mind blowing questions....@quant

[khanna_sumit]

Originally Posted by Eccentric

zero ! ! ! !

probability should be very small but not altogether zero........coz

prob= surface area of five spheres/
volume of the circumsphere

[Eccentric]

Originally Posted by khanna_sumit

probability should be very small but not altogether zero........coz

prob= surface area of five spheres/
volume of the circumsphere

He is talking about on surface.. so the denominator is /surface area of circumsphere.

Again probability of choosing a particular point on line/ surface is zero !!

[khanna_sumit]

Originally Posted by Eccentric

He is talking about on surface.. so the denominator is /surface area of circumsphere.

Again probability of choosing a particular point on line/ surface is zero !!

im sorry.....i took the question as this that if a point is chosan inside the circumsphere......yes if a point is chosen on the sueface of the circumphere in that case the prob will be zero

regards
Sumit

[jsmba]

Hi

Could u please post the answer for the odd ball (heavier/lighter) problem?? I could also figure it out in 4 weighings only


Thanks

[Eccentric]

Originally Posted by jsmba

Hi

Could u please post the answer for the odd ball (heavier/lighter) problem?? I could also figure it out in 4 weighings only


Thanks

Check out
12 Balls problem thread
http://www.pagalguy.com/cat/showthread.php?t=8438 (problem of 12 balls)

[Aarav]

Originally Posted by Aarav

Seem some geometry studs out there.

OK solve...

A point P inside the equilateral triangle is of length 3,4,5 from the vertices. Find the area of the triangle.

3 medians are of length 3,4,5. Find the area of the triangle.

Good problems these are. I solved 2nd in 4-5 lines.
Want a similar neat solution for 1st I did that in 2 pages

[khanna_sumit]

How many ordered triples of integers (

a, b, c) satisfy

|

a + b| + c = 19 and ab + |c| = 97?

[DesiGuru]

I dont know what ordered triples means... but the possible values are as follows...

12,8,-1
-12,-8,- 1
38, 2, -21
-38, -2, -21
Total 4.

Multiply by two for the reversibility of a,b.
If ordered triples means the interchanging of a,b is discounted then the answer maybe 4.
Otherwise 8 possible integer triples for a,b,c.

K

[khanna_sumit]

Originally Posted by DesiGuru

I dont know what ordered triples means... but the possible values are as follows...

12,8,-1
-12,-8,- 1
38, 2, -21
-38, -2, -21
Total 4.

Multiply by two for the reversibility of a,b.
If ordered triples means the interchanging of a,b is discounted then the answer maybe 4.
Otherwise 8 possible integer triples for a,b,c.

K

Ur answer is not sufficient...the values u found are correct but there are more integral values...please post the approach as well...and by ordered triples i mean that reversibility is to be taken into account......

[pinku]

other sets of ordered triplets are-

(116, 0, -97),
(0, 116, -97),
(-116, 0, -97),
(0, -116, -97)

-pinku