____________
1.what is the angle between teo hands of a clock when time is 8-30
ans:75(appro
2.a student is ranked 13th from right and 8th from left.how many are
there(similer)
3.a,b,c,d,e,f are arranged in a circle b is to right of c and so on(rs
aggrewal)
4.chain rule(work&time)
5.puzzle test.some data is given and he asked three qustions below.based on
data
we have to answer.
6.six questions on venn diagrams.
7.5 years ago sum of ages of father and son ans:40,10
8.assertion and reasoning 1.clouds 2.capital city of dweloped corentira ans
for
both is :a
9.man walks east & from turns to right & from to left &then 45degrees to
right.in which direction he went ans:north west
10.aa-b-bb-aaa-
11.96,85,....(a series with diff 11),3,15,45,....,total 3 questions are
given on
series
12.a student got 70% in one subject,80% in other.to get overall 75% how much
he
should get in third subject.
12.a news pwper must have a.news b.advertizements 3.editor 4.date 5.paper
13.if clouds are air, air are water and so on where birds fly?
14.press,cinema,tv ans:mass media
15.3 qustions on rs aggrewal relations i.e first few chapters of
aggrewal(english)
16.a man showed to a woman sitting in a pack & told to his friend.she is
the
daughter of my grand mother only son ans:daughter
group disscussion topic
1.if you become muncipal corporater what steps you take to develop your
area.
2.if you become administrative officer of this university what steps you
take to
improve standards.
3.tell what are the seven problems in calcutta and solutions to problems?
4.abortion is legal or illegal
the following two topics are omportant they asked the same topics to many
students by changing the numbers
5.in 2050 a nuclear disaster has ocurred and 50 persons are saved.which are
of
age above 15.of them 20 know 6 subjects and you have to choose only 3 of
six
subjects so that resticted resorces can used for future
subjects are 1.enggineering 2.medical 3.law 4.social sciences 5.life
sciences
6.----
6.in space ship 5men are going1.docter,drug asdditive 2.lady lawer has done
crime 3.teacher emotinally imbalanced
4.18.18 year old aeruonautical engineer 5. noble lauraut .suddenly some
thing
happend
and oxygen is avaliable to only three people to use which of three you
choose
interview
here they asked only personal.relating to what you have written in
sychometric
test
and some puzzles
one puzzle is 10 machines are there only one is defective iteams are coming
out
how tdo you find out which is defective
second: *
* * how do you reverse this in two transitions
* * *
* * * *
///////////////PATNI NEW OVER at jadavpur 99////////////
philips - 99
Section 1 - Electronics and Mathematics - 20 questions
section 2 - Computer Science
section 3 - Aptitude
Electronics and computer science
1. A circuit was given with lot of flip-flops etc and the operation of that
circuit was asked.
2. 15 software functions are there. It is known that atleast 5 of them are
defective. What is the probability that if three functions are chosen and
tested, no errors are uncovered.
Ans : 10 * 9 * 8
-- -- --
15 14 13
Computer Science
1.Java is
a) Multithreaded b) intrepreter c) compiler d) all of the above
Ans :d
2. The number of nodes in a k-level m-ary binary tree is :
Aptitude
A graph was given and questions regarding shortest path was asked.
For one question shortest path was asked and the answer is none of the
above.
Overall the question paper was easy.
I will try to mail u the entire question paper.
Interview:
HR questions:
1.What is your strengths and weaknesses
2.What are the values u respect
3.Site a reason why philips should hire u
4.What will u do if u are asked to manage a project which will definitely
skip its deadline.
Technical (for me)
1.What is runtime locatable code?
2.What is volatile, register definition in C
3.What is compiler and what its output.
negative marking be careful( + 5 for correct and -2 for wrong choice)
1. What is Java?
i. compiled language
ii. interpreted language
iii. Multi threaded
iv. Obj. Oriented.
Ans: i, ii, iii, iv
2. Suppose you want to create a file in a dir in UNIX. What should you have?
i. Read Permission
ii. Write ,,
iii. Exec. ,,
iv. Own the dir.
Ans : All the above
3. For all x, Roar(x) => Lion(x).
........
4. ASSERT b>0 c>0
if b > c
a = b/c /*path 1*/
else
a = c/b /*path 2*/
ASSERT a > 1
i. path 1 is ok
ii. path 2 is ok
iii. Both ok
iv. None ok.
Ans: i
5. A question concerning relations. (A is B's father. C is B's uncle etc.). Finally who is related to who.
6. A ques. on Data interpretation.
7. A set of page accesses is given. Max no of pages is 4. How many cache misses.
8. unsigned int x=(unsigned ) ~0;
shift x to the left till you get zero.
The program finds word length in a machine.
For K as large as possible, produce a K-digit integer M such that for each N=1,2,...,K, the integer given by the first N digits of M is divisible by N.
An example is K=4, M=7084, because 7 is divisible by 1; 70 is divisible by 2; 708 is divisible by 3; and 7084 is divisible by 4.
Answer:
K=25, M=3608528850368400786036725
Puzzle for July 2005. Solutions will not be solicited.
This puzzle was suggested by Alan O'Donnell.
We are not asking for solutions this month.
