Loan Interest Rate Equalization

Charles took out a loan of EUR 600 from Ivan 20.2 at a 20% annual rate. From January 10.5, a loan of 1100 Euros at a 14% rate. From Milo 15.9, a loan of 900 Euros at an 11% rate. All three return 30.12. Determine the annual interest rate for all three loans to pay the same total interest.

Final Answer:

p =  15.7162 %

Step-by-step explanation:

$$\begin{gathered} d_1=30.12.-20.2. \\ {d}_1=314 \\ {d}_2=30.12.-10.5. \\ {d}_2=235 \\ {d}_3=30.12.-15.9. \\ {d}_3=107 \\ {k}_1=\sqrt[365]{1+20/100}\doteq 1.0005 \\ {u}_1=600\cdot k_1^{d_1}-600=600\cdot 1.0005^{314}-600\doteq 101.8896\text{Eur} \\ {k}_2=\sqrt[365]{1+14/100}\doteq 1.0004 \\ {u}_2=1100\cdot k_2^{d_2}-1100=1100\cdot 1.0004^{235}-1100\doteq 96.8234\text{Eur} \\ {k}_3=\sqrt[365]{1+11/100}\doteq 1.0003 \\ {u}_3=900\cdot k_3^{d_3}-900=900\cdot 1.0003^{107}-900\doteq 27.9594\text{Eur} \\ u=u_1+u_2+u_3=101.8896+96.8234+27.9594\doteq 226.6724\text{Eur} \\ 600\cdot k^{d_1}-600+1100\cdot k^{d_2}-1100+900\cdot k^{d_3}-900=u \\ 600\cdot k^{314}-600+1100\cdot k^{235}-1100+900\cdot k^{107}-900=226.67 \\ k=1.0004 \\ k=(1+p/100{)}^{1/365} \\ p=100\cdot (k^{365}-1)=100\cdot (1.0004^{365}-1)=15.7162\% \end{gathered}$$

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