##### Lava Kumar Dukanam 4,158 karma
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.
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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

@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 .....

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

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 ...

###### CATOfficial Quant Thread for CAT 2011 [Part 6]
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
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###### CATOfficial Quant Thread for CAT 2011 [Part 5]
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.
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###### CATOfficial Quant Thread for CAT 2011 [Part 5]

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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###### CATOfficial Quant Thread for CAT 2011 [Part 5]
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
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###### CATOfficial Quant Thread for CAT 2011 [Part 5]
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
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###### CATOfficial Quant Thread for CAT 2011 [Part 5]
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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###### CATOfficial Quant Thread for CAT 2011 [Part 5]
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)
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