You have a container containing 1 liter of pure blue liquid dye. The container is cunningly designed in such a way that up to one extra liter of liquid may be added, but only the excess over 1 liter may ever be poured out. (If the container were filled to 1.5 liters only .5 liter could be poured out.)
If you have only one liter of pure water, what is the minimum concentration you could dilute the dye in the container to?
Assume the dye is completely soluble and mixes instantaneously.
Considering the initial concentration of the blue dye being t, let us glance at a simplified formula, which is very useful in the everyday of Pharmacy:
c1 x v1 = c2 x v2 (given pressure and temperature constants valid for both situations)
c1 being the concentration before the addition, v1 the initial volume of the container, c2 the concentration of the blue dye after the addition and v2 the resulting volume, using this formula we can come up with a few hints.
For instance, let us add half a liter of pure water and dispose of half a liter of the mix to see where it leads:
c1 x v1 = c2 x v2
t x 1 = c2 x 1.5, giving c2 = t/1.5 as the new concentration
Adding the remaining half liter of water, it's not hard to see we would come up with c2 = t/(1.5)^2 (or c2 = t/2.25). So, from this we take that the original concentration of the blue dye (t) is divided by (one liter plus an increment of pure water in volume) elevated to the nth power, where n equals the number of fractions the liter of pure water is divided into.
Thus, considering we can measure small quantities of water (such as one mililiter), we can come up with a formula such as this: c2 = t/(1.01)^100, which gives a final concentration of aproximately c2 = t/2.7048. This may not be the final answer, but I believe this is one of the paths to reach it.
Edited on June 8, 2006, 10:35 am
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