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In $\;2002,\, 2003, \, \text{and}\; 2004\;$ the total income of Jerry was $\$36,400$.

His income increased by $15\%$ each year. What was his income in $2004$?

Any hints or solution will be welcome.

Thanks in advance.

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I don't quite understand. Based on your previous questions, you should at least be at college level which means this kind of question should be breezy for you. Don't be offended. I'm just curious. –  Evariste Jun 2 '13 at 16:56
    
You are right but I'm curious to get more convenient way. :) –  Complex Guy Jun 2 '13 at 16:58
    
I was wondering what you meant by a more convenient way? I thought nothing would be easier than solving a linear equation of $x$. How do you reckon? –  Evariste Jun 2 '13 at 18:42
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3 Answers

up vote 2 down vote accepted

$ \begin{align} I_0 & : \quad \text{income earned in}\;2002.\tag{1} \\ \\ I_1 & = 1.15 I_0:\quad\text{income earned in}\; 2003.\tag{2} \\ \\ I_2 & = 1.15 I_1 = 1.15^2 I_0:\quad \text{income earned in} \; 2004.\tag{3} \end{align} $


$$\text{Income over $3$ years}:\;\;I_0 + I_1 + I_2 = \$36{,}400$$ $$ \iff I_0 + 1.15 I_0 + 1.15^2 = I_0\underbrace{(1 + 1.15 + 1.15^2)}_{\text{sum}\; =\; 3.4725} = 36400 $$

Solve for $I_0$, the income earned in $2002$, and then compute $I_2$ (income earned in $2004),\,$ using your computed solution for $I_0$ and the relation given by $(3)$ above.

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Even shows a generalization, which is nice! +1 –  Amzoti Jun 2 '13 at 20:49
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So let $x$ be his income in 2002. In 2003, his income was $x(1+0.15)$, in 2004, his income is $x(1+0.15)^2$. Therefore $x+x(1+0.15)+x(1+0.15)^2=36400$. Solve for x. Then the income in 2004 is just $x(1+0.15)^2$, where $x$ is what you just solved.

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Let $I$ be the income earned in 2004 (what you want to know). Express everything in terms of this. e.g. Income in $2003$ is $\cfrac {I}{1.15}=\cfrac {20I}{23}$. That should reduce you to a simple equation for $I$.

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