Wednesday, April 26, 2006

Water and Milk

You have two containers. One contains 1 gallon of milk and the other contains a gallon of water. If you were to take one cup of milk from the first one and pour it to the second one and after mixing, you were to pour one cup of the mixture from the second to the first one, which containers now have a pure concentration of their original contents?

Level: Medium

4 Comments:

Anonymous Vikram said...

Hi, nice of u to put up such problems.
Is the concentration the same in both the water and milk

11:49 AM, April 26, 2006  
Blogger Karthik Nagaraj said...

Vikram,
If you think the concentration of water and milk is the same even after the mixing then please explain how so?

12:06 PM, April 26, 2006  
Blogger Kala said...

Both of them are equally contaminated as they have been contaminated with the same amount of an external liquid. No matter how many many molecules of milk you put back into the container after it is mixed, you are still diluting it. You can put a mathematical formula to identify how much percentage of milk and water are left. However, to answer your question, none of them have a pure concentration of their original contents !

12:53 PM, April 26, 2006  
Blogger Karthik Nagaraj said...

Both Kala and Vikram are hitting hard on the right spot. When I say which one is pure - I mean which one has a greater %age of itself in the mixture after all the mix and match is done. Basically which one is more purer? Oops! missed that word in the typing. Anyway! To answer the Q, yes Kala is right, both are equally contaminated.

So the answer is "If the first container had 100% of milk, now it has some x% of milk in the mixture AND the second that had 100% of water, now has some x% of water. So both are equally pure and equally impure"

Excellent! - Lets do it the mathematical way to be sure we are correct. Lets use 1000 cc instead of gallon. Its easier. It doesn't matter:

[1] Start off with 1000 cc of milk and 1000 cc of water
[2] Let the cup measure 50 cc. Transfer 50 cc of milk to the container with water. So now %age of milk in the 1st is still 100% and the %age of water in the 2nd is 1000/1050 = 95.238%
[3] Transfer 50 cc of this mixture to the 1st. Now in that 50 cc - 95.238% is water, so that amounts to 47.619 cc of water and 2.381 cc of milk
[4] Now when transferred to the 1stm the %age of milk = (950 + 2.381)/1000 = 95.238% and the %age of water in the 1st = (1000 - 47.619)/1000 = 95.238%

SOLVED!

Good job guys and great thinking btw, both Vikram and Kala!

1:12 PM, April 26, 2006  

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