iisc

IISc- what’s different from IITs?

July 9, 2016 by arjun Leave a Comment

When I joined for IISc, I didn’t have a second thought. But nowadays people are really confused about IISc and IITB. So, just thought of sharing some information about IISc. Pros IISc is the premier research institute in India. So, expect the best faculties here though not everything that glitters is gold. Being research intensive, […]

23,998 total views

IISc. GATECSE Cutoff

April 8, 2016 by ibia Leave a Comment

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

18,870 total views

A GATE Preparation Went Wrong, But…

May 25, 2015 by Arjun Suresh 3 Comments

Preparation While during B.Tech as any normal student I also wanted to write GATE. As a natural step I thought of coaching. But due to lack of good coaching centers nearby, I didn’t join any. Instead, bought GATE forum materials. I tried to solve their questions, but couldn’t proceed more than 2 pages as they […]

31,465 total views, 2 views today

GATE Overflow
Aptitude Overflow

Consider the following code fragment executed by a process of PID=1055. int main() { int a=5; if (fork() == 0) { a = a + 5; b= getppid(); printf(“%d, %d \n”, a, b); } else { c = a - 5; d= getpid(); printf (“%d, %d \n”, c, d); } return 0; } Write the output of a, b, c, d? [...]

Answered: Self Doubt

Answered: Self Doubt

The answer is 20. See below image OR One way to select root. 10 / \ LST (3 elements) RST (2 elements) Now LST contains an element out of 5 (As the root is chosen) So Total ways to choose element $_{}^{5}\textrm{C}_{3}$ and RST contains 2 element So number of ways = $_{}^{2}\textrm{C}_{2}$ Let LST chosen { 20,30,40} Now 2 chosen as root of… [...]

Answered: Kenneth Rosen Edition 6th Exercise 2.2 Question 8 (Page No. 120)

i think in 7 rhs {{∅}, {∅}} both are same according to set definition only distinct are allowed so it will become {{∅}} now question is {{∅}} ⊂ {{∅}} is it proper subset so answer is false. please tell am i thinking right?? [...]

What is the admission procedure for IIT Madras MS in CSE?

Hello all, I wanted to know, what is the procedure for getting admission into MS in CSE in IIT Madras. Please also provide the GATE cutoff is possible. Thanks! [...]

Answered: GATE CSE 1988 | Question: 4ii

Answered: GATE CSE 1988 | Question: 4ii

The Ex-OR gate is used to differentiate b/w normal binary addition vs 2’s complement addition. ref:- https://www.ijert.org/efficient-design-of-2s-complement-addersubtractor-using-qca-2 [...]

Answered: GATE CSE 2021 Set 2 | Question: 20

Answer: 80000 bits per second Explanation Transfer Time(let Ty) given as 1 cycle. Let Preparation Time be Tx. Processor operating frequency given as 2 MHz, from this we can calculate Cycle Time = 1/2MHz = 0.5 micro seconds. %time CPU blocked = 0.5% (Given) For cycle stealing mode, %time CPU blocked = (transfer time/preparation time)*100, putting the values in this formula, 0.5% = 1 cycle/Tx Tx = 100 micro second. The Tx depends upon device data transfer rate. Now DMA is transferring 8 bit data, and to prepare 8 bits of data, time required by the device is 100 micro seconds, so we can… [...]

Answered: CAT2018-2: 92

Given that, $ 4^{n} . 17^{19} $ Taking the $ \log_{10}$ both sides. $ \log_{10}4^{n} . \log_{10} 17^{19} $ $ \Rightarrow n \log_{10}4 . 19 \log_{10}17 \; [ \because \log_{b}a^{x} = x \log_{b}a] $ $ \Rightarrow n > 19 \left( \frac{\log_{10}17}{ \log_{4}17} \right) $ $ \Rightarrow n > 19 ( \log_{4}17) \; \left[ \log_{b}a = \frac{ \log_{x}a}{ \log_{x}b} \right] $ For easy calculation, let us assume $17 \approx 16.$ Now, $ n > 19 ( \log_{4} 16)$ $ \Rightarrow n > 19 ( \log_{4} 4^{2})$ $ \Rightarrow n > 19 \times 2 ( \log_{4}4)$ $ \Rightarrow $ [...]

