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Santhosh Thiagarajan
@santhosh_163
4.0
k

Posted 07 Apr '12

> 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 a...

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

is there some proof ? or is there any way to solve ?

Santhosh Thiagarajan
@santhosh_163
4.0
k

Posted 07 Apr '12

> ^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 s...

^What if we tweak this question to

set A consists of 2011 integers and set B consists of 2010 integers. wat can be themaxno 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

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Santhosh Thiagarajan
@santhosh_163
4.0
k

Posted 07 Apr '12

> 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
...

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

Santhosh Thiagarajan
@santhosh_163
4.0
k

Posted 07 Apr '12

> 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 *
...

allan89 Saysset 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

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Santhosh Thiagarajan
@santhosh_163
4.0
k

Posted 06 Apr '12

> Sab logon ne ek dum durust farmaya :thumbsup: ::
*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 ...

Sab logon ne ek dum durust farmayaQ.) 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..

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Santhosh Thiagarajan
@santhosh_163
4.0
k

Posted 06 Apr '12

> 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....

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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Santhosh Thiagarajan
@santhosh_163
4.0
k

Posted 06 Apr '12

> Though easy one...bt please help me with the standard methodology behind such questions:
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...

Though easy one...bt please help me with the standard methodology behind such questions:

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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