# @santhosh_163

##### Santhosh Thiagarajan 4,031 karma IIM
Lost Warrior @naga25french
RIP Neil Armstrong :disappointed:
Santhosh Thiagarajan @santhosh_163

looks cool though i love the old one more :-/ guess will take time to adjust to this..

Lost Warrior @naga25french
RIP Neil Armstrong :disappointed:
Santhosh Thiagarajan @santhosh_163

wow, new PG looks like FB

###### CATOfficial Quant Thread for CAT 2012 [part 3]
f(x)*f(1/x) = f(x) + f(1/x)
f(2)*f(1/2) = f(2)+f(1/2)
9*f(1/2) = 9 + f(1/2)
8*f(1/2) = 9
f(1/2) = 9/8 = (1/2)^3 + 1

at x = 1
f(1)^2 = 2f(1)
f(1) = 2 = 1^3+1

f(2) = 9 = 2^3+1

observe
f(x) = x^3 + 1

f(4) = 4^3 + 1 = 65

is there some proof ? or is there any way to solve ?
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Lajwanti D'Souza @laj
_Dr Uday Karmarkar_ _*Dr Uday Karmarkar*, founder and director of the Business and Information Technologies Project (BIT) at the University of California's Anderson School of Management. BIT studies the impact of new o
Santhosh Thiagarajan @santhosh_163

what is the reason behind asking real name after every post ? Though you would suggest we take it through PM, I am sure this is going on in the minds of many users and so if you could post it once in public, it will help in clearing everyone's doubt? Santhosh T

###### CATOfficial Quant Thread for CAT 2012 [part 3]
^What if we tweak this question to

set A consists of 2011 integers and set B consists of 2010 integers. wat can be the max no of elements in another set C whose elements are given by all the possible sums of any 2 integers from sets A and B. i.e set C={a+b} , where a and b belong to set A and set B resp

max will be 2011*2010
• 1 Like
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###### CATOfficial Quant Thread for CAT 2012 [part 3]
no it was 4020.

My approach donno whether correct or not

take all 2010 elements of either set same. this would yield 2010 possible sums.

the extra 2011th element in set A, would yield another set of 2010 possible sums with each element of set B.

hence, total 4020 elements

na your answer is right but approach isnt.. if the 2010 elements of both set same.. for ex: 1 to 2010, then the sum can be 2 to 4020 which itself is 4019.. if the next element you take is say 0 or 2011, then you will get one more new sum either 1 or 4021.. hence 4020 is the minimum no of elements
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###### CATOfficial Quant Thread for CAT 2012 [part 3]
allan89 Says
set A consists of 2011 integers and set B consists of 2010 integers. wat can be the min no of elements in another set C whose elements are given by all the possible sums of any 2 integers from sets A and B. i.e set C={a+b} , where a and b belong to set A and set B resp

cant find any good approach.. is it 4010 by any chance ? :)

edit : sorry 4020.. i meant 2011+2010-1 but made a mistake in the addition
• 1 Like
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###### CATOfficial Quant Thread for CAT 2012 [part 3]
Sab logon ne ek dum durust farmaya

Q.) Which of the following cannot be the sum of 12 consecutive odd natural numbers?

a.) 2300...........b.) 2924..........c.) 3644.........d.) 4356........e.) All of these

@Sumeet bhai: Aapko bhi ______/\______. Bhai ye to AIMCAT ke ques hain, main to bas copy paste kar raha hoon ;)

d) 4356

rest three numbers have differences in multiples of 24

Edit : all of these..
• 1 Like
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###### CATOfficial Quant Thread for CAT 2012 [part 3]
when all question right.
5c5 =1 (only one candiate)
when 4 question right.
5c4* 4 (the last question can be 4 ways because one will be correct choice))
5+5c4*4+5c3*4*4 +5c2*4*4*4 = 825

mene kya golmal kardiya koi dekho

the question is atleast.. you have to go backwards.. the 5 questions can be solved in 5^5 ways which is 3125.. while there are only 3000 students.. so work backwards.. there is also a addition mistake.. it should be 821 which is exactly 125 more than 696 same way as 3125 is 125 more than 3000

821 is the no of students who can atmost be positive..
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###### CATOfficial Quant Thread for CAT 2012 [part 3]

1) In a mixture of 80 litres of milk and water, 25% of the milk. How much water should be added tp the mixture so that milk becomes 20% of the mixture?

I was able to do it by:

(60+x)/(80+x) = 80/100

Thanks

use the first statement alone.. figure out the water and milk content.. its 60 water and 20 milk.. now the second statement suggests that 20 milk needs to become 20% which is 1/5.. therefore total mixture has to be 20*5 = 100 litres.. already 60 litres water is present which means you need to add 20 litres.. this is how i solved it when i saw the question and i hope it helps.. this is just a logical explanation of the same equation that you have written. but logical approach worked better for me than equations
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