Answered: CAT2018-2: 100

Let, $ \text{S} = \frac{1}{ \log_{2}100} – \frac{1}{ \log_{4}100} +\frac{1}{\log_{5}100} – \frac{1}{\log_{10}100} + \frac{1}{\log_{20}100} – \frac{1}{\log_{25}100} + \frac{1}{\log_{50}100} $ $ \Rightarrow \text{S} = \log_{100}2 – \log_{100}4 + \log_{100}5 – \log_{100}10 + \log_{100}20 – \log_{100}25 + \log_{100}50 \left[ \because \log_{b}a = \frac{1}{\log_{a}b} \right] $ $ \Rightarrow \text{S} = \log_{100}2 – \log_{100}4 + \log_{100}5 – \log_{100}10 + \log_{100}(5 \cdot 4) – \log_{100}(5 \cdot 5) + \log_{100}(5 \cdot10) $ $ \Rightarrow \text{S} = \log_{100}2 – \log_{100}4 + \log_{100}5 – \log_{100}10 + \log_{100}5 +\log_{100}4 – \log_{100}5 – \log_{100}5 + \log_{100}5 + \log_{100}10 \; \left[ \because \log_{x}(ab) = \log_{x}a + \log_{x}b \right] $ $ \Rightarrow… [...]

Answered: CAT2018-2: 91

Given that, $ \text{T} = \{ 1,2,3,4\} $ $ \text{Q} = \{ 2,3,5,6 \} $ $ \text{R} = \{ 1,3,7,8,9\} $ $ \text{S} = \{ 2,4,9,10\} $ $ \boxed{ \text{A} \triangle \text{B} = ( \text{A} \cup \text{B}) – ( \text{A} \cap \text{B})} $ DIAGRAM $ \text{A} \triangle \text{B} = ({ \text{A} – \text{B}) \cup (\text{B} – \text{A})} $ Now, $ \text{P} \triangle \text{Q} = \{ 1,2,3,4\} \triangle \{ 2,3,5,6\} $ $ \Rightarrow \text{P} \triangle \text{Q} = \{ 1, 4, 5,6 \} $ And, $ \text{R} \triangle \text{S} = \{ 1,3,7,8,9 \} \triangle \{ 2,4,9,10 \} $ $ \Rightarrow \text{R} \triangle \text{S} = \{… [...]

Answered: CAT2018-2: 90

Let $x$ and $y$ be the two numbers. $ x^{2} + y^{2} = 97 \quad \longrightarrow (1) $ The geometric mean cannot exceed the arithmetic mean. $ \text{AM} \geqslant \text{GM} $ $ \Rightarrow \boxed{\frac{a_{1} + a_{2} + a_{3} + \dots + a_{n} } {n} \geqslant \sqrt[n]{a_{1} a_{2} \dots a_{n}}} $ Now, $ \frac{x^{2} + y^{2}}{2} \geqslant \sqrt{x^{2} \cdot y^{2}} $ $ \Rightarrow \frac{x^{2} + y^{2}}{2} \geqslant \sqrt{(xy)^{2}} $ $ \Rightarrow \frac{x^{2} + y^{2}}{2} \geqslant xy $ $ \Rightarrow x^{2} + y^{2} \geqslant 2xy $ $ \Rightarrow 97 \geqslant 2xy $ $ \Rightarrow 2xy \leqslant 97 $ $ \Rightarrow xy \leqslant… [...]

Answered: CAT2018-2: 81

Let say, $ p^{3} = q^{4} = r^{5} = s^{6} = k \quad \longrightarrow(1) $ Now, $ p^{3} = k \Rightarrow p = k^{\frac{1}{3}} $ $ q^{4} = k \Rightarrow q = k^{\frac{1}{4}} $ $ r^{5} = k \Rightarrow r = k^{\frac{1}{5}} $ $ s^{6} = k \Rightarrow s = k^{\frac{1}{6}} $ Now, $ \log_{s} (pqr) = \log_{k^{\frac{1}{6}}} \left( k^{\frac{1}{3}} \cdot k^{\frac{1}{4}} \cdot k^{\frac{1}{5}} \right) $ $ = \log_{k^{\frac{1}{6}}} \left[ k^{( \frac{1}{3} + \frac{1}{4} + \frac{1}{5})} \right] $ $ = \log_{k^{\frac{1}{6}}} \left[ k^{(\frac{ 20+15+12}{60})} \right] $ $ = \log_{k^{\frac{1}{6}}} \left( k^{\frac{47}{60}} \right) $ $ = \frac{47}{60} \; \log_{k^{\frac{1}{6}}} k \; [ \because \log_{b} a^{x} = x\log_{b} a ] $… [...]

Answered: CAT2018-2: 74

Given that, $ f(x) = \max \{ 5x, 52-2x^{2} \} \quad \longrightarrow (1) $ And, $x$ is a positive real number. The minimum value occurs when both the graphs intersect. Let $ y_{1} = 5x \quad \longrightarrow (2) $ And, $ y_{2} = 52 – 2x^{2} \quad \longrightarrow (3) $ DIAGRAM 1 From equation $(2),$ $ y_{1} = 5x$ is an equation of line. $ y = mx+c ; $ Where m = slope of the line , c= intercept on the y – axis. From equation $(3),$ $ y_{2} = 52 – 2x^{2}$ is a quadractic equation. $ y_{2}… [...]

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