Example 7 of 7 · Finance: savings with inflation
AI HL · Mathematics
Inflation is 2% per year. Lina finds the real value of her CHF 5000 after 10 years using the real rate 3% − 2% = 1%. The 3% is rounded to the nearest whole percent. Find the maximum percentage error in her estimate.
Smallest true value · 0.5%
| TI-Nspire · Finance Solver | |
|---|---|
| N | 10 |
| I(%) | 2.5 − 2 = 0.5 |
| PV | −5000 |
| Pmt | 0 |
| FV | 5255.70 |
| PpY | 1 |
| CpY | 1 |
\[\begin{aligned}T_2&=5255.70\\[6pt]\class{mjx-pe-s6}{\%\text{ error}}&\class{mjx-pe-s6}{{}=\left|\frac{5523.11-5255.70}{5255.70}\right|\times100}\\[6pt]&\class{mjx-pe-s6}{{}=\class{mjx-pe-key}{\boldsymbol{5.09\%}}}\end{aligned}\]
Approximate value · 1%
| TI-Nspire · Finance Solver | |
|---|---|
| N | 10 |
| I(%) | 3 − 2 = 1 |
| PV | −5000 |
| Pmt | 0 |
| FV | 5523.11 |
| PpY | 1 |
| CpY | 1 |
\[\begin{aligned}A&=5523.11\\[6pt]\class{mjx-pe-s5}{\%\text{ error}}&\class{mjx-pe-s5}{{}=\left|\frac{A-T}{T}\right|\times100}\end{aligned}\]
Largest true value · 1.5%
| TI-Nspire · Finance Solver | |
|---|---|
| N | 10 |
| I(%) | 3.5 − 2 = 1.5 |
| PV | −5000 |
| Pmt | 0 |
| FV | 5802.70 |
| PpY | 1 |
| CpY | 1 |
\[\begin{aligned}T_1&=5802.70\\[6pt]\class{mjx-pe-s7}{\%\text{ error}}&\class{mjx-pe-s7}{{}=\left|\frac{5523.11-5802.70}{5802.70}\right|\times100}\\[6pt]&\class{mjx-pe-s7}{{}=\class{mjx-pe-key}{\boldsymbol{4.82\%}}}\end{aligned}\]
Maximum percentage error = 5.09%. Compare with Example 6: the same rounding gives a slightly bigger percentage error once inflation is taken off.