Preparation

CMI Data Science Previous Year Papers with solution

January 27, 2021 by soujanya 2 Comments

CMI Offers the Data Science Programm including, Maths and CSE Basics. It has a unique environment to learn from leading researchers in the fields of computer science, mathematics and statistics. The M.Sc. in Data Science program offers an opportunity to study state-of-the-art syllabus from experts in the field. Eligibility: Undergraduate degrees such as B.A., B.Sc., […]

9,698 total views, 8 views today

CMI Mathematics Previous Year Papers with Solution

January 27, 2021 by soujanya 2 Comments

Chennai Mathematical Institute is a centre of excellence for teaching and research in the mathematical sciences. The main areas of research at CMI are Mathematics, Computer Science and Physics. In addition to a vibrant PhD programme, the Institute conducts BSc programmes in Mathematics and Computer Science as well as Mathematics and Physics, along with MSc […]

7,101 total views, 6 views today

CMI Previous Year Papers with Solution

January 27, 2021 by soujanya 2 Comments

CMI Mathematics Previous Year Papers with Solution CMI Data Science Previous Year Papers with Solution 9,101 total views, 16 views today

9,101 total views, 16 views today

TIFR Mathematics Previous Year Papers with Solution

October 18, 2020 by soujanya Leave a Comment

13,897 total views, 25 views today

13,897 total views, 25 views today

Spatial Aptitude Preparation Resources for GATE CSE

October 5, 2020 by soujanya 1 Comment

Spatial Aptitude is a subject for GATE and it usually comes for around 2 marks in GATE. Spatial Aptitude Syllabus: Transformation of shapes: translation, rotation, scaling, mirroring, assembling, and grouping. Paper folding, cutting, and patterns in 2 and 3 dimensions. Recommended Tests: GATE Overflow Test Series – GATE CSE 2022 GATE CSE Syllabus Best Books […]

10,122 total views, 16 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}… [...]

%d bloggers like this: