For part A, both option A and B are correct for part B, max subnets can be 8 ( as per option A) and 32 ( as per option B), since 8 is given so option C is correct [...]

Answered: IP addressing

option B is correct [...]

Answered: Subnet addresses

For 1st part option D is correct for 2nd part option C is correct [...]

Answered: GATE CSE 2005 | Question: 29, UGCNET-June2015-III: 9

Option C is the only FALSE statement. We can always have a lossless decomposition into $\textsf{BCNF}$ but not always we can have a lossless and dependency preserving decomposition. But this is always possible in the case of $\textsf{3NF}.$ Option A is true as the requirement of $\textsf{BCNF}$ required a relation schema to be in $\textsf{3NF}.$ Actually $\textsf{3NF}$ allows transitive dependency for prime attributes whereas $\textsf{BCNF}$ does not. Option D is true as shown below. Assume the two attributes to be $A$ and $B.$ Now, we can have three cases: Either $A$ or $B$ is the candidate key but not both. i.e., $A… [...]

Answered: ISI2018-PCB-CS10

SELECT ind, sum(val) FROM (SELECT * FROM U UNION ALL SELECT * FROM V ) GROUP BY ind [...]

Answered: GATE CSE 2021 Set 1 | Question: 27

Correct Answer: C Stub(isn’t concise & coherent) Whenever a Database Problem intimidating like the above one(maybe it’s just me) appears, it’s often worth to Dissect the statements for Individual components and build up your arguments from there rather than attempting it head-on by some random example/argument only to get swayed by your hidden biases and choose the wrong answer. Couple of Basic Ideas: $\rho_{r1(x,y...)}$ is the rename operation which changes the name of the $empAge’s$ attributes to $empNo1,age1$, it’s done here to resolve the conflicts that will arise while referring the relations’(the table) attributes(column) when using relations that might… [...]

Answered: CAT2019-1: 92

Correct Answer: D. [...]

Answered: CAT2019-1: 95

Correct Answer: C. [...]

Answered: CAT2014-27

Important steps to solve this question: The perpendicular from center (O) to C will divide the chord AB into equal halves. Hence BC = AC = 3 metres. Now, in △ OCB, ∠ OCB = 90°. Hence the sides of this triangle will obey the Pythagorean rule. Hence, OB² = OC² + BC² If BC is 3 metres, the only possible Pythagorean triplet would be 3, 4 and 5. Hence OB = 5 metres. [...]

Answered: CAT2019-1: 71

Given that, corners are cut from an equilateral triangle to produce a regular convex hexagon. The smaller triangles cut are equilateral triangles (If we cut off corners to create a regular hexagon, then each angle of the hexagon is $120^{\circ}$, and so each angle of every removed triangle is $60^{\circ}$ making these triangles equilateral) Let the original triangle side be $3a.$ The area of regular convex hexagon $H = \frac{3\sqrt{3}}{2}s^{2},$ the area of equilateral triangle $ T= \frac{\sqrt{3}}{4}s^{2},$ where $s$ is the side length. Now, $H_{1} = \frac{3\sqrt{3}}{2}a^{2},T_{1} = \frac{\sqrt{3}}{4}(3a)^{2} = \frac{9\sqrt{3}}{4}a^{2}$ The required ratio $ = \dfrac{H_{1}}{T_{1}} = \dfrac{\frac{3\sqrt{3}}{2}a^{2}}{\frac{9\sqrt{3}}{4}a^{2}} = \frac{2}{3}.$ $\therefore$ The ratio of the area of… [...]

Answered: CAT2019-1: 89

Correct Answer: 6144. [...]

Answered: CAT2019-1: 100

Correct Answer: B. [...]

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