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IITB GATECSE Cutoff

April 9, 2016 by ibia 1 Comment

Cutoffs for M.Tech, PhD admissions in IIT Bombay for CSE

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IISc. GATECSE Cutoff

April 8, 2016 by ibia Leave a Comment

Cutoffs for M.Tech./Research admissions in IISc.

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GATE Overflow
Aptitude Overflow

Answered: NTA NET 2020 (DISK)

5 platters ,each platter has 2000 tracks each track 50 sectors with size 512 bytes so T=50x512 /1024 =25 K S=2000x25K =50000K Hence option 3) is right ans [...]

Answered: UGCNET-Oct2020-I: 32

Option D)a,b,d only is correct ans Simply put, cloud computing is the delivery of computing services—including servers, storage, databases, networking, software, analytics, and intelligence—over the Internet (“the cloud”) to offer faster innovation, flexible resources, and economies of scale. Few service providers Amazon Web Services. Highly scalable, complete cloud platform. ... Microsoft Azure. IaaS and PaaS computing for development, deployment, and management. ... Google Cloud Platform. Developer products and cloud technologies hosted by Google. ... IBM Cloud. ... Rackspace https://www.ibm.com/blogs/cloud-computing/2014/01/31/cloud-computing-defined-characteristics-service-levels/ [...]

Answered: UGCNET-Oct2020-I: 50

Ans is option A effect improvement through a joint review of achievement of performance goals within a given timeframe The effective functioning of this system is an agreement between a manager and an employee about that employee’s groups performance goals during a stated time period. [...]

Answered: UGCNET-Oct2020-I: 49

option B ) By identifying and executing performance targets in a collaborative mode The effective functioning of this system is an agreement between a manager and an employee about that employee’s groups performance goals during a stated time period. These goals can emphasise either output variables of intervening variables or some combination of both. The important thing is that goals are jointly established and agreed upon in advance [...]

Answered: gatebook-os

Answer : 4 Exp : Fetching lw : CPU has address, say A1, from where to fetch lw. We’ll look into page table, in worst case, page table may not be present in the memory as it is given in the question that “Page tables are not locked in memory and may be swapped to disk”. So, we bring page table. 1 page fault. Now, it could happen that the page containing lw word is not in memory, so we bring it, 1 page fault. So, fetch cycle, in worst case, may have 2 page faults. Decoding lw :… [...]

Answered: UGCNET-Oct2020-II: 35

Answer is D The process of removing detail from a representation is called abstraction. [...]

Answered: NIELIT 2019 Feb Scientist C - Section C: 6

$\textrm{According to the question $E+H=100$,}$ $\textrm{where E=number of students who can speak English, H=number of a student who can speak Hindi.}$ $\textrm{there are 10 students who don’t speak any one of the languages.}$ $\therefore$ $H+E=90$ $n(E)=20,n(H)=80,n(H\cup E=90)$ $\because$ $n(H\cup E)=n(E)+n(H)-n(H\cap E)$ $\implies$ $90=20+80-n(H\cap E)$ $\implies$ $n(H\cap E)=10$ $\textrm{The number of students knowing both Hindi and English is 10}$ [...]

Answered: NIELIT 2019 Feb Scientist C - Section C: 2

$\textrm{according to the question}$ $\frac{area of 1^{st}square}{area of 2^{nd}square}=\frac{9}{1}$ $\because$ $\textrm{area of square is $a^2$,where a is side of square.}$ $\textrm{assume that side of square 1 is $a_1$ & side of square 2 is $a_2$.}$ $\implies$ $\frac{(a_1)^2}{(a_2)^2}=\frac{9}{1}$ $\implies$ $\frac{a_1}{a_2}=\frac{3}{1}$ $\therefore$ $\textrm{side of square 1 is 3 and side of square 2 is 1}$ $\because$ $\textrm{perimeter of square is 4a,so there ratio is;}$ $\implies$ $\frac{4*a_1}{4*a_2}$ $\implies$ $\frac{4*3}{4*1}$ $\implies$ $3:1$ $\textrm{Option b is correct.}$ [...]

Answered: NIELIT 2019 Feb Scientist C - Section C: 18

$\textrm{let’s assume one part of 50 is $x$,then other will be $(50-x)$}$ $\textrm{Now according to the question sum of reciprocal is $\frac{1}{12}$}$ $\implies$ $\frac{1}{x}+\frac{1}{(50-x)}=\frac{1}{12}$ $\implies$ $\frac{50-x+x)}{x*(50-x)}=\frac{1}{12}$ $\implies$ $50*12=x*(50-x)$ $\implies$ $600=50x-x^2$ $\implies$ $x^2-50x+600=0$ $\implies$ $x^2-30x-20x+600=0$ $\implies$ $x(x-30)-20(x-30)=0$ $\implies$ $(x-30)(x-20)$ $\implies$ $x=30,20$ $\textrm{option a is correct.}$ [...]

