Programming Preparation Resources for GATE CSE

July 2, 2015 by Arjun Suresh 6 Comments

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Answered: GATE CSE 1987 | Question: 1-V

Position Independent Code: This code can be placed anywhere in the primary memory, executes properly regardless of its absolute address. This differs from relocatable code which requires a program loader to modify program before execution so it can run from a particular memory location. Relative Addressing mode is suitable for Program-Independent-Code (PIC) because Effective Address can be calculated by adding displacement to the PC. # Indexed mode can not be used because Base Address is provided in the instructions itself. Whenever position of code changes, address in all instructions needs to be changed. However, Base Register Mode (BRM) can be used to write… [...]

Answered: Sliding Window Protocol Question 1

Suppose A sent B a frame that arrived correctly, but there was no reverse traffic. After a while A would time out and retransmit. B would notice that the sequence number was incorrect, since it would be below FrameExpected. Consequently, it would send a NAK, which carries an acknowledgement number. Each frame would be sent exactly two times. [...]

Answered: UGCNET-Oct2020-II: 36

The correct answer is option 4. Heuristic search is a search technique that seeks to solve a problem by iteratively enhancing the solution using a heuristic function or a cost analysis. Heuristic search is two types. Direct Heuristic Search Techniques in AI( Breadth-First Search (BFS) and Depth First Search (DFS).) and Weak Heuristic Search Techniques in AI (Best-First Search, A* Search, etc) Every AI program has to do the process of searching for the solution steps that are not explicit in nature. knowledge plays a vital role in intelligence as well as creating artificial intelligence. ∴ Hence the incorrect statement… [...]

Answered: UGCNET-Oct2020-II: 31

Ans) Option A For correcting, the d-bit error hamming distance should be 2d+1. So, 2d+1=5 d=2 It means the error bits between the codeword 0000000111 and given option should be 2-bits. Only Option 1 matches. Option verify: 0000000111 (xor) 0000011111 = 0000011000 (2-bit error) 0000000111 (xor) 0000000000 = 0000000111 (3-bit error) 0000000111 (xor) 1111100000 = 1111100111 (8-bit error) 0000000111 (xor) 1111111111 = 1111111000 (7-bit error) 0000011111 is minimum with the 2-bit error. ∴ Hence the correct answer is 0000011111. [...]

Answered: MadeEasy Subject Test: Digital Logic - Circuit Output

Option B is the answer, Caption [...]

Answered: GATE CSE 2004 | Question: 75

Using a graph analogy, this would be easier to understand. Suppose, we have three colours , possible color combination pairs we can have is $\binom{n}{2}+n$, Self Loops represent when both prints can be colored by the same color. Three colours Red,Blue,Green. here $n=3$ Number of possible pairwise colorings which can be done is $\binom{3}{2}+3=6$ Now to color 26 alphabets, sufficient number of colors required. is $\min_{k} \binom{k}{2} \geq 26$. Solving it we get $k=7$. [...]

Answered: CAT2019-2: 91

Given that, total work $=172$ hours. Let be $x$ regular work hours, then overtime work will be $(172-x)$ hours. John gets Rs.$57$ per hour of regular work. Then regular income of John $=x\times57$ John gets Rs.$144$ per hour of overtime work. Then overtime income of John $=(172-x)\times114$ His income from overtime hours is $15\%$ of his income from regular hours. Then, $\Rightarrow (x \times 57) \times \frac{15}{100} = (172-x) \times114$ $\Rightarrow x \times \frac{15}{100} = (172-x) \times 2$ $\Rightarrow \frac{3x}{20} = (344-2x)$ $\Rightarrow 3x = (344-2x) \times 20$ $\Rightarrow 3x = 6880-40x$ $\Rightarrow 43x = 6880$ $\Rightarrow x = 160$… [...]

Answered: CAT2019-2: 83

Given that, in $2010$ total books of two categories (fiction and non-fiction) are $11500.$ Let be $x$ fiction categories books, then non-fiction books will be $11500 – x.$ And, in $2015,$ the library contained a total of $12760$ books in these two categories. In $2015,$ there was $10\%$ increase in the fiction category while there was $12\%$ increase in the non-fiction category. Then, $\Rightarrow x\times\frac{110}{100}+(11500-x)\times\frac{112}{100}=12760$ $\Rightarrow \frac{11}{10}x+(11500-x)\frac{28}{25}=12760$ $\Rightarrow \frac{55x+(11500-x)56}{50}=12760$ $\Rightarrow 55x+644000-56x=638000$ $\Rightarrow 644000-x=638000$ $\Rightarrow -x=63800-644000$ $\Rightarrow -x=-6000 $ $\Rightarrow x=6000$ $\therefore$ The number of fiction books in the library in $2015 = x \times \left(\frac{110}{100}\right)$ $\qquad = 6000 \times… [...]

Answered: CAT2019-2: 73

Let the score of $B=100\%$ Then, the score of $A = 100\%-10\times \left(\frac{100}{100}\right)\% = 90\%$ The ratio of $A$ and $B=90\%:100\%$ $A:B=9:10 \quad \longrightarrow (1)$ Let the score of $C=100\%$ Then the score of $B=100\%+100\times \left(\frac{25}{100}\right)\% = 125\%$ The ratio of $B$ and $C=125\%:100\%$ $B: C=5: 4 \quad \longrightarrow (2)$ Let the score of $D=100\%$ Then the score of $C =100\%-100\times \left(\frac{20}{100}\right)\% = 80\%$ The ratio of $C$ and $D=80\%:100\%$ $C: D=4: 5 \quad \longrightarrow (3)$ Now, from the eq$^{n} \; (2)$ and $(3)$, combining the ratios, we get $B: C: D = 5: 4: 5$ $B: C: D =… [...]

Answered: CAT1994-44

bca [...]

Answered: CAT2019-2: 75

Given that, the average of $30$ integers is $5 \Rightarrow$ Sum of $30$ integers $= 30\times5=150$ Among these $30$ integers, there are exactly $20$ which do not exceed $5$.$\Rightarrow 20$ $|\leq5|$ But $10$ integers might exceed $5$. We need the maximum possible average for $20$ integers, so for the remaining $10$ integers, we can take the minimum values so that the average get balanced. So, minimum possible sum of remaining $10$ integers $= 10$ $|>$ $5|$ $ \Rightarrow 10 \times6 = 60$ So, sum of remaining $20$ integers $\Rightarrow$ $150-60=90 $ Let the average of $20$ integers be $x.$ Integers… [...]

Answered: CAT2005-29

Let Englishmen be E1, E2 and E3 and Frenchmen be F1, F2 and F3. Presuming E3 knows French, so 5 calls are required so that E3 knows everybody’s secret and everybody knows E3’s secret. One call would from E1 to E2. Besides this, 3 calls i.e. F1 to F2, F2 to F3 and F1 to F3. Hence total number of calls = 5 + 1 + 3 = 9. So answer is 3rd option. [...]

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