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Problem 6 on page 47 of Marcel B. Finan's A Basic Course in the Theory of Interest and Derivatives Markets: A Preparation for the Actuarial Exam FM/2 is:

Fund A is invested at an effective annual interest rate of 3%. Fund B is invested at an effective annual interest rate of 2.5%. At the end of 20 years, the total in the two funds is 10,000. At the end of 31 years, the amount in Fund A is twice the amount in Fund B. Calculate the total in the two funds at the end of 10 years.

This question appears after a section all about compound interest, so I'm assuming the funds yield compound interest. So, calling the principals $P_A,P_B$, the given facts are

  • $P_A1.03^{20}+P_B1.025^{20}=10000$ and
  • $P_A1.03^{31}=2P_B1.025^{31}$.

Recasting the first of those yields $P_B=(10000-P_A1.03^{20})/1.025^{20}$, and using that in the second equation yields $$P_A1.03^{31}=2\frac{10000-P_A1.03^{20}}{1.025^{20}}1.025^{31}=2(10000-P_A1.03^{20})1.025^{11}$$ i.e. $$P_A(1.03^{31}+2\cdot1.03^{20}1.025^{11})=20000\cdot1.025^{11}$$ i.e. $$P_A=\frac{20000\cdot1.025^{11}}{1.03^{31}+2\cdot1.03^{20}1.025^{11}}.$$ Using that in the recasting of the first equation yields $$P_B=\left(10000-\frac{20000\cdot1.025^{11}}{1.03^{31}+2\cdot1.03^{20}1.025^{11}}1.03^{20}\right)/1.025^{20}\\=\frac{10000\cdot1.03^{31}+20000\cdot1.03^{20}1.025^{11}-20000\cdot1.025^{11}1.03^{20}}{(1.03^{31}+2\cdot1.03^{20}1.025^{11})1.025^{20}}\\=\frac{10000\cdot1.03^{11}}{1.03^{11}1.025^{20}+2\cdot1.025^{31}}.$$ We wish to find $P_A1.03^{10}+P_B1.025^{10}$: using these last values for $P_A,P_B$, we have $$\frac{20000\cdot1.025^{11}}{1.03^{31}+2\cdot1.03^{20}1.025^{11}}1.03^{10}+\frac{10000\cdot1.03^{11}}{1.03^{11}1.025^{20}+2\cdot1.025^{31}}1.025^{10}$$ which a calculator tells me is $7569.073\ldots$. I don't doubt the answer, but find it hard to believe that I was expected to do all that. Is there a shorter way?

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1 Answer 1

I do not think that conceptually there is a shorter way. I would probably let $Q_A$ be the amount in fund A after $10$ years, and similarly for $Q_B$. Then we want $Q_A+Q_B$.

After $10$ years we have $10000$, so $$Q_A(1.03)^{10}+Q_B(1.025)^{10}=10000.\tag{$1$}$$

After $21$ years, fund A is twice as big as fund B, so $$Q_A(1.03)^{21}=2Q_B(1.025)^{21}.\tag{2}$$

These are your equations, simplified a bit.

Now instead of keeping everything in symbols, I think it is time to compute say $Q_B$. Using Equation $(2)$, we find that $Q_A(1.03)^{10}=\dfrac{2Q_B(1.025)^{21}}{(1.03)^{11}}$.

Shove this into the calculator. Now we can use Equation $`$ to find $Q_B$, and then $Q_A$.

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