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swatz,

did you try tracking the speed post using Welcome to the Indiapost Web Site ? Use the code given in the receipt of speed post to track it.

Also call the admissions dept. to see whether they got your letter.

- 1 Like

What we need to understand that is thatthese two are very different in their nature and their culture.

The choice should be based on which b-school is morefor you, and not just which b-school is perceived as'suitable'or just ranked'better'in some XYZ b-school rankings.'higher'

Thanks for the points mentioned!

I also want to ask about refund policy of VGSOM. It's not mentioned anywhere on website. And when you call, they're not giving the answer.

Somebody please shed some light on this. Will they deduct Rs. 1000 or Rs. 10,000? What are the time limits to withdraw admission?

- 1 Like

Guys, I have converted SIBM-P and IIT-Kgp VGSOM. Waitlisted for IIT-Kanpur.

I am confused as to what I should do?

Also, isn't the fee of VGSOM on the higher side compared with other IITs? It's 1.5 lac, where as it's 40K for IIT-K (per semester).

Please comment.

- 2 Likes

I have a doubt. My graduation marks are in the form of GPA. How shall I fill the graduation marks field?? It's not accepting any decimal values.

It asks for marks obtained and maximum marks; and calculates %age automatically.

What is to be done?

What a relief.........................

I made it(gd/pi score 86.83 out of 150)

cant stop dancing,1st convert of the season, waiting nervously 4 the big bad cat.

Congrats to others also who hav converted. And rest dont lose hope.

Hey I scored 91.3/150 and still was rejected. Is there sectional (GD/PI/GT) cut-off or what? :shocked:

I am from general category. Could you please elaborate...?

the thing u're getting wrong is that the age of the employee who left was 59 years in the previous year and in the current year, his age will be 60. so when u'll decrease the age from the sum of ages in last year, it must be 59.

so the equation that u'll get will be:

40*n = 46*n - 59 + 25 + (n-1)

=> 7n = 35

=> n = 5

I hope this is correct.

ohhh... i assumed that the person who retired is 60 years old. (i.e. at the time of retirement)

But if your logic works in all cases, i guess you might be right.

Thanks.

Commenting on this post has been disabled.

Hi,

Ok now I understand that you added the increase in the ages. But why x-1? It should be only x. Because you are subtracting 60 ie. in the total you should keep the increased years for each of the x members. And then subtract 60 and add 25.

Hope I am clear.

Cheers.............:cheers:

Hi..

(x-1) because the 60 yr old person who left, is not counted anymore; because increase in his age now does not contribute to the sum and/or average. What I am doing is equating the sum of ages of employees in the next year.

(New average * New no. of people) = (Old average * Old no. of people) - (age of person who left/retired) + (age of person who newly joined) + (sum of increase in the ages of the rest of the employees)

So, what is going wrong...?

Commenting on this post has been disabled.

Hi... could anyone crack the employee caselet (Sales, Accounts, Admin)...?

I got all fractional values while calculating no. of employees in any particular dept.

Please correct my method if wrong.

e.g. Avg age in a year = 46 ; No. of employees that year = x

Avg age in the next year = 40.

Suppose we assume a 60yr old retired in that year, and a new 25 yr old man joined. So eqn is:

40*x = 46*x - 60 + 25 + (x-1)

which gives 7*x = 36

anything wrong with the eqn??

Commenting on this post has been disabled.

Answer is 1-(125/216) = 0.42

Probability That you will win = 1 - (Your Number would not come) = 1 - (5*5*5/6*6*6) = 91/216 = 0.42

But the problem asks what is the probability of winning

So we have to calculate the probability that the man earns more money than he loses.

Probability of getting the no. on all 3 dice = 1/6* 1/6 * 1/6 =

Probability of getting the no. on only 2 dice = (1/6* 1/6 * 5/6) + (1/6* 5/6 * 1/6) +(5/6* 1/6 * 1/6) =

Probability of getting the no. on only 1 die = (1/6* 5/6 * 5/6) + (5/6* 1/6 * 5/6) +(5/6* 5/6 * 1/6) =

1/216 + 15/216 + 75/216 = 91/ 216 = 1 - (125/216)

The point is, in 216 trials the man (ideally) wins 3*1 + 2*15 + 1*75 =

but he loses 1*125 =

Can someone help?

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