Candidates applying to the programmes are required to visit FMS website (www.fms.edu) and apply online. Applications will be received only through the online registration process.

Announcement in major National English Dailies: 05th September, 2010

Online registration opens on 6th September 2010 and closes on 15th October 2010

FMS Admissions Office will begin despatching the Entrance Test Admit Cards
for MBA (FT) and PhD candidates from the 2nd week of November
2010, and for MBA (PT) and MBA (PT) HCA candidates from the last
week of December 2010.

All candidates applying to any of the FMS
programmes are required to send back the acknowledgement form to
FMS Office on or before 15th October 2010.

Admission Test for MBA (FT) and PhD - 5th December 2010

Admission Test for MBA (PT) and MBA (PT) HCA - 23rd January 2011.

Each student comes to FMS with unique professional
and personal reasons for pursuing a masters degree in business
administration. You have two years to explore, to be bold,
creative and learn as much as you can. The curriculum includes
a range of courses that take advantage of new teaching methods
and the reinforcement of basic skills and concepts. The first
year comprises of core subjects. In the second year you take
a decision to specialize in a particular discipline like Finance,
Marketing or Systems. You can even decide upon a general management
degree.

The intake of this programme for 2011 is 226 participants .

Eligibility Criteria

1. Candidates applying for admission to the first year of
MBA (Full-Time) Programme must have pursued at least a 3-year
Bachelor’s Degree Programme after 12 years of formal
schooling in any of the following disciplines. The minimum
requisite percentage of marks in different disciplines is
as follows :-

Arts, Commerce or Social Sciences - 50%

Sciences - 55%

Mathematics or Statistics - 60%

Medicine, Engineering or Technology - 60% / CGPA of at
least 6.00 in a scale of 10.00

OR

Post Graduate Degree or 2nd Degree examination after 10+2+3
scheme, securing at least 60% marks.

Note : Candidates appearing for the final
examination of Bachelor’s / Post Graduate Degree examinations
may also apply.
2. There is no reservation whatsoever for the graduates of
Delhi university or the residents of Delhi.

Selection Procedure

Applications for admission to various programmes are invited
through advertisements in major national newspapers in September/October
every year. Application forms are distributed to interested
candidates on payment of the application fees. The filled-in
application forms are to be sent to the Administrative Officer,
Faculty of Management Studies, University of Delhi, Delhi-110007
for further processing. Eligible candidates are be called
for a written test and/or an interview.

The eligible applicants will be called for a written test,
and on the basis of their performance in the test, they will
be called for an interview, an ex-tempore and a group discussion.
The candidates will be required to produce their original
certificates and mark-sheets at the time of the interview.
The final list of candidates selected to the program will
be announced after conducting all the interviews.

Fee Structure
The fee payable at the time of admission is
Rs. 10, 450/-.

1) 2^2n-1 is always divisible by 3 2^2n-1 = (3-1)^2n -1 = 3M +1 -1 = 3M, thus divisible by 3

2) What is the sum of the divisors of 2^5.3^7.5^3.7^2? ANS : (2^6-1)(3^8-1)(5^4-1)(7^3-1)/2.4.6 Funda : if a number 'n' is represented as a^x * b^y * c^z .... where, {a,b,c,.. } are prime numbers then

Quote:

(a) the total number of factors is
(x+1)(y+1)(z+1) ....
(b) the total number of relatively prime numbers less than the number is n *
(1-1/a) * (1-1/b) * (1-1/c)....
(c) the sum of relatively prime numbers less than the number is n/2 * n *
(1-1/a) * (1-1/b) * (1-1/c)....
(d) the sum of factors of the number is {a^(x+1)} * {b^(y+1)} *
...../(x*y*...)

http://hallosushant.blogspot.com/ 3) what is the highest power of 10 in 203!

