Be careful. It's careless
There are two containers named
A
and
B
. Container
A
has milk in it, container
B
has water (same amount as of milk in container
A
) in it. Some amount of milk is transferred from container
A
to container
B
and then , the same amount is transferred from the container
B
to the container
A
.
Let container A has x amount of water in it and container B has y amount of milk in it.
Which one option is correct?
Note : Finally the amount in both the containers is same as it was initially.
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2 solutions
Discussions for this problem are now closed
May I ask something? Is the viscosity of the liquids or let's just say that, are the characteristics of the liquids to be considered in this problem because I consider it in solving this and I get the answer of X < Y even though I know already at the first glance that the answer would be X = Y but I rather chose X < Y. Just curious to know what is the answer to my question.. Please if someone knows, just answer my question...
I think it's assumed that the liquids involved are miscible, and that they are thoroughly mixed after each transfer. If the liquids were, say, oil and water, then there would be issues with immiscibility, but that's not the case here. :)
Ahhh,,, Ok I understand but still I have doubts on it. Maybe, I could just advice you to put a note or assumptions when making questions like this because some solvers have different level and fields of thinking and they may apply something to it which is not a case to your problem just like me but you've got nice problems though...! Looking forward for another problems.
You are still taking to the assumption that both liquids are perfectly mixed and that a perfect percentage of the mixture is transferred which would Never be the case, unless the liquids are the same type. Say you have 1l of water, and transfer .25l of mil to It. You would have to assume that the mixture is perfect and that the .25l transferred after mixing is 80%water 20%milk, or .2l water and .05l milk.by forcing that assumpution, the statement that x=y is true at .2l each. But as it is an assumption you must take, it is an equally valid assumption to say the milk is heavier and thus the solution combination is not equal as dependant on where you draw your solution from, would give you different percentage combinations. Drawing from the top would be more watery, drawing from the bottom would be more milky. The only way to valid this is to consider you have to take two assumptions one way and only one the other. Its ultimately a very badly worded problem.
@Michael Ryan Curths – No; no matter how well you mix, no matter how you try to take the second one (try to take pure water or pure milk or some combination of the two), the result is guaranteed to hold. Try proving it.
What if the initial amounts in each container weren't equal, but the amount of each transfer was equal?
The same result still holds. Say A has N liters of milk initially and B has M liters of water. After the two transfers are complete, A has N liters of liquid and B has M liters of liquid, but now the compositions have changed.
If A now has x liters of water in it, then it must have N − x liters of milk, while the remaining x liters of milk stays in container B . The volume of B never comes into play in this particular calculation.
Things get more interesting when we look at a different calculation. With the two containers starting with the same volume as in this question, if we were to keep transferring back and forth, (mixing thoroughly after each transfer), ad infinitum, we would end up with both containers holding mixtures that are half water and half milk. In general, if they started with N liters of milk in A and M liters of water in B , then after the infinite transfer scenario both containers would hold fluids that are a fraction N + M N milk and N + M M water. So there would be N ∗ N + M M = N + M N M liters of water in A and M ∗ N + M N = N + M N M liters of milk in B , i.e., the same.
I guess this is like the Zeroth Law of Thermodynamics, only with mixing. :)
I invite you to check my problem "More Coffee or Tea #2"
@Guiseppi Butel – I checked out that problem and found that I had answered it before. :)
I also checked out your "Pushing the boundaries: Codicil" question. I find that x and y are the solutions to the set of equations
(i) x 2 + ( y − 2 ) 2 = ( x + 2 ) 2 and
(ii) x 2 + ( y + 1 ) 2 = ( x + 3 ) 2 ,
which give two real solutions, the one in the first quadrant being
x = 2 0 + 1 2 3 , y = 8 + 4 3 .
When I add these coordinates together I get x + y = 2 8 + 1 6 3 = 5 5 . 7 1 2 8 1 3 . . . . , which is 5 5 . 7 1 3 rounded to 3 decimal places, but this was considered incorrect. I've triple-checked and can't find a mistake in my reasoning, so I'm not sure why we're getting different answers. :(
Here is the graph, which appears to have all the required features.
@Brian Charlesworth – I may have erred. Must check over my scribbled notes.
@Guiseppi Butel – I powered it through Excel spreadsheet and I concur with your answer, Brian.
@Brian Charlesworth – What am I missing? When I solve (i) and (ii) I get x - 3y = -4. How did you proceed from there?
@Guiseppi Butel – So after subtracting (i) from (ii) and simplifying we get x = 3 y − 4 . Now substitute this back into (i) to find that
( 3 y − 4 ) 2 + ( y − 2 ) 2 = ( 3 y − 2 ) 2 ,
which after expanding and simplifying yields
y 2 − 1 6 y + 1 6 = 0 ,
which has solutions y = 2 1 6 ± 1 6 2 − 4 ∗ 1 6 = 8 ± 4 3 .
Now y = 8 − 4 3 implies that x = 3 y − 4 = 2 0 − 1 2 3 , which is less than 0 so can be discarded. That leaves us with
y = 8 + 4 3 and thus x = 2 0 + 1 2 3 .
@Brian Charlesworth – Thanks, Brian. I have started proceedings to have the answer corrected.
@Guiseppi Butel – Great. I'm presently enjoying the challenge of "The Shrinking Square". Well, at least I'm playing around with it in my mind for now; I'll get down to brass tacks later. :)
At Start Assume
A has milk = m
B has water =m
First Transfer from A to B is 'i' milk
Now
A has milk = m-i
B has water =m and milk=i
Then Next transfer from B to A is i=a(milk)+b(water)
Now
A has milk =m-i+a=m-a-b+a=m-b
and water =b
B has water =m-b
and milk=i-a=a+b-a=b
Let container A has x amount of water in it and container B has y amount of
milk in it.
so x=b and y=b
so x=y=b
Suppose we start with N liters of liquid in each bottle. After the two transfers are complete, there are still N liters of liquid in each bottle, although now as mixtures.
If A now has x liters of water in it, then it must have N − x liters of milk in it. Thus, since the total amount of milk must be preserved, there must be N − ( N − x ) = x liters of milk in bottle B , and so x = y .