 Engineering Maths for JMET,IIMC-PGDCM,tech. PIs

Hi guys, Im starting this thread to discuss engineering maths questions,fundas etc. which might be useful for exams like jmet,iimc-pgdcm and might also help in PIs of instis like iims,nitie etc. Since im blessed with loads of free time (quit j…

Hi guys,
Im starting this thread to discuss engineering maths questions,fundas etc. which might be useful for exams like jmet,iimc-pgdcm and might also help in PIs of instis like iims,nitie etc.
Since im blessed with loads of free time (quit job for cat ) i thot instead of keeping on eating "haram ka khana" ill help out by posting some qs and solns.i am putting forth a disclaimer that im no funda at math.Only intention is to start a discussion and a thread to consolidate the knwoledege.
as a start ill try to post solutions to the sample q paper of iimc pgdcm which has quite a few qs in matrics,limits,continuity etc.Along with the solns ill add some relevant fundas,techniques i know and i expect the real math geniuses to help me out with qs i cant answer,clear any errors that i may make,suggest better solns etc.
Andy.

Hi again,
source:iimc-pgdcm sample q paper.
1.A regular polygon with n sides is circumscribed on a circle of radius r.Let t(n) be the ratio of the area of the polygon to that of the circle.
1.t(n) equals
a)(rn/pi)tan(pi/n)
b)(rn/pi)tan(2pi/n)
c)(n/pi)tan(pi/n)
d)(n/pi)tan(2pi/n)

2.The limit of t(n) as n tends to infinity
a)exists and is equal to 1
b)exists and equals 2
c)exists and equals r
d)does not exist.

The soln.
the answer to the 2nd question is obvious (choice a) cos as the no the sides tends to infinity the polygon ends up coinciding with the circle.
I attempted to find the answer to the first question strictly based on trignometry fundas but i think none of the answers are right.
Then i checked out which one of the answers wud tend to 1 as n tended to infinity.

answer choice (c) fits this (and is give as the soln. in the answer key)
if u check it for n=4 i think the soln is right ..not sure abt others.

3.The number of 2*2 matrices A,for which AA equals identity matrix,is
a)1
b)2
c)3
d)infinity

soln:

spose matrix A is a1 a2|
a3 a4
now AA is (just to refresh matrix multiplitcation)
a1 a2| |a1 a2| is | a1*a1+a2*a3 a1*a2+a2*a4
a3 a4| |a3 a4| a3*a1+a4*a3 a3*a2+a4*a4
i.e to get the 1st row,1st column element in the product
multiply terms in the first row of the first set with corresponding terms in the first column and sum all these.
returning to problem.
identity matrix:matrix with diagonal elements(a11,a22,a33...) 1 rest all 0.
diagonal matrix:only diagonal elements are non-zero.
now look at the product since all non diagonal elemets have to be 0 either
a2 and a3 are 0 or (a1+a4) is 0.
a2,a3 zeros=>for diagnoal elements to be 1 ..a1^2 and a4^2 shud be 1..
so already we get 4 solutions..
now if (a1+a4)=0 =>a1^2=a4^2
only conditions to be satisfied are
a1^2+a2*a3=1 and
a4^2+a2*a3=1
this offers infinite solns (since whatever value u choose for a1^2(or a4^2) u will always be able to find combinations of a2*a3 that satisfy the eqns.

q no 13.
Two functions h(x) and f(x) are defined as follows
h(x)=1/(1+x), x>0
=1/(1-x) , x
f(x)=h(h(x)),for all x.
Which of the following statements is true?

1.f(x) is discontinuous at x=0.
2.f(x)3.f(x) decreases in value as x increases for all x > 0
4.The max value attained by f(x) in the interval -3
soln.we can eliminate the first three statements.

some fundas:to check whether a function is contiinuous at a point.
the method i rember is to to see the limit as x takes an infinitesimal value greater
than the point at which we want to check continuity and compare with the results if we take a infinitesimal value less.both have to be same for continuity .
in this case .the point in question is 0.
spose we have a point del(x) which is a very small positive no:
so h(del1(x))=1/1+del1(x) which is 1/1+a small positive qty
so f(del1(x))=1/1+(h(del1(x))
again if we have del2(x) which is slightly less that 0
so h(del2(x))=1/1-del2(x) this is also of the form 1/1+small positive qty..('cos del2(x) is negative.
=>f(del2(x))=1/1+h(del2(x)) ...
as x tends to 0..put both dels as 0 ..u will get same value in both cases..
so f(x) is continuous at x=0
2 and 3 can be verified by substitution.
4 is the only possible one.(im not able to find a logic for getting 4/5 as the maximum
..im open for help)

so to summarize:to check continuity evaluate the limits of the fn on both sides of the potential discontinuity ..if they are same the function is continuous.

thats it for today.im eagerly waiting for fback from ppl out there b4 continuing furthur.Hope ive been of some help.
MrAnderson.

