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Thomas invests X into Fund $1$ at the beginning of each year for $10$ years. Fund $1$ pays interest annually into Fund $2$. Fund $1$ earns $7$% annually while Fund $2$ earns $6$% annually. After $10$ years, Thomas has a total of $50000$. Calculate X.

Interest earned is given by $0.07 X$

Amount at the end of $10$ years

$50000=X+0.07X(\frac{1.06^{10}-1}{0.06})$

This gives $X=26 005.69$, when answer is $3416.80$

Tosh
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  • I got the same answer as you did. – barak manos May 17 '16 at 06:56
  • @barakmanos, thank you for trying. It could be that the answer given by the author of this question is wrong. I am using this manual http://faculty.atu.edu/mfinan/actuarieshall/mainf.pdf, you could have a look at problem 32.5 – Tosh May 17 '16 at 07:09
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    Oh, I think I see the problem now. Thomas invests this amount ($X$) at the beginning of each year (and not only once, at the beginning of the $10$-year period). – barak manos May 17 '16 at 07:15
  • This means the equation is now $X[10+\frac{0.07}{0.06}(\frac{1.06^{10}-1}{0.06}-10)]=50000.$ I get $X=3646.73$ – Tosh May 17 '16 at 07:30
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    I think you get those $6%$ only for $9$ years (since the first payment occurs at the second year). – barak manos May 17 '16 at 07:52
  • Glad to have helped! – barak manos May 17 '16 at 08:14
  • Btw, have you checked if your term (last comment) is right ? – callculus42 May 17 '16 at 08:54
  • @callculus, yes I checked it. Yours is also very interesting. There is a link in the second comment, do check it. You shall get your answer. – Tosh May 17 '16 at 09:06
  • I think this is the equation and w.a. gives a different result: http://www.wolframalpha.com/input/?i=X(11%2B0.07%2F0.06*((1.06%5E11-1)%2F(0.06)-11))%3D50000 $\texttt{You have to copy the adress}$ – callculus42 May 17 '16 at 09:20
  • I wrongly used the calculator. Yes, you are right, it gives different result. We should go yr way. – Tosh May 17 '16 at 09:31
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    You can go my way for a deeper understanding. But the "formula" of the linked document is right. You have input not the right values. This is the right interpretation: $X(10+0.07/0.06((1.06^{11}-1)/(0.06)-11))=50000$ and the calculation of w.a.: http://www.wolframalpha.com/input/?i=X(10%2B0.07%2F0.06((1.06%5E11-1)%2F(0.06)-11))%3D50000 – callculus42 May 17 '16 at 09:38
  • @callculus Thank you for clearing my doubts further. – Tosh May 17 '16 at 10:04

1 Answers1

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The equation is

$$10X+0.07\cdot X\cdot \sum_{k=1}^{10} \frac{1-1.06^k}{1-1.06}=50,000$$

You have already found out that $X$ is invested ten times.

The last period one interest payment is made due the last investment, but not compounded ($k=1$, the fraction is 1).

The period before another investment is made. The interest for the last period are $0.07X$ (not compounded). The interest of the interest of the period before are $0.07X\cdot 1.06$ ($k=1$). You can go on backward like this.

The equation can be simplified to

$$10X+\frac{0.07}{0.06}\cdot X\cdot (1.06\cdot \frac{1.06^{10}-1}{0.06}-10)=50,000$$

callculus42
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