1$) n is collision number as A has Successfully Transmit 1 packet and B has undergone first collision when A and B collide again for B second collision happened as for A new packets ! collision occurred A can wait between (0 or 1) before retransmission B can wait between (0-3) before retransmission (as it… [...]

Answered: NIELIT 2019 Feb Scientist D - Section D: 30

$\textrm{Given that:}$ $log_{10}(2^x+x-41)=x(1-log_{10}5)$ $\because log_aa=1$ $\textrm{We can use this property here, $log_{10}10=1$}$ $\implies$ $log_{10}(2^x+x-41)=x(log_{10}10-log_{10}5)$ $\implies$ $log_{10}(2^x+x-41)=x(log_{10}(\frac{10}{5}))$ $(\because log_ax-log_ay=log_a(\frac{x}{y}))$ $\implies$ $log_{10}(2^x+x-41)=x(log_{10}2)$ $\implies$ $log_{10}(2^x+x-41)=log_{10}2^x$ $(\because log_ax^n=n*log_ax)$ $\textrm{compair both side we get:}$ $\implies$ $2^x+x-41=2^x$ $\implies$ $x=41$ $\textrm{Option b is correct.}$ [...]

Answered: NIELIT 2019 Feb Scientist C - Section D: 19

$\textrm{Given that:}$ $x=\frac{\sqrt10+\sqrt2}{2},y=\frac{\sqrt10-\sqrt2}{2}$ $\textrm{first find $x^2+xy+y^2$}$ $\implies$ $x^2+xy+y^2+xy-xy$ $\implies$ $x^2+2xy+y^2-xy$ $\implies$ $(x+y)^2-xy….(1)$ $\textrm{Let find x+y first:}$ $\implies$ $\frac{\sqrt10+\sqrt2}{2}+\frac{\sqrt10-\sqrt2}{2}$ $\implies$ $\frac{2*\sqrt10}{2}$ $\implies$ $x+y=$ $\sqrt10...(2)$ $\textrm{Let find x*y:}$ $\implies$ $\frac{\sqrt10+\sqrt2}{2}*\frac{\sqrt10-\sqrt2}{2}$ $\implies$ $\frac{(10-2)}{4}$ $\implies$ $x*y=$$2...(3)$ $\textrm{Substitute eq (2) &(3) into (1), we get:}$ $\implies$ $(\sqrt{10})^2-2$ $\implies$ $8$ $\textrm{Now find l$og_2(8)$,which is equal to 3}$ $\textrm{Option D is correct.}$ [...]

Answered: NIELIT 2019 Feb Scientist C - Section C: 19

$\textrm{Given that: }$ $a^x=b ….(1)$ $b^y=c ….(2)$ $c^z=a….(3)$ $\textrm{taking log in above equations we get:}$ $\implies$ $xlog_2a=log_2b$ $\implies$ $x=\frac{log_2b}{log_2a}….(4)$ $\textrm{in the same way}$ $y=\frac{log_2c}{log_2b}….(5)$ $z=\frac{log_2a}{log_2c}…..(6)$ $\textrm{multiply equations (4),(5),(6).}$ $\implies$ $x*y*z=\frac{log_2b}{log_2a}*\frac{log_2c}{log_2b}*\frac{log_2a}{log_2c}$ $\implies$ $x*y*z=1$ $\textrm{Option C is correct.}$ [...]

Logical reasoning

Each question below contains six statements followed by four sets of combinations of three; Choose the set in which the statement are logically related. A. Final year students would like a good career. B. All final year students are eligible as candidates for MBA entrance examination. C.Final year students are eligible for a good career. D. Some of those who are candidates for an MBA entrance examination are final year students. E. All those eligible as candidates for an MBA entrance examination are eligible for a good career. F. All those who would like a good career are entitled to… [...]

Answered: NIELIT 2019 Feb Scientist D - Section C: 21

Now, if 6th reading is x, the x+x+3+x+8= 93.8 after solving, x=27.6 [...]

Answered: NIELIT 2019 Feb Scientist C - Section C: 15

Difference = 500 5% = 500. 100% = (500 *100)/5 = 10000. 100% is the labeled price. So, TV price = 10000 - 20% of 10000 = 80% of 10000 = 8000. [...]

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