ANS : express 10 as product of primes; 10 = 2*5

divide 203 with 2 and 5 individually 203/2 = 101 101/2 = 50 50/2 = 25 25/2 = 12 12/2 = 6 6/2 = 3 3/2 = 1 thus power of 2 in 203! is, 101 + 50 + 25 + 12 + 6 + 3 + 1
= 198

divide 203 with 5 203/5 = 40 40/5 = 8 8/5 = 1

thus power of 5 in 203! is, 49

so the power of 10 in 203! factorial is 49

4) x + y + z = 7 and xy + yz + zx = 10, then what is
the maximum value of x? ( CAT 2002 has similar question ) ANS: 49-20 = 29, now if one of the y,z is zero, then the
sum of other 2 squares shud be equal to 29, which means, x can take a max value
at 5

5) In how many ways can 2310 be expressed as a product
of 3 factors? ANS: 2310 = 2*3*5*7*11 When a number can be expressed as a product of n distinct
primes, then it can be expressed as a product of 3 numbers in
(3^(n+1) + 1)/2 ways

6) In how many ways, 729 can be expressed as a
difference of 2 squares? ANS: 729 = a^2 - b^2 = (a-b)(a+b), since 729 = 3^5, total ways of getting 729 are, 1*729, 3*243, 9*81, 27*27. So 4 ways Funda is that, all four ways of expressing can be used to
findout distinct a,b values, for example take 9*81 now since 9*81 = (a-b)(a+b) by solving the system a-b = 9
and a+b = 81 we can have 45,36 as soln.

7) How many times the digit 0 will appear from 1 to
10000 ANS: In 2 digit numbers : 9, In 3 digit numbers : 18 + 162 = 180, In 4 digit numbers : 2187 + 486 + 27 = 2700, total = 9 + 180 + 2700 + 4 = 2893

8 ) What is the sum of all irreducible factors between
10 and 20 with denominator as 3? ANS : sum = 10.33 + 10.66 + 11.33 + 11.66 + 12.33 + 12.66 +
13.33 + 13.66……. = 21 + 23 + …… = 300

9) if n = 1+x where x is the product of 4 consecutive
number then n is, 1) an odd number, 2) is a perfect square SOLN : (1) is clearly evident (2) let the 4 numbers be n-2,n-1,n and n+1 then by multing
the whole thing and adding 1 we will have a perfect square

10) When 987 and 643 are divided by same number 'n' the
reminder is also same, what is that number if the number is a odd prime number? ANS : since both leave the same reminder, let the reminder
be 'r', then, 987 = an + r and 643 = bn + r and thus 987 - 643 is divisible by 'r' and 987 - 643 = 344 = 86 * 4 = 43 * 8 and thus the prime is 43
hence 'r' is 43

11) when a number is divided by 11,7,4 the reminders
are 5,6,3 respectively. what would be the reminders when the same number is
divided by 4,7,11 respectively? ANS : whenever such problem is given, we need to write the numbers in top row and rems in the
bottom row like this

11 7 4 | \ \ 5 6 3

( coudnt express here properly )
now the number is of the form, LCM ( 11,7,4 ) + 11*(3*7 +
6) + 5 that is 302 + LCM(11,7,4) and thus the rems when the same
number is divided by 4,7,11 respectively are,

302 mod 4 = 2 75 mod 7 = 5 10 mod 11 = 10

12) a^n - b^n is always divisible by a-b

13) if a+b+c = 0 then a^3 + b^3 + c^3 = 3abc EXAMPLE: 40^3-17^3-23^3 is divisble by since 40-23-17 = 0, 40^3-17^3-23^3 = 3*40*23*17 and thus,
the number is divisible by 3,5,8,17,23 etc.
http://hallosushant.blogspot.com/ 14) There is a seller of cigerette and match boxes who
sits in the narrow lanes of cochin. He prices the cigerattes at 85 p, but found
that there are no takers. So he reduced the price of cigarette and managed to
sell all the cigerattes, realising Rs. 77.28 in all. What is the number of
cigerattes?

a) 49 b) 81 c) 84 d) 92

ANS : (d) since 77.28 = 92 * 84, and since price of cigarette is
less than 85, we have (d) as answer

Quote:

i have given this question to make the funda clear

15) What does 100 stand for if 5 X 6 = 33 ANS : 81 SOLN : this is a number system question, 30 in decimal system is 33 in some base 'n', by solving we
will have n as 9 and thus, 100 will be 9^2 = 81

