Ratio of Anti-Freeze to WaterDate: 02/05/2003 at 10:53:37 From: Shelly Subject: Ratio word problem A barrel is filled with 45 gallons of anti-freeze. 9 gallons are removed and the barrel is refilled with water. 9 gallons of the mixture are removed and again, the barrel is refilled with water. What is the final ratio of anti-freeze to water in the barrel? Date: 02/06/2003 at 16:24:49 From: Doctor Ian Subject: Re: Ratio word problem Hi Shelly, The problem with using numbers is that it's tempting to simplify things too soon. Also, it's easy to get frustrated when the numbers don't work out nicely. So let's avoid those problems and say that the original amount of antifreeze is A, and at each step we remove some fraction (1/p) of the mixture, to be replaced by water. In this particular case, the numbers are A=45 and p=5 (because 9 gallons is 1/5 of 45 gallons), but let's forget about that for the moment. It will be easier to understand what follows if you have a solid grip on the fact that when you remove 1/p of something, the amount that remains is ((p-1)/p). For example, if you remove 1/10 of something, you're left with 9/10 of it. Okay? Then let's go. You start out with Antifreeze Water ---------- ----- A 0 (Start) We take out 1/p of the 'mixture', Antifreeze Water ---------- ----- A 0 ((p-1)/p)A 0 (Remove 1/p of mixture) and replace it with water: Antifreeze Water ---------- ------------- A 0 ((p-1)/p)A 0 ((p-1)/p)A (1/p)A (Replace with water) Again, we remove 1/p of the mixture: Antifreeze Water ----------------- --------------- A 0 ((p-1)/p)A 0 ((p-1)/p)A (1/p)A ((p-1)/p)((p-1)/p)A ((p-1)/p)(1/p)A (Remove) And replace it with water: Antifreeze Water ----------------- ------------------------ A 0 ((p-1)/p)A 0 ((p-1)/p)A (1/p)A ((p-1)/p)((p-1)/p)A ((p-1)/p)(1/p)A ((p-1)/p)((p-1)/p)A ((p-1)/p)(1/p)A + (1/p)A (Replace) At this point, you're probably thinking: "Hey, I thought he said it would be simpler!" But now we start reaping the benefits of our patience. The ratio of antifreeze to water is ((p-1)/p)((p-1)/p)A ------------------------ ((p-1)/p)(1/p)A + (1/p)A The first thing we can do with this is combine the terms in the denominator, using the distributive property, ac + bc = (a+b)c to get ((p-1)/p)((p-1)/p) * A --------------------------- [(p-1)/p)(1/p) + (1/p)] * A which means that we can just get rid of the A's altogether. In other words, the answer doesn't depend on how much antifreeze we started with. So now we have ((p-1)/p)((p-1)/p) --------------------- (p-1)/p)(1/p) + (1/p) (p-1)^2 / p^2 = -------------------------- (p-1) / p^2 + p / p^2 Using the distributive property again, we have (p-1)^2 / p^2 = ----------------- (p - 1 + p) / p^2 (p-1)^2 = ----------- (p - 1 + p) (p-1)^2 = ------- (2p - 1) We said that p was 5, so this comes out to (5-1)^2 = ---------- (2(5) - 1) = 16/9 Which is kind of cool, that everything cancels out that way, isn't it? Now, I have to admit, if you just had to solve this particular problem one time, and you knew you'd never see one like it again, you'd be better off doing the same steps, but using 45 and 1/5 and 4/5 instead of A and 1/p and ((p-1)/p). But the nice thing about solving it this way is that if I want to change the numbers around - say, I want to start with 36 gallons, and replace 3 gallons each time, or start with 42 gallons, and replace 7 gallons each time - I can get the answer in about two seconds. And now I'm kind of curious to try a couple of generalizations: (1) What if the fraction is q/p, instead of 1/p? (2) What if I continue for k steps, instead of just two? Will it continue to work out nicely? Who knows? That's what makes it interesting. Anyway, I hope this helps. Write back if you'd like to talk more about this, or anything else. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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