@abhishek.2011 Then we can't be sure of the exact ans..?? Because I got 18.18% and obviously marked none of these and must have been correct, since you are saying so! ( I didn't check )
@falcao
the best fit arrangement I believe would be when each of the bangles are touching each other, with their centres as vertices of an imaginary regular hexagon of side = 2*radii = 2*2 = 4
So the radius of the circle encircling these bangles= radius of the circumcircle + radius of bangle
= 4 + 2 = 6
=>Min. Diameter of plate = 12
@Crysis @abhishek.2011 -i don't have the oa with me.i got 12 as the answer using same arrangement as @Crysis . can u explain ur solution @abhishek.2011 .i would like to see if u have a better solution we might have overlooked
@falcao said:6 Bangles each of 4cm in diameter,what is the minimum diameter of plate required so that each bangles are kept without overlapping(bangles touching each other)?p.s:don't have oa.answer with a solution shall be appreciated.
Place the circles on vertices of a pentagon and one in the centre, from here, diameter = 12
@falcao said:6 Bangles each of 4cm in diameter,what is the minimum diameter of plate required so that each bangles are kept without overlapping(bangles touching each other)?p.s:don't have oa.answer with a solution shall be appreciated.
let the bangles centre point be vertice of a hexagon
length = 2a+4 where a =4
so 12
@abhishek.2011 @Crysis @ScareCrow28 -so looks like we all agree on one solution.the answer should be 12 then
@falcao said:6 Bangles each of 4cm in diameter,what is the minimum diameter of plate required so that each bangles are kept without overlapping(bangles touching each other)?p.s:don't have oa.answer with a solution shall be appreciated.
12 ?
place the bangles on the vertices of the hexagon
so diameter = 2*a + 4 = 2*4 + 4 = 12...a = side of hexagon
@ScareCrow28 said:Place the circles on vertices of a pentagon and one in the centre, from here, diameter = 12
I think this arrangement will not be possible as the circle that can come in a regular polygon is smaller that can be put on its vertices...
In this case, the circumcircle of the pentagon = 3.4
=> the radii of the inner circle = 3.4 - 2 = 1.4
so the 6th bangle wont fit in the the imaginary pentagon 😃
@Crysis said:I think this arrangement will not be possible as the circle that can come in a regular polygon is smaller that can be put on its vertices...In this case, the circumcircle of the pentagon = 3.4 => the radii of the inner circle = 3.4 - 2 = 1.4 so the 6th bangle wont fit in the the imaginary pentagon
Thought so! 
Thanks for correcting sir.. Symmetry is beautiful! Unfortunately, there's none in a pentagon!
Thanks for correcting sir.. Symmetry is beautiful! Unfortunately, there's none in a pentagon!if the roots of the eqn a(b-c)X^2 + b(c-a)X + c(a-b) = 0 are equal, then a,b ,c are in
AP
HP
GP
No of roots of the eqn x - 2/(x-1) = 1 - 2/(x-1) is
@vbhvgupta said:No of roots of the eqn x - 2/(x-1) = 1 - 2/(x-1) is
non?
x(x - 1) - 2 = x - 1 - 2
x^2 - 2x + 1 = 0
(x - 1)^2 = 0
x = 1
so ... x - 2/(x - 1) = 1 - 2/0..which is undefined!
@vbhvgupta said:No of roots of the eqn x - 2/(x-1) = 1 - 2/(x-1) is
bhai 2/(x-1) toh cancel out ho rh hai..is question correct??
@vbhvgupta said:if the roots of the eqn a(b-c)X^2 + b(c-a)X + c(a-b) = 0 are equal, then a, ,c are inAPHPGP
hp
assumed 1/(x-1) as y and thn approcahed the question...
at final step got 1=y-2y^2=y-2*y^2
can not solve this.. roots CBD??
@vbhvgupta said:if the roots of the eqn a(b-c)X^2 + b(c-a)X + c(a-b) = 0 are equal, then a, ,c are inAPHPGP
x1-x2=0 (where x1,x2 are roots)
4c(a-b)=b^2(c-a)^2/a(b-c)
then tried with option AP and GP are not satisfying so HP???
cant fing better approach..:(
@Vipul24 said:x1-x2=0 (where x1,x2 are roots)4c(a-b)=b^2(c-a)^2/a(b-c)then tried with option AP and GP are not satisfying so HP???cant fing better approach..
buddy assume the value of a,b,c for AP, GP and HP then solve....the roots shd come as equal...
easier to solve.
Hi though this question is very easy but the calculation involved is bothering me a li'l....please suggest an easy way to do this:
Q) P borrowed Rs 5000 at the rate of 12% for 3 years compounded quarterly. What is the amount he will have to pay back finally?
Here the calculation (1.3)^12 is difficult to solve. Is there some other way?
Thank you