Question

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?

Answer 1

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(1+r/(100n))^(nt)`

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

1.(a) Hence amount becomes `5000(1+4.5/1200)^84`

One can calculate it using a scientific calculator as this is

`5000(1+9/2400)^84=5000(1+3/800)^84`

= `5000xx1.00375^84=5000xx1.369452257=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(1+r/100)^10` or `(1+r/100)^10=14000/7000=2`

Hence taking log (base 10), `10log(1+r/100)=log2=0.301003`

or `log(1+r/100)=0.030103`

or `1+r/100=10^0.0301=1.07177`

hence `r/100=0.07177`

and `r=7.177%`