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If a quantity is successively increased by 20% for two times and another quantity is successively decreased by 10% for three times, the first quantity is 80% of the second. What percentage is the original first quantity as compared to the original second quantity?

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Why do you want to know? What have you tried? –  Mariano Suárez-Alvarez Oct 15 '10 at 15:48
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I'm guessing it's homework and have tagged it as such. –  Arturo Magidin Oct 15 '10 at 16:00
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Well, exams can be considered to be homework without great loss of generality. @Tretwick: What have you tried? Don't you have any idea? –  Mariano Suárez-Alvarez Oct 15 '10 at 16:13
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By the way, Tretwick, the title of your question is in no way related to the content of the question. Maybe you could edit into something more helpful? –  Mariano Suárez-Alvarez Oct 15 '10 at 16:38
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@T: All of them, namely none. I do not have any objections to answering homework problems; but I prefer not to answer homework questions if the OP does not show even a hint of having thought about the problem. –  Mariano Suárez-Alvarez Oct 15 '10 at 19:55
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1 Answer

Say your quantity is $Q$. To get 80% of that, you multiply by $0.8$, so $0.8Q$. How do you get 80% of that? Well, you multiply by $0.8$ again, so you get $0.8(0.8Q) = (0.8)^2Q$.

How do you take 80% successively three times? Four times? Five times? $n$ times?

If you first take $k$% ($k$ between $0$ and $100$), then you multiply by $\frac{k}{100}$. So if you started with $Q$, you now have $\frac{k}{100}Q$. If you then take $\ell$% (again, $\ell$ between $0$ and $100$), then you need to multiply that by $\frac{\ell}{100}$. So you would have $\left(\frac{\ell}{100}\right)\left(\frac{k}{100}\right)Q$. Etc.

So, say that instead of your problem, I wanted to first decrease by 10%, then the result by 20%, then the result of that by 45%, and finally the result of that by 15%. How would I do it? I would get $(.85)(.55)(.8)(.9)Q$.

If you are increasing by 20%, then instead of multiplying by 0.8, you multiply by $1.2$ (since adding 20% means taking 120% of the original). Same idea.

Now, you have two quantities, $Q_1$ and $Q_2$; you are doing different things to them. At the end you will have some multiple of $Q_1$ and some multiple of $Q_2$, say $aQ_1$ and $bQ_2$. But you are told that "the first quantity is 80% of the second". That means that you can write $aQ_1$ as a multiple of $bQ_2$, from which you can "solve for $Q_1$" in terms of $Q_2$. That should tell you exactly what you want to know.

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