I too received an offer ... I had applied for IM...
The course would start in Jan 2012... but online program would start in Nov 2011... Does that mean I can continue my job till Dec last week?
Hi johhnybravo11, I have received a similar email. I had applied for 2 year course at SPJ 2 years back.
Seniors, I have a question here. What would be the procedure if I transfer my profile submitted for 2year course to PGPM? Will I be required to submit CAT/XAT/GMAT scores again?
Thanks a lot in advance.
At the time of second interviews, SPJ asked us "What other calls do you have".
What is the significance of asking other calls. Can this be a deciding factor in declaring 2nd and 3rd list. They might want to ensure that a lot of final converters in 2nd list does not opt for other colleges. Just a thought.. Seniors plz answer...
Seems SP-Jain is loading the results ... I can see a message on right under Announcements saying "List of the Selected Candidates for the 2 year PGDM 2009-11 batch".
But this is a text and not a link... The link might be activated anytime now...
All the Best everyone...
Hi... Can you please tell me what is the GMAT score one should target to get a call for PGSEM course???
Also can you please tell me what are the rough no of students being called for Interviews by IIMB for PGSEM ???
Also what kinds of roles I can expect to be doing in the software industry after I complete PGSEM course.
In how many ways can the letters of the word JUPITER be arranged in a row so that the vowels appear in alphabetic order?
(1) 736 (2) 768 (3) 792 (4) 840 (5) 876
Vowels E,I,U needs to appear in alphabetical order...
so E I U is the only possible way for vowels...
These vowels leave 4 spaces for first consonant to be arranged...
Adding first consonant leaves 5 spaces for next consonant...
Which leaves 6 spaces for third consonant and finally 7 for the last consonant.
So total no of ways = 4*5*6*7 = 840
so answer is (4) 840
By Adding all three equations we get
log(4 * a^2 * b^2 * c^2) = loga*logb + logb*logc + logc*loga
8log(4abc) = loga*logb + logb*logc + logc*loga
By raising LHS and RHS to the power of e, this can be written as
(e ^ log(4abc))^8 = (e^loga)^logb * (e^logb)^logc * (e^logc)^loga
Simplifying we get
(4abc)^8 = a^logb + b^logc + c^loga
Suppose there exists a solution which has a=b=c=X
so the above eqn can be written as
(4X^3)^8 = X^(3*logX)
4^8 * X^24 = X^(3logX)
simplifying we get
X^(8-logX) = 2^(-16/3)
taking log on both sides we get a quadratic equation in (logX)
(logX)^2 - 8logX - (16/3)log2 = 0
solving it we get
X = 4 + sqrt(16 + (16/3)log2)
as the other option with -ve sign will yield -ve X which is not possible.
So there exists atleast 1 value of pair (a,b,c) such that a=b=c
There can be more solutions possible