Example 6 of 7 · Finance: savings with interest
AI HL · Mathematics
Lina invests CHF 5000 for 10 years at 3% per year compounded annually. The rate is rounded to the nearest whole percent. Find the maximum percentage error in her estimate of the final balance.
Smallest true value · 2.5%
| TI-Nspire · Finance Solver | |
|---|---|
| N | 10 |
| I(%) | 2.5 |
| PV | −5000 |
| Pmt | 0 |
| FV | 6400.42 |
| PpY | 1 |
| CpY | 1 |
\[\begin{aligned}T_2&=6400.42\\[6pt]\class{mjx-pe-s6}{\%\text{ error}}&\class{mjx-pe-s6}{{}=\left|\frac{6719.58-6400.42}{6400.42}\right|\times100}\\[6pt]&\class{mjx-pe-s6}{{}=\class{mjx-pe-key}{\boldsymbol{4.99\%}}}\end{aligned}\]
Approximate value · 3%
| TI-Nspire · Finance Solver | |
|---|---|
| N | 10 |
| I(%) | 3 |
| PV | −5000 |
| Pmt | 0 |
| FV | 6719.58 |
| PpY | 1 |
| CpY | 1 |
\[\begin{aligned}A&=6719.58\\[6pt]\class{mjx-pe-s5}{\%\text{ error}}&\class{mjx-pe-s5}{{}=\left|\frac{A-T}{T}\right|\times100}\end{aligned}\]
Largest true value · 3.5%
| TI-Nspire · Finance Solver | |
|---|---|
| N | 10 |
| I(%) | 3.5 |
| PV | −5000 |
| Pmt | 0 |
| FV | 7052.99 |
| PpY | 1 |
| CpY | 1 |
\[\begin{aligned}T_1&=7052.99\\[6pt]\class{mjx-pe-s7}{\%\text{ error}}&\class{mjx-pe-s7}{{}=\left|\frac{6719.58-7052.99}{7052.99}\right|\times100}\\[6pt]&\class{mjx-pe-s7}{{}=\class{mjx-pe-key}{\boldsymbol{4.73\%}}}\end{aligned}\]
Maximum percentage error = 4.99%. Over 10 years the rounding in the rate compounds, so the error is far bigger than in the loan.