Hi again,
source:iimc-pgdcm sample q paper.
I attempted to find the answer to the first question strictly based on trignometry fundas but i think none of the answers are right.
Then i checked out which one of the answers wud tend to 1 as n tended to infinity.

answer choice (c) fits this (and is give as the soln. in the answer key)
if u check it for n=4 i think the soln is right ..not sure abt others.

using limits to arrive at the answer for the first question was simply brilliant.
no need to go thru the slightly lengthier method of trigonometry.

but using trig.....
first divide the polygon into n equal triangles by joining the vertices to the centre.since the polygon circumscribes the circle,the height of each of the n triangles is equal to the radius of the circle.
the included angle between the line joining the vertex to the centre and the altitude is (pi/n).
so the side of the polygon is 2*r*tan(pi/n) where r is the radius of the circle.so the area of the polygon is n*(0.5*2*r*tan(pi/n)*r) and the ratio of the area of the polygon with that of the circle comes out to be (n/pi)*tan(pi/n).
using limits to arrive at the answer for the first question was simply brilliant.
no need to go thru the slightly lengthier method of trigonometry.

but using trig.....
first divide the polygon into n equal triangles by joining the vertices to the centre.since the polygon circumscribes the circle,the height of each of the n triangles is equal to the radius of the circle.
the included angle between the line joining the vertex to the centre and the altitude is (pi/n).
so the side of the polygon is 2*r*tan(pi/n) where r is the radius of the circle.so the area of the polygon is n*(0.5*2*r*tan(pi/n)*r) and the ratio of the area of the polygon with that of the circle comes out to be (n/pi)*tan(pi/n).

Hey,..i had mistakenly considered the polygon as inscribed in the circle ..so thot the answer choices were wrong..read the q again after ur post and realized my mistake.
thnx for the hlp.the limit approach works either way..so was lucky andy.

In case you want to refresh the basic formulas for differentiation , go to http://www.utexas.edu/student/utlc/handouts/864.html

For diff with limits, go to
http://www.ping.be/~ping1339/diff.htm

Srini
p.s
can somebody thro light as to the exact type of q's asked in JMET. I am totally in the dark.

Can anybody plz throw some light on the topics to be covered for quant in JMET.,...???????

plz...

cheers

rids

these are al i can think of at present.....calculus, matrices, and all the CAT topics

thanks for the links yaar log

Hi again,
source:iimc-pgdcm sample q paper.
1.A regular polygon with n sides is circumscribed on a circle of radius r.Let t(n) be the ratio of the area of the polygon to that of the circle.
1.t(n) equals
a)(rn/pi)tan(pi/n)
b)(rn/pi)tan(2pi/n)
c)(n/pi)tan(pi/n)
d)(n/pi)tan(2pi/n)

2.The limit of t(n) as n tends to infinity
a)exists and is equal to 1
b)exists and equals 2
c)exists and equals r
d)does not exist.

The soln.
the answer to the 2nd question is obvious (choice a) cos as the no the sides tends to infinity the polygon ends up coinciding with the circle.
I attempted to find the answer to the first question strictly based on trignometry fundas but i think none of the answers are right.
Then i checked out which one of the answers wud tend to 1 as n tended to infinity.

answer choice (c) fits this (and is give as the soln. in the answer key)
if u check it for n=4 i think the soln is right ..not sure abt others.

Even a siple logic could be applied of eliminatio....as it is asked for the ratio...so forget the option a) now coming to the second elimiation.....as ths angle would be half of the angle subtended by one side...so onjly option left is c) which should be the answer....
Regards

Hi friends,

1)can u send me some examples of application of bayes theorem.

2)concept of multinomial theorem and its applications.

3)this is a problem on multinomial theorem

four dice are thrown.what is the probability that the sum of the numbers appearing on the dice is 18.

What if the dice are not thrown simultaneously?

What if the dice are not cubes but of different shapes? (tetrahedron , pyramid etc)

What if dice are of different shapes and thrown one after the other ?

