Let we have 8 coins named (for simplicity) x1, x2, x3, y1, y2, y3, a, b

According to your question "3 of them weigh x units" : x1+x2+x3 =x

"3 of them weigh y units": y1+y2+y3 =y

now consider the case when i grouped  (x1, x2 x3) in single group

(y1,y2 ,y3) in one group and (a,b) in one group.

Now we will measure weight of 1st group (x1, x2, x3) :It will give x then
we conclude that all elements in this group are those element whose total
weight is x :Type 1 element.

similarly for y.

and for the last group we will pick up any element and measure its weight
..its weight will be either a or b. Depending upon outcome we will
categorize them a and b.

So 3 weighing is required.

On Tue, Jul 10, 2012 at 7:05 PM, payal gupta <[email protected]> wrote:

> @navin...Sorry didnt get you how come u were able to segregate all the
> coins by the proposed method??
>
>
> On 7/10/12, Navin Kumar <[email protected]> wrote:
> > Minimum no. weighings:
> >
> > Divide 8 coins in group of 3, 3 and 2.
> >
> > For minimum weighsing group1 's total weight is x units(say) -->FIrst
> > weighing
> > Groups 2nd total weights is y units         --------> Second weighing.
> > Lastly one more weighing among a unit and b unit coins.--->3 rd weighing
> >
> >
> > So minimum 3 weighing is required.
> >
> >
> >
> > On Tue, Jul 10, 2012 at 11:03 AM, payal gupta <[email protected]>
> wrote:
> >
> >> You have 8 coins. 3 of them weigh x units, 3 y units, 1 a units and 1 b
> >> units. They are all mixed and look identical. What are the minimum no of
> >> weighings reqd to seperate the for types of coins???
> >>
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