# 1(a). How much will €5000 will become, if they are invested for 7 years at 4.5% per annum compounded every month? (b). If an amount €7000, invested for 10 years compounded every year doubles, at what rate was it invested?

Sep 23, 2017

1.(a) $6847.26$ euros and 1.(b) r=7.177%

#### Explanation:

Let $P$ be the amount invested at annual interest rate of r% for $t$ number of years compounded $n$ times a year - note that compounded every year means $n = 1$; compounded every half-year means $n = 2$; compounded every quarter means $n = 4$ and compounded every month means $n = 12$.

then the amount after $t$ years becomes $P {\left(1 + \frac{r}{100 n}\right)}^{n t}$

We have $P = 5000$ euros, r=4.5%, $t = 7$ years and $n = 12$

1.(a) Hence amount becomes $5000 {\left(1 + \frac{4.5}{1200}\right)}^{84}$

One can calculate it using a scientific calculator as this is

$5000 {\left(1 + \frac{9}{2400}\right)}^{84} = 5000 {\left(1 + \frac{3}{800}\right)}^{84}$

= $5000 \times {1.00375}^{84} = 5000 \times 1.369452257 = 6847.261285$

or $6847.26$ euros.

1.(b) Here we have $P = 7000$ euros, $r$ is not known, $t = 10$ years and $n = 1$ and amount becomes $14000$, as it doubles and hence

$14000 = 7000 {\left(1 + \frac{r}{100}\right)}^{10}$ or ${\left(1 + \frac{r}{100}\right)}^{10} = \frac{14000}{7000} = 2$

Hence taking log (base 10), $10 \log \left(1 + \frac{r}{100}\right) = \log 2 = 0.301003$

or $\log \left(1 + \frac{r}{100}\right) = 0.030103$

or $1 + \frac{r}{100} = {10}^{0.0301} = 1.07177$

hence $\frac{r}{100} = 0.07177$

and r=7.177%