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?