# Paint mixing

1. Jun 8, 2004

### ceptimus

A painter has a gallon tin (eight pints) full of yellow paint and another gallon tin half full of blue paint. He wants to mix the paints together so he can paint a room green. He reckons he will need at least 10 pints of paint to cover the walls.

Unfortunately, he doesn't have a mixing vessel large enough to hold all the paint - the only other container he has, besides the tins, being a pint glass.

Is it possible for the painter to mix the paint correctly using just these containers? What is the quickest method he can use?

2. Jun 8, 2004

### Gokul43201

Staff Emeritus
Does he need to mix them in the ratio 1 : 1 (Y:B) or in the ratio that he has them (2:1) ?

Edit : Ignore that.

3. Jun 20, 2004

### Moonbear

Staff Emeritus
Quickest? Okay, you take the can of blue paint, quickly invert it over the can of yellow paint, hold them tightly together, shake well, and then quickly turn them both right side up again. Sure, some will wind up sloshing onto the floor while doing this, but he has 12 pints and only needs 10, so can be a little sloppy about it. :-D

4. Jun 21, 2004

### Njorl

pour 1 pint of yellow into pint can

pour 4 pints from yellow to blue, mix

pour 5 pints from blue can to yellow can, mix

pour 5 pints from the yellow can to the blue can, mix

pour 24/19 pints from blue can to yellow can. Pour 1 pint of yellow into blue can. Mix both.

Done

5. Jun 21, 2004

### Gokul43201

Staff Emeritus
How do u do that ? You can only pour integral numbers of pints.

6. Jun 21, 2004

### vikasj007

i am not sure, but does he have to mix the two colours in the ratio of 1:1.

if this is so, then, i dont think that their is any real solution to the problem, because, he needs 10 pints to paint the whole room, that means that he needs 5 pints of yellow paint and 5 pints of blue paint, but he has only half gallon of blue paint i.e. 4 pints.
so how can he mix them in equal poportions to get 10 pints.

7. Jun 21, 2004

### Gokul43201

Staff Emeritus
No, 1:1 is not required...but all the paint needs to be the same color. You can't have - for instance - 5 pints of 2:1 and 5 pints of 3:1.

8. Jun 21, 2004

### Njorl

That makes it much harder.

Let's see...

You need to end up with 2 to 1 ratios in each container

The only amounts you can have in your containers are:
8,4,0
8,3,1
7,5,0
7,4,1
6,6,0
6,5,1
(and permutations)

You can transfer :

1,2 or 4 from 8 to 4
5 from 8 to 3
1 2 or 3 from 7 to 5
1 from 5 to 7
3 from 7 to 4
1 from 4 to 7
1 or 2 from 6 to 6
3 from 6 to 5
2 from 5 to 6
1 from anything to zero

Is there a clever way to do this, or is it just awful trial and error?

Njorl

9. Jun 21, 2004

### Gokul43201

Staff Emeritus
I was thinking more along the lines of saving a pint of yellow; then keep intermixing till the difference (in color) between the two pails can be negated by pouring the pint of yellow into one of them.

10. Jun 21, 2004

### ceptimus

Nice try but the two cans are different diameters, so this won't work.

And the cans and glass are also unevenly tapered so you can't tip either on the diagonal to get it exactly half full either (someone always tries that dodge )

11. Jun 21, 2004

### Gokul43201

Staff Emeritus
Ceptimus,

I'd like to repeat Njorl's question : "Is there a clever way to do this, or is it just awful trial and error?"

If it is the latter, please tell us that it can be done in 3 or 4 steps...else the number of possibilities get too large.

Finally, do we really need the whole gallon of yellow ? When I was walking to luch yesterday, I "thought" I figured a way in my head, that involved throwing out 2 pints of yellow in the beginning. Can't seem to recreate it since, to know if it was really correct. Only recall having one pint that was either 2/3 or 4/5 blue and transfering this at the end...extremely annoying !