16) In any number system 121 is a perfect square, SOLN: let the base be 'n' then 121 can be written as n^2 + 2*n + 1 = (n+1)^2 hence proved

17) Most of you ppl know these, anyways, just in case

Quote:

(a) sum of first 'n' natural numbers -
n*(n+1)/2
(b) sum of the squares of first 'n' natural numbers - n*(n+1)*(2n+1)/6
(c) sum of the cubes of first 'n' natural numbers - n^2*(n+1)^2/4
(d) total number of primes between 1 and 100 - 25

18 ) See Attachment to
know how to find LCM, GCF of Fractions

Quote:

CAT 2002 has 2 questions on the above
simple concept

19) Converting Recurring Decimals to Fractions

let the number x be 0.23434343434........

thus 1000 x = 234.3434343434...... and 10 x = 2.3434343434......... thus, 990 x = 232 and hence, x = 232/990

20) Reminder Funda

(a) (a + b + c) % n = (a%n + b%n + c%n) %n EXAMPLE: The reminders when 3 numbers 1221, 1331, 1441 are
divided by certain number 9 are 6, 8, 1 respectively. What would be the reminder
when you divide 3993 with

9? ( never seen such question though ) the reminder would be (6 + 8 + 1) % 9 = 6

(b) (a*b*c) % n = (a%n * b%n * c%n) %n EXAMPLE: What is the remainder left when 1073 * 1079 *
1087 is divided by 119 ? ( seen this kinda questions alot ) 1073 % 119 = ? since 1190 is divisible by 119, 1073 mod 119 is 2 and thus, "the remainder left when 1073 * 1079 * 1087
is divided by 119 " is 2*8*16 mod 119 and that is 256 mod 119 and that is
(238 + 18 ) mod 119 and that is 18

Glossary : % stands for reminder operation

find the number of
zeroes in 1^1* 2^2* 3^3* 4^4.............. 98^98* 99^99* 100^100

the expresion can be
rewritten as (100!)^100 / 0!* 1!* 2!* 3!....99!

Now the numerator has 2400 zeros

the formular for finding number of zeros in n! is

[n/5]+[n/5^2]...[n/5^r] where r is such that 5^r<=n<5^(r+1)

and [..] is the grestest integer function

for the numerator find the number of zeros using the above
formulae..

for 0!...4! number of zeros ..0 5!...9!.number os zeros ..1 9!...14!... 2 15!..19!..................3 20!..24!..................4! now at 25! the series makes a jump to 6 25!...29!.................6 30!...34!.................7 this goes on and again makes a jump at 50! and then at 75!

so the number of zeros is...

5(1+2....19) + 25+ 50+ 75

the last 3 terms 25 50 and 75 are because of the jumps..

this gives numerator has 1100 zeros

now total number of zeros in expression is no of zeros in
denominator - no of zeros in numerator 2400 - 1100

If an equation (i:e f(x)=0 ) contains all positive co-efficient of any powers of x , it has no positive roots then.

eg: x^4+3x^2+2x+6=0 has no positive roots . ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

For an equation f(x)=0 , the maximum number of positive roots it can have is the number of sign changes in f(x) ; and the maximum number of negative roots it can have is the number of sign changes in f(-x) .

Hence the remaining are the minimum number of imaginary roots of the equation(Since we also know that the index of the maximum power of x is the number of roots of an equation.) ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

For a cubic equation ax^3+bx^2+cx+d=o

sum of the roots = - b/a sum of the product of the roots taken two at a time = c/a product of the roots = -d/a ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

For a biquadratic equation ax^4+bx^3+cx^2+dx+e = 0

sum of the roots = - b/a sum of the product of the roots taken three at a time = c/a sum of the product of the roots taken two at a time = -d/a product of the roots = e/a +++++++++++++++++++++++++++++++++++++++++++++++++++++++++

If for two numbers x+y=k(=constant), then their PRODUCT is MAXIMUM if

x=y(=k/2). The maximum product is then (k^2)/4 +++++++++++++++++++++++++++++++++++++++++++++++++++++++++

If for two numbers x*y=k(=constant), then their SUM is MINIMUM if

x=y(=root(k)). The minimum sum is then 2*root(k) . +++++++++++++++++++++++++++++++++++++++++++++++++++++++++

|x| + |y| >= |x+y| (|| stands for absolute value or modulus )

(Useful in solving some inequations) ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Product of any two numbers = Product of their HCF and LCM .

