This is in continuation of
Which is Which.
You have ten balls  five red and five black, having precisely the same shape and size. All the red balls weigh differently, i.e. one of them is
very heavy, the other
heavy, another is
medium, another is
light and the fifth
very light. Each red ball
has a black twin of the same weight. It is known that:
(i) A
very heavy and a
medium ball put together weigh as much as two
heavy balls.
(ii) A
heavy and a
light ball put together weigh as much as two
medium balls and:
(iii) A
medium and a
very light ball put together weigh as much as two
light balls.
Determine the least number of weighings required on a balancing scale to determine which is which, given that:
(i) Each ball can only be paired with a ball of like color before weighing, but pairing a given ball with its twin is not allowed.
(ii) Each ball can also be paired with its twin (in addition to a ball of like color) before weighing.
The system of equations provided implies (from a lightest, to f heaviest):
a+d=b+c; a+f=b+d; b+f=c+d; and a+b+c=d+f.
Other than that I have some sympathy with Dej's comments, particularly that once, say, {a,b,c,d,f} black are correctly ordered, then the problem is solved, with no need to deal with the red balls. That can be done in 5 weighings, possibly less, though I don't see a quicker way at the moment.
This may also help (KEY: LHS is L=lighter, H= heavier, == both sides equal:)
LHS RHS
{a,b} {c,d} L
{a,b} {c,f} L
{a,b} {d,f} L
{a,c} {b,d} L
{a,c} {b,f} L
{a,c} {d,f} L
{a,d} {b,f} L
{a,d} {c,f} L
{a,f} {c,d} L
{b,c} {a,f} L
{b,c} {d,f} L
{b,d} {c,f} L
{a,d} {b,c} ==
{a,f} {b,d} ==
{b,c} {a,d} ==
{b,d} {a,f} ==
{b,f} {c,d} ==
{c,d} {b,f} ==
{a,f} {b,c} H
{b,d} {a,c} H
{b,f} {a,c} H
{b,f} {a,d} H
{c,d} {a,b} H
{c,d} {a,f} H
{c,f} {a,b} H
{c,f} {a,d} H
{c,f} {b,d} H
{d,f} {a,b} H
{d,f} {a,c} H
{d,f} {b,c} H
Edited on February 16, 2012, 7:15 am

Posted by broll
on 20120216 06:34:13 