iisc

IISc- what’s different from IITs?

July 9, 2016 by arjun Leave a Comment

When I joined for IISc, I didn’t have a second thought. But nowadays people are really confused about IISc and IITB. So, just thought of sharing some information about IISc. Pros IISc is the premier research institute in India. So, expect the best faculties here though not everything that glitters is gold. Being research intensive, […]

27,457 total views, 12 views today

IISc. GATECSE Cutoff

April 8, 2016 by ibia Leave a Comment

Cutoffs for M.Tech./Research admissions in IISc.

20,161 total views

A GATE Preparation Went Wrong, But…

May 25, 2015 by Arjun Suresh 3 Comments

Preparation While during B.Tech as any normal student I also wanted to write GATE. As a natural step I thought of coaching. But due to lack of good coaching centers nearby, I didn’t join any. Instead, bought GATE forum materials. I tried to solve their questions, but couldn’t proceed more than 2 pages as they […]

35,240 total views, 8 views today

GATE Overflow
Aptitude Overflow

Database Architecture

if an unpipelined processor with a cycle time of 25 ns is evenly divided into 5 pipeline stages using pipeline latches with 1-ns latency,what is the total latency of the pipeline? how about if the processor is divided into 50 pipeline stages? [...]

Answered: NIELIT 2016 MAR Scientist C - Section C: 54

Option - (C) In a circular linked list, insertion of a record involves modification of two pointer.A circular linked list is a linked list in which the last node points to the head or front node making the data structure to look like a circle. There is no NULL at the end.For insertion of node in the beginning we need traverse the whole list. Also, for insertion and the end, the whole list has to be traversed. If instead of start pointer we take a pointer to the last node then in both the cases there won’t be any… [...]

Answered: GATE IT 2005 | Question: 1

P (no two marble have the same colour) = 1 - P(Two or more marble have the same colour) Case I --> P( Two marble same colour) = 3C2*P(both Red ,blue or Green) = 3*{(1/2*1/2*1/2)+(1/6*1/6*5/6)+(1/3*1/3*2/3)} ==> 24/36 Case II--> P(Three marble same colour) = 3C3*P(All Red, Blue or Green) = 1*{(1/2*1/2*1/2)+(1/3*1/3*1/3)+(1/6*1/6*1/6)} ==> 6/36 Therefore Probability of no two drawn marble have the same colour ==> { 1 - (30/36)} = 1/6 . [...]

Answered: justify the answer

Answered: justify the answer

The Program call $3$ fork() system call . How the execution happen ? After the first fork() call the we create one child process (C1) . Then parent (P) and Child (C1) both start the execution of the program form the end of first fork() call. Now after the first fork() call we have currently two process one child(C1) and the parent process(P) both start executing concurrently. Now both C1 and P call the second fork() system call . Now when C1 call the second fork() it will create one child say C2 and when P execute the second fork() it will… [...]

Answered: KVS Dec 2018

Natural join is used to join two relations having any number of attributes. It is denoted by a symbol ( ⨝). It is the combination of projection and filtered cartesian product. Ref: https://stackoverflow.com/questions/14127306/relational-algebra-cartesian-product-vs-natural-join [...]

Answered: NIELIT 2016

In relational databases, a tuple is one record (one row). A single entry in a table is called a Tuple or Record or Row. Ref: https://web.csulb.edu/colleges/coe/cecs/dbdesign/dbdesign.php?page=tables.php [...]

Answered: In Excel, how the contents and attributes of a cell or range of cell have to be erased

Option$: \text{A}$ Select the cells, choose Edit , and select Clear, the All [...]

Answered: CAT 1996 | Question: 49

abcs [...]

Answered: CAT 2001 | Question: 52

AE BF CG DH [...]

Answered: CAT 2010 | Question: 33

AE BF CG DH [...]

Answered: NIELIT 2019 Feb Scientist C - Section A: 22

Inverse relation for mother-in-law – either son-in-law or daughter-in-law. Option A is the only possible answer. [...]

Answered: NIELIT 2016 MAR Scientist D: 16

We have $(D\to E), (\neg E \to A), (\neg B \to \neg C), (F \to \neg H), (\neg A \to B) $ Now, we have $(D,C)$ inside the hideout. So, $E$ must also be inside (due to $D \to E),$ and $B$ must also be inside due to $([\neg B \to \neg C] \leftrightarrow [C \to B]).$ This eliminates options A, B and C. Correct answer: D. [...]

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