Last edited: Jun 21, 2004
12. Jun 22, 2004

### ceptimus

There are at least two relatively simple solutions that both only involve a small number of steps. There is no trick or wordplay involved. As long as you end up with at least 10 pints of uniformly mixed paint, you have solved the puzzle, so you can throw away up to two pints of paint if you wish.

13. Jun 22, 2004

### The Bob

I think I have it.

Right this is what you do:
1. Pour 1 pint of blue paint into the glass and throw it away. (Yellow:8 pints - Blue :3 pints)
2. Then pour another pint of blue paint into the glass and throw it away. (Y:8 - B: 2)
3. Then pour one blue pint into the glass and keep it (Y:8 - B:1 - Glass:1 (Blue))
4. Then pour 4 pints of yellow paint into the blue tin (I assume you can see if it is half or not by using your eye) [Y:4 - B:5 - G:1]
5. Then pour the blue paint from the glass into the yellow tin and that is it. (Y:5 - B:5 - G:0) and the concentrations are 1 part blue to 4 parts yellow. There is also 10 pints of paint.

Hope it is right, if not I will have to try harder.

14. Jun 22, 2004

### Gokul43201

Staff Emeritus
No, you can not !

15. Jun 22, 2004

### Njorl

You know, I completely missed that "10 pints" bit. I've been trying to get 12 pints.

Njorl

16. Jun 22, 2004

### Gokul43201

Staff Emeritus
Still doesn't change that fact that it is basically a trial-and-error problem.

17. Jun 22, 2004

### Njorl

Discard 1 pint yellow with pint glass, repeat.

use pint glass to transfer 1 pint yellow into blue can (1y 4b)
pour one pint yellow into glass and hold in reserve.

We now have (4y) in one can (1y,4b) in another and (1y) in the glass

Now dump all you can from blue can to yellow can.
You get (4.8y,3.2b) (.2y,.8b) (1y)

Dump pint glass into blue can to get 60/40 ratios in both cans.

It was just work, for the most part.

I figured we either dump 2 yellow, or 2 blue. We are also going to be dumping back and forth winding up with 8 pints in the receiving can, so we need a 4.8 to 3.2 ratio or a 6.4 to 1.6 ratio. Either way we need fifths of pints. After that, I just plugged away.

Was there some clever way to get the answer?

Njorl

18. Jun 22, 2004

### BobG

I just can't see this truly converging.

However,

If I filled the pint glass with beer, and drank the beer while I thought about it, I might get out of painting, altogether.

Or, I could drink the beer while I mixed the paint. I'd pour yellow paint into the blue until the 'blue' can was full. Then I'd take a sip of beer. Then I'd stir the 'blue' can. Then another sip of beer. Then pour mixed paint from the 'blue' can into the 'yellow' can. Sip of beer. Stir. Sip of beer. Back into the 'blue' can. Etc.

After seven shuffles .... I mean redistributions... each mixture would be within 1% of each other. If I hadn't finished my beer yet, I could even shuffle a few more times. After 14 repititions, each mixture would be within 0.1%.

19. Jun 22, 2004

### The Bob

Sorry I missed that point but I will continue to try. I have been close though.

The Bob

P.S. I am going to claim that it is impossible (just to clear this up).

20. Jun 22, 2004

### ceptimus

Here's another, that gives a 3:1 Yellow:Blue ratio (Njorl's gave 3:2)

1. Transfer one pint of blue paint to the glass. Throw it away.
2. Use the empty glass to transfer three pints of yellow paint to the blue tin.
3. Mix well - the second tin now contains 6 pints of a 1:1 yellow/blue mix.
4. Transfer one pint of the mixture to the glass. Throw it away.
5. Transfer a second pint of the mix to the glass. Keep it safe, for step 8.
6. The first tin now contains 5 pints of yellow, and the second tin 4 pints of 1:1
7. Pour yellow paint into the mixture till the tin is full. The mixture is now 3:1
8. Add the pint of 1:1 mix to the remaining pint of yellow in the first tin.