Upon a rectangular table of finite dimensions L by W, we place n identical, circular coins; some of the coins may be not entirely on the table, and some may overlap. The placement is such that no new coin can be added (with its center on the table) without overlapping one of the old coins. Prove that the entire surface of the table can be covered completely by 4n coins.
Answer:
Start with the n "small" coins in the rectangle. Replace each one with a "big" coin of twice the radius, at the same center.
Claim that these completely cover the rectangle; otherwise, a point not covered could serve as the center of a new "small" coin that would not have overlapped with any of the original "small" coins. Now since n "big" coins cover a rectangle of dimension L by W, then n "small" coins will cover a rectangle of dimension L/2 by W/2 (just shrink the whole picture), and four copies of that -- 4n "small" coins -- will cover the original L by W rectangle.
Puzzle for June:
This puzzle is based on a suggestion by Aditya K Prasad,
(Further attribution will be given with the solution.)
Consider a string S of N symbols, selected from the set {A,B}.
In any consecutive substring of S,
the number of A's differs from the number of B's by at most 3.
How many such strings S are there (as a function of N, in closed form)?
This puzzle (but not the solution) was adapted from a problem (B-5) in the 1996 Putnam exam.
For any string S, let f(k) be the number of A's among the first k symbols, minus the number of B's among the first k symbols. We know f(0)=0, |f(k)-f(k-1)|=1. The sequence f(0),f(1),...,f(n) has to be confined to one of the following intervals: either [0,3] or [-1,2] or [-2,1] or [-3,0]. This is because the confining interval must contain f(0)=0, and if the interval is of length 4 or greater, say |f(j)-f(k)|=4 with j
But the sequence might also be confined to a smaller interval, either [0,2] or [-1,1] or [-2,0], in which case it will be counted twice, so we must subtract it. Or it might be confined to an even smaller interval: [0,1] or [-1,0], in which case it will have been counted three times and subtracted twice, so that we're okay. Letting M(i,j)=M(i,j;n) be the number of strings of length n confined to the interval [i,j], the principle of inclusion and exclusion tells us we want M(0,3)+M(-1,2)+M(-2,1)+M(-3,0)-M(0,2)-M(-1,1)-M(-2,0).
To calculate M(i,j;n) we use recurrence relations. We have the initial conditions M(0,3;0)=M(-1,2;0)=1. M(0,3;n)=M(-3,0;n) and M(-1,2;n)=M(-2,1;n), by symmetry. M(0,3;n)=M(-1,2;n-1): the first element must be "A", and the ensuing string must be one of those counted by M(-1,2;n-1). M(-1,2;n)=M(-2,1;n-1)+M(0,3;n-1): the first element can be "A", and the rest of the string would be one of those counted by M(-2,1;n-1), or the first element can be "B", and the rest of the string would be one of those counted by M(0,3;n-1). Substituting the earlier identities into the last one, we find M(-1,2;n)=M(-1,2;n-1)+M(-1,2;n-2), a relation satisfied by the Fibonacci numbers. Along with the initial conditions, we find M(0,3;n)=M(3,0;n)=F(n+1), M(-1,2;n)=M(-2,1;n)=F(n+2), where we take as the Fibonacci numbers
F(0)=0, F(1)=1, F(k)=F(k-1)+F(k-2), F(k)={ ((1+sqrt(5))/2)**k - ((1-sqrt(5))/2)**k } / sqrt(5).
M(0,2;n)=M(-1,1;n)=M(-2,0;n)=1.
M(0,2;n)=M(-2,0;n) by symmetry.
M(0,2;n)=M(-1,1;n-1) as before.
M(-1,1;n)=M(0,2;n-1)+M(-2,0;n-1) as before.
The solution is
M(0,2;n)=2**[n/2] where [x] is the greatest integer in x.
M(-1,1;n)=2**[(n+1)/2].
Putting it together, our answer is:
M(0,3)+M(-1,2)+M(-2,1)+M(-3,0)-M(0,2)-M(-1,1)-M(-2,0)
= F(n+1) + F(n+2) + F(n+2) + F(n+1) - 2**[n/2] - 2**[(n+1)/2] -2**[n/2]
= 2F(n+3) - 2**[(n+2)/2] - 2**[(n+1)/2].
Max Alekseyev pointed out that the sequence is in OEIS:
http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A027557
Puzzle for July 2005. Solutions will not be solicited.
This puzzle was suggested by Alan O'Donnell.
We are not asking for solutions this month.
Upon a rectangular table of finite dimensions L by W, we place n identical, circular coins; some of the coins may be not entirely on the table, and some may overlap. The placement is such that no new coin can be added (with its center on the table) without overlapping one of the old coins. Prove that the entire surface of the table can be covered completely by 4n coins.
july 2005:
Start with the n "small" coins in the rectangle. Replace each one with a "big" coin of twice the radius, at the same center.
Claim that these completely cover the rectangle; otherwise, a point not covered could serve as the center of a new "small" coin that would not have overlapped with any of the original "small" coins. Now since n "big" coins cover a rectangle of dimension L by W, then n "small" coins will cover a rectangle of dimension L/2 by W/2 (just shrink the whole picture), and four copies of that -- 4n "small" coins -- will cover the original L by W rectangle.
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