aptitude

Aptitude Overflow Book

December 13, 2015 by Arjun Suresh 1 Comment

Updated April 2016 April 2016 (excluding General Awareness and Computer Knowledge) 96,912 total views

96,912 total views

Verbal Aptitude Preparation Resources for GATE CSE

July 29, 2015 by Arjun Suresh 4 Comments

Materials for GATE preparation in Verbal Ability

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GATE Overflow
Aptitude Overflow

It is homework

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

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

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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