The first tin contains one-and-a-half pints of yellow and half-a-pint of blue.
The second tin contains six pints of yellow and two pints of blue.
Both tins together now contain ten pints of green paint with the same 3:1 yellow/blue ratio.

I don't know if there is an elegant way of seeing the solution. I made up this puzzle, so I suppose it's easier to make a puzzle than to solve it afterwards. :)

21. Jun 23, 2004

### NateTG

Neat puzzle

Here's a family of solutions that gives a range of ratios including the two that were presented:

1. Measure 1 pint of yellow paint, and put it into the blue tin.
2. Measure 2 pints of yellow paint from the yellow paint, and pour some amount (anywhere from 0 to 2 pints) into the blue paint, and dicard the balance.
3. Fill the pint glass with yellow paint and set it aside.
You should now have a tin with four pints of yellow paint, a tin with 5 (or more) pints of green(1), and a 1 pint glass of yellow paint.
4. Fill the yellow tin from the green tin.
Since the tin holds 8 pints, there will be equal parts of green(1) and yellow mixed to make green(2).
5. Paint the walls using the first 8 pints of green 2.
6. Decant the pint of yellow paint into the now empty tin.
7. Add one pint of green(1).
This will give 2 more pints of green(2) which is sufficient to finish painting the room.

This solution can produce controlled ratios of 2:3, 1:2 and 1:3, and uncontrolled ratios of any intermediate value.

22. Jun 23, 2004

### NateTG

Admittedly this only occured to me in retrospect, but I think the key is in being able to perform 2 proportional mixings:
4p in one tin + 4 p from the other
and
1p from the glass + 1 p from the other

Working backwards, this means that it's possible to create 10p from any situation where there is one tin with exactly 5p, and one tin with at least 5p of paint. (Exactly 5p in each tin, if mixing in the middle of painting is not allowed.)

Recognizing that, it's possible to get a ratio of 1 blue to 4 yellow (the theoretical minimum blue content):
Pour off 2p of blue paint.
Move 3p of yellow paint into the blue paint container creating green(1)
Put 1p of yellow paint into the glass
Fill the yellow paint tin with green(1) paint mixing to create green(2).
Paint using the first 8 pints.
Mix the remaining pint of pure yellow with the remaining pint in the blue paint tin to get two more pints of green(2).

23. Jun 23, 2004

### Gokul43201

Staff Emeritus
Now here's a solution outside of Nate's family

- Throw out 2Y(2 pints of yellow) from YT(yellow tin). YT now has 6Y.
- Pour 1Y into BT. BT now has 4B,1Y & YT has 5Y
- Take out 1 from BT and save. This has 4/5B,1/5Y & the BT has 16/5B,4/5Y.
- Pour 4 from YT to fill BT. So BT now has 16/5B, 24/5Y(B:Y=2:3) and YT has 1Y
- Pour the 1 (4/5B, 1/5Y) from the pint glass into the YT. YT now has 4/5B, 6/5Y (B:Y=2:3)

I'm almost certain this is what I had found before.

24. Jun 25, 2004

### vikasj007

well, one way of trying this is-
first of all throw away 2 pints of yellow paint using the 1 pint glass(for those who didn't get it, use it 2 times to remove a pint of yellow paint at a time)
So we have 4 pints of blue paint and 6 pints of yellow paint.
Now what we do is pour some yellow paint in the blue paint.after this we pour the paint from the second tin(which originally had blue paint)into the first tin, this process is repeated for some time and after that all the paint will have the same composition as liquids have a tendancy to spread evenly.

we can also throw away 2 pints of blue paint instead of yellow paint, if somebody wants it that way.

well this was the only solution i could come up with till now.

25. Jun 25, 2004

### Gokul43201

Staff Emeritus
For a finite number of intermixings, the color fractions in the two tins will always be different from each other. Only in the limiting case of an infinitely large number of intermixings, will they get the same color.