Hence product of two numbers = LCM of the numbers if they are prime to each other

For an isosceles trapezium , sum of a pair of opposite sides is equal in length to the sum of the other pair of opposite sides .(i:e AB+CD = AD+BC , taken in order) .

If a+b+c+d=constant , then the product a^p * b^q * c^r * d^s will be maximum

if a/p = b/q = c/r = d/s . ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Consider the two equations

a1x+b1y=c1 a2x+b2y=c2

Then , If a1/a2 = b1/b2 = c1/c2 , then we have infinite solutions for these equations. If a1/a2 = b1/b2 <> c1/c2 , then we have no solution for these equations.(<> means not equal to ) If a1/a2 <> b1/b2 , then we have a unique solutions for these equations.. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

For any quadrilateral whose diagonals intersect at right angles , the area of the quadrilateral is

0.5*d1*d2, where d1,d2 are the lenghts of the diagonals. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Problems on clocks can be tackled as assuming two runners going round a circle , one 12 times as fast as the other . That is ,

the minute hand describes 6 degrees /minute the hour hand describes 1/2 degrees /minute .

Thus the minute hand describes 5(1/2) degrees more than the hour hand per minute .

The hour and the minute hand meet each other after every 65(5/11) minutes after being together at midnight. (This can be derived from the above) . ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

If n is even , n(n+1)(n+2) is divisible by 24

If n is any integer , n^2 + 4 is not divisible by 4

Given the coordinates (a,b) (c,d) (e,f) (g,h) of a parallelogram , the coordinates of the meeting point of the diagonals can be found out by solving for

1/2*base*altitude = 1/2*a*b*sinC = 1/2*b*c*sinA = 1/2*c*a*sinB = root(s*(s-a)*(s-b)*(s-c)) where s=a+b+c/2 =a*b*c/(4*R) where R is the CIRCUMRADIUS of the triangle = r*s ,where r is the inradius of the triangle .

In any triangle a=b*CosC + c*CosB b=c*CosA + a*CosC c=a*CosB + b*CosA ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

If a1/b1 = a2/b2 = a3/b3 = .............. , then each ratio is equal to

(k1*a1+ k2*a2+k3*a3+..............) / (k1*b1+ k2*b2+k3*b3+..............) , which is also equal to (a1+a2+a3+............./b1+b2+b3+..........) ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

In any triangle

a/SinA = b/SinB =c/SinC=2R , where R is the circumradius ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

x^n -a^n = (x-a)(x^(n-1) + x^(n-2) + .......+ a^(n-1) ) ......Very useful for finding multiples .For example (17-14=3 will be a multiple of 17^3 - 14^3)

If Q be the volume of a vessel q qty of a mixture of water and wine be removed each time from a mixture n be the number of times this operation be done and A be the final qty of wine in the mixture

then , A/Q = (1-q/Q)^n +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Area of a trapezium = 1/2 * (sum of parallel sids) * height = median * height

where median is the line joining the midpoints of the oblique sides. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

when a three digit number is reversed and the difference of these two numbers is taken , the middle number is always 9 and the sum of the other two numbers is always 9 .

Let a be the side of an equilateral triangle . then if three circles be drawn inside this triangle touching each other then each's radius = a/(2*(root(3)+1)) ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Let 'x' be certain base in which the representation of a number is 'abcd' , then the decimal value of this number is a*x^3 + b*x^2 + c*x + d

Many of u must b aware of this formula, but the ppl who don't know it must b useful for them.

a+b+(ab/100)

this is used for succesive discounts types of sums. like 1999 population increses by 10% and then in 2000 by 5% so the population in 2000 now is 10+5+(50/100)=+15.5% more that was in 1999

and if there is a decrease then it will be preceeded by a -ve sign and likeiwse

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