MaskedMenace

@MaskedMenace

> My scores: OA______45A____34C____91 Sec1____20A____16C____44 Sec2____25A____18C____47 PS: Can anyone tell me how to view the poll result, as in names of guys with >120 score ?? The guy who started the thread should have made it public while creating a poll, then its possible t...
My scores:

OA______45A____34C____91
Sec1____20A____16C____44
Sec2____25A____18C____47

PS: Can anyone tell me how to view the poll result, as in names of guys with >120 score ??

The guy who started the thread should have made it public while creating a poll, then its possible to see the names of guys who polled.
I think, now a moderator can change it.
  • 1 Like  
laj
Lajwanti D'Souza @laj
Shaishav Amrut Solanki, is the Indian Institute of Management (IIM) Indore's first author - could very well be the first on-campus author among all IIMs. Having just released a book few days ago, the young author is all smiles seeing it do brisk business. The book, the author claims is as much fi...
MaskedMenace
Lava Kumar Dukanam @MaskedMenace
@Nijesh: Does it really matter that guy choosing to do MBA for whatever purpose? He is not at all ruining chances of others, the bare fact is that people who have it in them,they will get through.All others name it fate,destiny,luck et all .....
laj
Lajwanti D'Souza @laj
Shaishav Amrut Solanki, is the Indian Institute of Management (IIM) Indore's first author - could very well be the first on-campus author among all IIMs. Having just released a book few days ago, the young author is all smiles seeing it do brisk business. The book, the author claims is as much fi...
MaskedMenace
Lava Kumar Dukanam @MaskedMenace
Hmmmmm....One more to my list of "why do MBA" !! To become famous eh .... This is the most honest answer,usually people cover up that with i love managing things , management is my DNA et all...When everything outside is so saturated then we fall back on the most trivial things trying to do them differently and call it innovation. just a thought ...
> ***How many 3 digit numbers are divisible by 3 or 5 but not by 7?* *1\. 51* *2\. 300* *3\. 360* *4\. 180* *5\. 128* We need to find numbers divisible by 3,5 and eliminate those divisible by 21,35,15 and add those divisible by 105. divisible by 3 => 102 to 999 => 897/3 +...
How many 3 digit numbers are divisible by 3 or 5 but not by 7?

1. 51
2. 300
3. 360
4. 180
5. 128


We need to find numbers divisible by 3,5 and eliminate those divisible by 21,35,15 and add those divisible by 105.

divisible by 3 => 102 to 999 => 897/3 + 1 => 300
divisible by 5 => 105 to 995 => 895/5 + 1 => 180
Divisible by 21 => 21*47 is near 999 so , 47 - 4 = 43 such numbers
Divisible by 35 => 35*28 is near 999 so, 28 - 2 = 26 such numbers
Divisible by 15 => 105 to 990 => 885/15 + 1 => 60
Divisible by 105 are 9 numbers

so, Answer is 300 + 180 - 129 + 9 => 360
  • 4 Likes  
> emanresu Says > > *P is a group of four numbers 1, 2, 3 and 1. In every step, 1 is added to any two numbers in group P. In how many such steps is it possible to make all the four numbers in group P equal? * Its not possible to make all of them equal.
emanresu Says
P is a group of four numbers 1, 2, 3 and 1. In every step, 1 is added to any two numbers in group P. In how many such steps is it possible to make all the four numbers in group P equal?


Its not possible to make all of them equal.
  • 2 Likes  
> * Find the sum of the last three digits of S=871x873x875x878x881x883? * 129*127*125*122*119*117 Last digit is 0 now, Last two digits of 29*27*25*61*19*17 is 25 > h.kagda Says > > *How many ordered pairs (x,y) of integers are possible for xy/(x+y) = 196?* 1/x + 1/y ...

Find the sum of the last three digits of S=871x873x875x878x881x883?


129*127*125*122*119*117
Last digit is 0
now, Last two digits of
29*27*25*61*19*17 is 25
h.kagda Says
How many ordered pairs (x,y) of integers are possible for xy/(x+y) = 196?


1/x + 1/y = 1/196
(x+196)(y+196) = 196^2
RHS takes all the divisors of 196^2 = 14^4 => 7^4 * 2^4 => 25
(Oops Ordered pairs i ignored)

@RC89 : Wish you a Happy Birthday
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> yeh wala try kariye zara bhai log .. mera frnd planet of the apes dekh ke aaya aur mujhe pucha .. he said usi movie me dekha tha ... *there are 4 wts . a,b,c,d wts = a>b>c>d* *there are 3 pans . the wts are kept in the first pan such that the heaviest is at the bottom. u need to re-...
yeh wala try kariye zara bhai log .. mera frnd planet of the apes dekh ke aaya aur mujhe pucha .. he said usi movie me dekha tha ...

there are 4 wts . a,b,c,d
wts = a>b>c>d

there are 3 pans . the wts are kept in the first pan such that the heaviest is at the bottom.
u need to re-shuffle them and place them back again using all the 3 pans such that the same formation is obtained ( each wt. needs to be moved atleast once ).
rule = at no point can u keep the heavier wt on a lighter one


d
c
b
a
_ _ _
1 2 3

question is , minimum how many moves are needed ?


Its 2^n - 1, n is number of weights
(For more info search for towers of Hanoi)
  • 1 Like  
> not sure if am right .. jus approximating the calculation number of 5 will be 2*67 = 134 number of 5^2 will be 1*13 = 13 number of 5^3 will be 1*6 = 6 number of 5^4 will be 1*3 = 1 so i think its around 154 It will be 168
not sure if am right .. jus approximating the calculation

number of 5 will be 2*67 = 134

number of 5^2 will be 1*13 = 13

number of 5^3 will be 1*6 = 6

number of 5^4 will be 1*3 = 1

so i think its around 154

It will be 168
  • 2 Likes  
> Ten tickets are numbered 1,2,3,...,10. Six tickets are selected at random one at a time with replacement. The probability that the largest number appearing on the selected ticket is 7 is??? 7^6 is the number of cases in which the 6 tickets chosen are numbered between 1 to 7 i.e., wher...
Ten tickets are numbered 1,2,3,...,10. Six tickets are selected at random one at a time with replacement. The probability that the largest number appearing on the selected ticket is 7 is???


7^6 is the number of cases in which the 6 tickets chosen are numbered between 1 to 7 i.e., where there is no 8, 9 or 10.
6^6 is the number of cases in which the 6 tickets are numbered between 1 to 6 i.e. where there is no 7, 8, 9 or 10.
7^6 - 6^6 is the no of cases in which 7 is there and all are numbered between 1 to 7.

So, Probability is (7^6 - 6^6)/10^6
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> can u post the solution for this prob of why it is 5 ? i am getting just two solutions y takes value 44 and 60 Here are the solutions (19,684) (21,252) (27,108 ) (45,60) (99,44)
can u post the solution for this prob of why it is 5 ?

i am getting just two solutions

y takes value 44 and 60


Here are the solutions
(19,684)
(21,252)
(27,108 )
(45,60)
(99,44)
  • 4 Likes