Mix This
Bottle A has Water
Bottle B Has Milk
Take a spoon full of Milk from Bottle B and Mix with Bottle A
Take the same spoon full of Liquid from Bottle A, and mix it with Bottle B
Now tell Me does Bottle A have More Water or Bottle B Has more Milk
Assume both the Bottles have same dimensions
and Strictly Speaking Pure Milk does not have any water.
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2 solutions
Here is one more try, Call it a solution if you feel.
Can we exchange the idea of bottle and spoon and call it residual R and transfer T component ( so Bottle A = Bottle B = R) and SPOON going from B to A and SPOON going Back from A to B as T, now applying boundary conditions T = 0 (no transfer) till T = R ( full Transfer) , Note this process will always occur within these limits,
we obviously know for both the cases, final state will have same amount of parent liquid , in First case 100% parent liquid in each bottle ( no changes) and the last case 50% parent liquid each ( since we are mixing equal and total amount of milk and water) , hence for all the in between cases also we can deduce each bottle will have the same parent liquid.
I have a doubt??? that in first case when we mix milk in water, it will become diluted so when we take another spoon full of liquid from A it would have less quantity of milk which we initially mix in it. So answer should be ' Bottle A has more Water'.
Nicely done, Siddhartha. I focussed on tracking each step of the process, while you focussed on just the beginning and end states, resulting in a more elegant solution than mine. :)
nice one siddhartha, although I had done it based on induction rather than algebra or logic, i took 100 ml on both sides and spoon of 1 ml, 10ml 50 ml and 99 ml did the maths and induced it :) seems like a lot of hard work now
Suppose we start out with x spoonfuls in each bottle.
After transferring one spoonful from B to A there are x − 1 spoonfuls of milk in B and x + 1 spoonfuls of liquid in A , x of which are water and 1 of which is milk. Once the liquids in A are mixed thoroughly, the fraction of liquid in (every portion of) A that is water is x + 1 x , and the fraction that is milk is x + 1 1 .
When a spoonful of the mixed liquid in A is transferred to B , the fraction of liquid in A that is water remains x + 1 x .
Now since the spoonful that is transferred back to B has x + 1 1 spoonfuls of milk in it, the fraction of liquid in B that is milk will now be
( x − 1 ) + 1 x − 1 + x + 1 1 = ( x + 1 ) ∗ x x 2 = x + 1 x .
Thus once the process is complete and both bottles once again each contain x spoonfuls of liquid, bottle A will have the same amount of water in it as bottle B has milk.
Brilliant Answer Brian
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Thanks. I'm interested to see if other solvers have different solution methods to share.
Here is my solution , although i marked the wrong option by mistake
let us assume that there are 10 parts of each milk and water now let us transfer 1 part of milk now in bottle a there are 10 parts of water and 1 milk and in b there are 9 parts of milk now if we again transfer only 1/11 parts of milk are returning to bottle B so now it has 9+1/11 = 100/11 parts of milk whereas the other bottle has 10-10/11 parts of water=100/11 parts of water so hence both bottles have same amount of parent liquid .
Suppose there are L liters in each bottle. After the doing the required steps, we see that both bottles still contain L liters of liquid. Let there be x liters of water remaining in bottle A. Therefore there is L − x liters of milk in bottle A.(Since Bottle A contains L liters of liquid.)
Now in bottle B, There is L − x liters of water and L − ( L − x ) = x liters of milk, ( Since originally there was L liters of both liquids. ). Hence the amount of water in Bottle A is equal to the amount of Milk in bottle B