4)another sum.

There are 4 pens each of a different colour and four caps each of a different colour but same as that of one of the the pens.in how many ways can these caps be put on the pens so that a cap doesnot go to the pen of same colour?

i)if there are L objects of one kind , m objects of second kind , n objects of 3rd kind and so on ,then number of ways of choosing r objects out of these (l+m+n+.)objects is given by the coefficient of x^r in the expansion of

(1+ x + x^2 +.+ x^L) (1+ x + x^2 +.+ x^m) (1+ x + x^2 +.+ x^n)

= / (1-x) / (1-x) / (1-x)

HOW DIFFERENT ISTHIS QUESTION FROM GROUPING OF 5 APPLES,4 ORANGES AND 3 BANANAS SO THAT EACH GROUP HAS ATLEAST ONE FRUIT?

PLEASE CONVERT THIS SUM IN THE ABOVE FORM IF POSSIBLE.

ii) in case one object of each kind is to be included in such a collection , the number of ways of choosing r objects is given by the coefficient of x ^ r in

(x+ x^2 + x^3 +.+ x^l) (x+ x^2 + x^3 +.+ x^m) (x+ x^2 + x^3 +.+ x^n)

some more sums:

1)in an examination ,max marks for each of the three papers are 50 each ..max marks for the fourth paper is 100.find no of ways in which a candidate can secure 60% marks in the aggregate?

2)a professor has 8 scholars working under him.he takes them three at a time to conferences as often as he can without taking the same three scholars more than once .

a)how often will each scholar go?

b)how often will the professor go.?

3)in how many ways can a pack of 52 cards be divided equally among four players in order?
In how many ways can you divide these cards in 4 sets ,three of them having 17 cards each and the fourth one just one card.

4) show that the number of different selections of 5 letters from 5 As' , 4 Bs' , 3 Cs' , 2 Ds' , and 1 E is 71.

5)in how many ways can an examiner assign 30 marks to 8 questions giving not less than 2 marks to any question?

PLEASE GIVE ME SOLUTIONS IN DETAILS SO THAT I CAN UNDERSTAND.

THANX.

REGARDS.

Hi friends,
There are 4 pens each of a different colour and four caps each of a different colour but same as that of one of the the pens.in how many ways can these caps be put on the pens so that a cap doesnot go to the pen of same colour?

some more sums:

1)in an examination ,max marks for each of the three papers are 50 each ..max marks for the fourth paper is 100.find no of ways in which a candidate can secure 60% marks in the aggregate?

5)in how many ways can an examiner assign 30 marks to 8 questions giving not less than 2 marks to any question?

PLEASE GIVE ME SOLUTIONS IN DETAILS SO THAT I CAN UNDERSTAND.

THANX.

REGARDS.

for the first qn...not very sure ...
but this is how i see it..
spose u have 2 sets of four each {abcd} and {ABCD}
now u have to relate each member in one set to on in the other...
and aA bB cC dD are not possible....
B C or D (spose B)
after thts done for B u can chosse frm ACD ..again 3 ways..(spose A
now for C and D there is only one way of assigning ..so total no of ways
3*3*1 = 9

summarizing the possibilites
all 4 in correct envelope ..1 way
3 in correct envelopes.not possible
2 in correct envelopes.4C2 ..6
1 in correct envelope 4C1*2 = 8
0 in correct envelope = 9
total = 1+6+8+9 = 24...so the ans is correct.
(this is the same as 4 letters 4 envelopes prob...so forgive me for the change) for the 3rd qn:
assign 2 marks to all 8 qns first...
so total 16 mrks already given remaining 14 marks have to be distributed
across 8 qns.
u can consider this scenario as distributing 14 (i.e,x) identical 1's (mark) between 8(ie n) separate qns.
the formula for this is (n+x-1)C(x-1)...(dunno how this reslt is obtained..but is correct..)
so for this case it is 8+14-1 C 14-1 ...21C13 ...is the answer.

another appln of this method....
spose there are 3 bus stops and 4 students.How many ways can these students get down if we consider all of them as identical...
so here 4 identical students have to be distributed across 3 diff stops so
3+4-1 C 4-1 ..6C3

for the second qn u can work it out using this method..but its a crazy answer...if the qn was how he can get more than 60% in aggregate its fairly straight forward...

Andy.

to andy,

i am really thankful to you for your reply.please also help me out with the dice problem if you can.

regards,
gauri. 😁