Answered: NIELIT 2019 Feb Scientist C - Section C: 22

$\textrm{according to the question it structures look like $n*n$ square.}$ $\textrm{there are $n^{th}$ row and each row is dividing into $n^{th}$ coloum.}$ $\textrm{now it’s given that $5$ men's are extra here.}$ $\implies$ $n*n=17429-5$ $\implies$ $n^2=17424$ $\implies$ $n=132$ $\textrm{so the number of men in front row is 132.}$ [...]

Answered: NIELIT 2019 Feb Scientist D - Section D: 24

$\textrm{let us assume $sin^{-1}(\frac{3}{5})=\theta$}$ $\implies$ $\sin\theta=\frac{3}{5}$ $\because \sin\theta=\frac{perpendicular}{base}$ $\textrm{by using pythagoras theorem we can find the base of right angle triangle }$ $\implies$ $H^2=B^2+P^2$ $\textrm{base =4}$ $\therefore$ $\tan\theta=\frac{perpendicular}{base}$ $\implies$ $\tan\theta=\frac{3}{4}$ $\implies$ $\theta=\tan^{-1}\frac{3}{4}$ $\therefore$ $tan^{-1}\frac{3}{4}+tan^{-1}\frac{1}{7}$ $\because \tan^{-1}(A+B)=tan^{-1}(\frac{A+B}{1-AB})$ $\implies$ $\tan^{-1}(\frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{4}*\frac{1}{7}})$ $\implies$ $tan^{-1}(1)$ $\because tan^{-1}(1)=\frac{\pi}{4}$ $\textrm{option A is correct.}$ [...]

Answered: NIELIT 2019 Feb Scientist C - Section D: 16

$\textrm{Given that:}$ $\csc \theta-\sin\theta =l$ $\sec\theta-\cos\theta=m$ $\textrm{ substitute l & m value in given equation;}$ $\implies$ $(\csc\theta-\sin\theta)^2*(\sec\theta-\cos\theta)^2*\left((\csc\theta-\sin\theta)^2+(\sec\theta-\cos\theta)^2+3\right)$ $\because \sin\theta=\frac{1}{\csc\theta},\cos\theta=\frac{1}{\sec\theta}$ $\implies$ $(\frac{1}{\sin\theta}-\sin\theta)^2*(\frac{1}{\cos\theta}-\cos\theta)^2*\left((\frac{1}{\sin\theta}-\sin\theta)^2+(\frac{1}{\cos\theta}-\cos\theta)^2+3\right)$ $\implies$ $(\frac{1-\sin^2\theta}{\sin\theta})^2*(\frac{1-cos^2\theta}{\cos\theta})^2*\left((\frac{1-sin^2\theta}{\sin\theta})^2+(\frac{1-cos^2\theta}{\cos\theta})^2+3\right)$ $\because sin^2\theta+cos^2\theta=1$ $\therefore$ $\implies$ $(\frac{\cos^2\theta}{\sin\theta})^2(\frac{sin^2\theta}{\cos\theta})^2\left((\frac{\cos^2\theta}{\sin\theta})^2+(\frac{sin^2\theta}{\cos\theta})^2+3\right)$ $\implies$ $\frac{\cos^4\theta}{\sin^2\theta}*\frac{sin^4\theta}{\cos^2\theta}* \left(\frac{cos^4\theta}{\sin^2\theta}+\frac{sin^4\theta}{cos^2\theta}+3\right)$ $\implies$ $\sin^2\theta\cos^2\theta*\left(\frac{\cos^6\theta+\sin^6\theta+3\sin^2\theta\cos^2\theta}{\sin^2\theta\cos^2\theta}\right)$ $\implies$ $(\cos\theta^2)^3+(\sin\theta^2)^3+3\sin^2\theta cos^2\theta$ $\because a^3+b^3=(a+b)^3-3ab(a+b)$ $\therefore$ $\implies$ $(sin^2\theta+cos^2\theta)^3-3sin^2\theta cos^2\theta(sin^2\theta+cos^2\theta)+3sin^2\theta cos^2\theta$ $\implies$ $1-3sin^2\theta cos^2\theta+3sin^2\theta cos^2\theta$ $\implies 1$ $\textrm{option a is correct.}$ [...]

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