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Painter's Pesky Problem (Posted on 2002-06-29) Difficulty: 2 of 5
A painter needed 6 pints of grey paint, so he mixed 4 pints of black paint with 2 pints of white paint. This turned out to be too dark a grey and he realised that what he really needs is a mixture that has 4 pints of white with 2 of black.

The painter understands that he will have to throw away some of the paint, but wants to get the correct amount of the right mixture with as little waste as possible. How should he do it?

See The Solution Submitted by Dulanjana    
Rating: 3.5556 (9 votes)

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Solution re(2): chocolate milkshake? | Comment 3 of 4 |
(In reply to re: chocolate milkshake? by levik)

Well, all attemps using the schema I descibed would have to because they would be a combination of the pour and dilute sceme that resulted in 3 pints waste, and the dilute and pour scheme that wasted 6.

But your point is well taken. There might be methods that use different schema. How do we know that they will waste at least three pints?

First note that there are two more pints of black in the mix than there should be. Those two pints need to be accounted for in the the waste, so there cannot be less than 2 pints waste.

Is it possible to come up with a scheme to keep the waste at 2 pints? Only if you can separate 2 pints of black out of the dark gray, which is impossible.

Then, can we salvage more than one pint of the white from the mix? This would still require separating the paint into two batches one of which had a higher black ratio than the original dark gray, and would still require "unmixing" the original dark gray.

So the best we can do for the waste is the same 2:1 dark gray as the bad batch. since we need to remove 2 pints of black, 1 pint of white must come with it. So 3 pints is the least amount of total waste possible.
  Posted by TomM on 2002-06-29 17:17:10

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