gatecse discussion

Computer Networks Preparation Resources for GATE CSE

July 29, 2015 by Arjun Suresh 9 Comments

Materials for GATE preparation in Computer Networks

52,638 total views, 2 views today

CO & Architecture Preparation Resources for GATE CSE

July 29, 2015 by Arjun Suresh 20 Comments

Materials for GATE preparation in Computer Organization and Architecture

57,942 total views

Compiler Design Preparation Resources for GATE CSE

July 29, 2015 by Arjun Suresh 12 Comments

Materials for GATE preparation in Compilers

42,658 total views, 2 views today

Digital Logic Preparation Resources for GATE CSE

July 29, 2015 by Arjun Suresh 8 Comments

Materials for GATE preparation in Digital Logic

57,033 total views, 1 views today

Algorithms Preparation Resources for GATE CSE

July 16, 2015 by Arjun Suresh 9 Comments

Materials for GATE preparation in Algorithms

88,946 total views

GATE Overflow
Aptitude Overflow

Answered: NIELIT 2021 Dec Scientist B - Section B: 94

To find the complement of any boolean expression following steps are there: Convert $0$ to $1$ and vice versa. Convert $+$ to $*$ and vice versa Take variables complement, $A\leftrightharpoons \bar A$ So complement of the given expression is as follow: $Y=ABC+AB\bar C+\bar A\bar BC+\bar ABC$ $Y=AB(C+\bar C)+\bar AC(\bar B+B)$ $Y=AB+\bar AC$ $Y^c=(\bar A+\bar B)*(A+\bar C)$ Option $(A)$ is correct. [...]

Answered: NIELIT 2021 Dec Scientist B - Section B: 62

We have 9000 4 digit numbers. now we need 4 digit number contain at most 2 different digit which is equal to total number – all 4 digits different – 3 digits different. =9000- 9*9*8*7 -9*9*8*6 [3 different digit can place 6 ways that’s why multiply by 6 in 3 different digit] = 576 (option A) [...]

Answered: NIELIT 2021 Dec Scientist B - Section B: 66

Dijkstra Algorithm cannot be applied on graph which has negatives edges as well as graph with negative cycle. If the graph has negative edge it still can give you wrong results. correct ans B. PS. https://www.quora.com/What-are-the-limitations-of-Dijkstras-algorithm-greedy-approach [...]

Answered: NIELIT 2021 Dec Scientist B - Section B: 69

page number offset m-n bit n nit logical address space =2^m. so m bit require to represent the address space page size=2^n. so n bit require for offset. so page number =m-n. Ans is option B. [...]

Answered: NIELIT 2021 Dec Scientist B - Section B: 87

Answered: NIELIT 2021 Dec Scientist B - Section B: 87

$(503201)_6\rightarrow(39601)_{10}\rightarrow(21222301)_4$ Base $6$ to base $10$ conversion: $(1*6^0+0*6^1+2*6^2+3*6^3+0*6^4+5*6^5=(39601)_{10}$ Now base $10$ to base $4$ conversion: Option $(C)$ is correct. [...]

Answered: NIELIT 2021 Dec Scientist B - Section B: 70

correct answer option B. A bottom up parser generates Reverse of Rightmost derivation . PS:-https://www.d.umn.edu/~rmaclin/cs5641/Notes/Lecture8.pdf [...]

Answered: CAT2020-1: 54

Given that, area of rectangular metal sheet $ = 135 \; \text{sq inches}.$ Now, we can draw the diagram. Diagram Let the length and breadth of the rectangle be $l$ and $b$ respectively. Let the radius of a circle be $r \; \text{inches}.$ $lb = 135$ Let the painted area be $3x.$ Then left unpainted area will be $2x.$ So, $3x + 2x = 135$ $\Rightarrow 5x = 135$ $ \Rightarrow \boxed{x = 27}$ The painted area (area of circle) $ = 3 \times 27 = 81 \; \text {sq inches}.$ The left unpainted area $ = 2 \times 27… [...]

Answered: CAT2020-1: 53

Let the volumes of $ A, B,$ and $C$ be $3x, 4x$ and $7x.$ Weights of the same volume of the metals $ A, B,$ and $C$ are in the ratio $5:2:6.$ So, the ratio of the weights in the overall alloy. $A:B:C = (3x \times5) : (4x \times 2) : (7x \times 6)$ $ \Rightarrow A:B:C = 15 : 8 : 42 $ $\therefore$ The weight of metal $C = 130 \times \frac{42}{65} = 84 \; \text{kg}.$ Short Method$:$ $\qquad \qquad A \quad B \quad C$ Volume$: \quad3 : 4 : 6$ Weight$: \quad 5 : 2 : 6 \; (1… [...]

Answered: CAT2020-1: 52

Given that, Veeru$:$ Principle $(P_{1}) = \text{Rs} \; 10000$ Rate $(R_{1}) = 5 \%$ Joy$:$ Principle $(P_{2}) = \text{Rs} \; 8000$ Rate $(R_{2}) = 10 \%$ Let the time after Joy’s investment, when the balances become equal be $n$ years. i.e., principal plus accumulated interest $ P_{1} + \left[ \dfrac{P_{1} \times R_{1} \times (n+2)}{100} \right] = P_{2} + \left[ \dfrac{P_{2} \times R_{2} \times (n)}{100} \right] $ $ \Rightarrow 10000 + \left[ \dfrac {10000 \times 5 \times (n+2)}{100} \right] = 8000 + \left[ \dfrac{ 8000 \times 10 \times (n)}{100} \right] $ $ \Rightarrow 2000 + 500 (n+2) = 800n $ $ \Rightarrow 2000 + 500n + 1000 = 800n… [...]

Answered: CAT2020-1: 51

Given that, the product of $3- \text{digit}$ numbers is more than $2$ but less than $7.$ Let the $3- \text{digit}$ number be $xyz.$ Then, $ 2< x \times y \times z < 7 $ $ \Rightarrow x \times y \times z = 3 (or) 4 (or) 5 (or) 6 $ Case$1:$ When $x \times y \times z = 3$ Three possibilities. Case$2:$ When $x \times y \times z = 4$ Six possibilities. Case$3:$ When $x \times y \times z = 5$ Three possibilities. Case$4:$ When $x \times y \times z = 6$… [...]

Answered: CAT2020-3: 75

Let us assume that the total marks is $100x$ Then Bishnu marks =$52x$ Asha marks = $64x$ Now according to the question, if $y$ is the mark obtained by Ramesh, $52x=y-23 \longrightarrow (1)$ $64x=y+34 \longrightarrow (2)$ Equation $(2)-(1)$ $\Rightarrow 12x=57 $ $\Rightarrow x= \frac{57}{12}$ $\therefore 84x=\frac{57*84}{12}=399$ Note: $84x$ means scored obtain by Gita that is $84 \%$ So option $B$ is correct. [...]

Answered: CAT2020-3: 52

Let us assume Tom age is $x$ then: Dick’s age= $3x$ Harry’s age=$2.3x=6x$ Now according to the given question, Dick’s age is 1 year less than the average age of all three, $\therefore 3x=\left (\frac{3x+6x+x}{3} \right )-1$ $\Rightarrow 3x=\left (\frac{10x}{3} \right )-1$ $\Rightarrow 1=\frac{10x}{3}-3x$ $\Rightarrow 1=\frac{10x-9x}{3}$ $\Rightarrow 1=\frac{x}{3}$ $\Rightarrow x=3$ So Tom age is $3$ year, Harry’s age is $=6x=6*3=18$ year. $\therefore$ Harry’s age is $18$ year. [...]

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