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We have two white, two red and two blue balls. For each color, one ball
is heavy and the other is light. All heavy balls weigh the same. All
light balls weigh the same. How many weighing on a beam balance are
necessary to identify the three heavy balls?
You need only 2 weighings! Click here to know how!
What was the task given?
Actually, we need only 2 weighing.
Weigh 1 red & 1 white ball against 1 blue & 1 white ball. Weighing result between these balls helps us to deduce the relative weights of other balls.
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Case 1 :
If this weighing is equal then the weight of red ball and blue ball are different.
That's because the weighing is equal only when there is combination of
H (Heavy) + L (Light) against L (Light) + H (Heavy).
Since, one white ball must be heavier than the other white ball placed in other pan, the red & blue balls place along with them must be of 'opposite' weights with respect to the white balls place along with them.
So weigh this red against blue will give us the heavier red (or blue) leaving other red (or blue) as lighter.
If red (or blue) weighs more in this weighing then the white ball placed with this red (or blue) in first weighing must be lighter than the other white ball placed with blue (or red) ball.
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Case 2 : Red + White > Blue + White.
The white ball in Red + White must be heavier than the white ball in Blue + White.
The reason is if this white was lighter then even heavier red in the Red + White can't weigh more than Blue + White having heavier white. That is H + L can't beat H + H or obviously would equal with H + L!
So we have got the heavier and lighter white balls for sure.
Now, weigh red from Red + White and blue from Blue + White against other red and blue balls left.
Case 2.1/2.2 : Red + Blue > or < Red + Blue
Obviously, red and blue balls in left pan are heavier (or lighter) than those in other.
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Case 2.3 : Red + Blue = Red + Blue
Obviously, that's because of H + L = L + H.
But the red is taken from Red + White which was heavier than the Blue + White.
Since, L + H (white is heavier as concluded) can't weigh more than H + L
(other white is lighter as concluded) or L + L, the red in Red + White
must be heavier making H + H combination in that pan in first weighing (case 2).
So, we got heavier red and lighter blue obviously leaving lighter red
and heavier blue in other pan.
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Case 3 : Red + White < Blue + White.
Just replace Red with blue & vice versa in the deduction made in case 2.
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