[catathalon]

i cudn't do the circumradius one...just after the xam realised
1/2 h b = abc/4R, so no need to find the third side!! missed out!!:crazyeye:

[chekuthan]

Here is the detailed solutions for QA section

[prvineeth]

seed(n) question

solution=>
n= 9------(single digit no)
18,27,36,45,54,63,72,81,90 --------(9 two digit nos)
108,117,126,...180,189,198 --------(11 ,three digits starting with1)
207,216,...,270,279,288,297 -----------(11,three digits starting with 2)
306,315,324,....,360,369,378,387,396,333---(12,three digits starting with 3)
405,414,423,..,450,459,468,477,486,495----(11,three digits starting with 4)

so total= 1+9+11+11+12+11 = 55

[theharpooner]

But then a^2 + b^2 < c^2 because it is an obtuse angled triangle. So the maximum of x can go upto 12 because after that it will become acute as 15 squared is 225, and 13 squared plus 8 squared is greater than that. So the possible values are 8,9,10,11,12. Answer is 5.

[yogsconnect]

Originally Posted by vibhu_kk

and the question on Seed ???

hey the seed q is just about finding the no. of multiple from 1 - 499. cos only for multiples of 9, the digit sum adds up to 9 ( its same as the divisibility 9 rule).. so we get quotient of (499/9) = 55

[atshubha]

Originally Posted by MaskedMenace

yups tht ans is 2008 - 1/2008

seed ans is 55

Seed of a number is also the divisibility test for 9 , this was a sitter if you knew just that. i guess its just [500/9] = 55

[nagarajuiitkgp]

. . .

[nagarajuiitkgp]

Originally Posted by deep_agrawal

This would still make f(1) = 1. Please be careful about such elementary errors.

if f(1)=0 ; then f(2x1)=f(2)f(1) which forces f(2) to be 0.
But given f(2)=4. So f(1) can't be 0.
so we shud take f(1)=1.

[laxminarayanan_]

Thats correct..
even i got the same answer...

The options should have been -10 or none of the above..
but unfortunately "Cant be determined" is given ... :(

Originally Posted by venkatesh629

as we know 3 and -4 are the roots of the eqn we have (x-3)(x+4)=0

i.e x^2+x-12=0

so a+b+c = -10

Thats how I got the

Please correct me if I am wrong

[nagarajuiitkgp]

Originally Posted by laxminarayanan_

Thats correct..
even i got the same answer...

The options should have been -10 or none of the above..
but unfortunately "Cant be determined" is given ... :(

No. Ans is not -10.
ax^2+bx+c=0.
3 is a root. and -4 is other root.
then f(x)=a(x^2+x-12)
so a+b+c=a(-10)
=-10a and we don't know 'a'.
so can't be determined. []