You have two beakers, one that contains water and another that contains an equal

Question

You have two beakers, one that contains water and another that contains
an equal amount of oil. A certain amount of water is transferred to the oil
and thoroughly mixed. Immediately, the same amount of the mixture is
transferred back to the water. Is there now more water in the oil or is there
more oil in the water?

(a) Before any computations are done, use common sense to guess the
solution to this problem.

(b) Choose numbers for the above problem and solve the problem with
these numbers. Use this computation to test your guess|and possibly
make a new guess.

(c) Use algebra to solve the problem.

Explain your reasoning in each step above.

Explanation / Answer

Let's see: (A) You should answer this based on your own opinion. My guess was since we transfer pure water to the oil and mixture back (not pure oil), I would say it is either even mixtures or more water in the oil. (B) To keep it simple, take 1/2 the water and put it in the oil, mix well. Now this beaker is 2/3 oil and 1/3 water. Now take the same amount, which would now be 1/3 of the mixture. That 1/3 mixture has the same volume as the initial half of 1/2 and is 1/3 water and 2/3 oil.. So we left behind 2/3 of 1/2 of water which is 2/6 or 1/3 of the original volume. And the original oil beaker is 2/3 oil and 1/3 water. When we add to the water beaker the return mix, we end up having 1/2 o water, and 1/2 mixed bringing again 1/3 of 1/2 or 1/6 water, and 2/3 of 1/2 or 1/3 oil. So the final mix is the sqame on both sides (reversing water and oil) 2/3 of the original liquid and 1/3 of the other. Is that because we took 1/2 of the water to start ? (C) The volume of liquid in each beaker is V. Let us say beaker 1 has water and beaker 2 has oil. We take a portion of the V of water in beaker 1, say a percentage W and transfer it into the oil beaker. Now beaker 1 contains V.(1-W) of water, and beaker 2 mixed volumes V of oil and V.W of water. We mix well and take out the same volume, but since the beaker now contains more liquid: V+VW or V.(1+W), we take equal volume but a smaller percentage. W% of V is [W/(1+W)]% of the new volume V.(1+W). The "return" mix will contain water and oil totalling in volume W.V and in the following ratios: Water ratio will be W / (1+W) Oil ratio will be 1 / (1+W) (c.2) We are leaving behind in beaker 2 a total volume V divided as follows: Water quantity of (W.V)/(1+W), and Oil quantity of (V)/(1+W) (c.3) Now we pour the return mixture, total volume VW, into beaker 1, which contains V.(1-W) water and to which we add the volume VW of mixture as detaled above in (c.1). So beaker 1 will end up containing: Oil brought in mixed (W.V).1/(1+W) So beaker 1 contains WV/(1+W) of oil, which is the same volume of water left behind in beaker 2 (see (c.2)). (c.4) we can verify also the other quantities. We left in beaker 2 an oil quantity (see (c.1)) of (V)/(1+W). Does this match the quantity of water in beaker 1 ? Beaker 1 water is: Water which was there, volume (1-W).V and aater brought back (W.V).W/(1+W) --- total final water in b.1: T = (1-W).V + (W.V).W/(1+W) T = (1-W).V.(1+W) / (1+W) + (W.V).W/(1+W) T = (1-W).(V+V.W) / (1+W) + (W.V).W/(1+W) T = [ (1-W).(V+V.W) + (W.V).W ] /(1+W) T = [ (V+V.W-W.V-W.V.W)) + W.V.W ] /(1+W) Which simplifies V.W with -W.V and the -W.V.W with the last W.V.W an: T = V /(1+W) which maches the oil in beaker 2. So we can confirm the mixtures will be similar with oil and water roles reversed.

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