Chapter 3: Problem 1
Calculate (a) \(10 \%\) of \(\$ 2.90\) (b) \(75 \%\) of \(\$ 1250\) (c) \(24 \%\) of \(\$ 580\)
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Chapter 3: Problem 1
Calculate (a) \(10 \%\) of \(\$ 2.90\) (b) \(75 \%\) of \(\$ 1250\) (c) \(24 \%\) of \(\$ 580\)
These are the key concepts you need to understand to accurately answer the question.
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Find the value of the geometric series $$ 1000+1000(1.03)+1000(1.03)^{2}+\ldots+1000(1.03)^{9} $$
A firm decides to invest in a new piece of machinery which is expected to produce an additional revenue of \(\$ 8000\) at the end of every year for 10 years. At the end of this period the firm plans to sell the machinery for scrap, for which it expects to receive \(\$ 5000\). What is the maximum amount that the firm should pay for the machine if it is not to suffer a net loss as a result of this investment? You may assume that the discount rate is \(6 \%\) compounded annually.
(Excel) The sum of \(\$ 100\) is invested at \(12 \%\) interest for 20 years. Tabulate the value of the investment at the end of each year, if the interest is compounded (a) annually (b) quarterly (c) monthly (d) continuously Draw graphs of these values on the same diagram. Comment briefly on any similarities and differences between these graphs.
A person requests an immediate bank overdraft of \(\$ 2000\). The bank generously agrees to this, but insists that it should be repaid by 12 monthly instalments and charges \(1 \%\) interest every month on the outstanding debt. Determine the monthly repayment.
If a principal, \(P\), is invested at \(r \%\) interest compounded annually then its future value, \(S\), after \(n\) years is given by $$ S=P\left(1+\frac{r}{100}\right)^{n} $$ (a) Use this formula to show that if an interest rate of \(r \%\) is compounded \(k\) times a year then after \(t\) years $$ S=P\left(1+\frac{r}{100 k}\right)^{k t} $$ (b) Show that if \(m=100 k / r\) then the formula in part (1) can be written as $$ S=P\left(\left(1+\frac{1}{m}\right)^{m}\right)^{r t / 100} $$ (c) Use the definition $$ \mathrm{e}=\lim _{m \rightarrow \infty}\left(1+\frac{1}{m}\right)^{m} $$ to deduce that if the interest is compounded with ever-increasing frequency (that is, continuously) then $$ S=P \mathrm{e}^{r / 100} $$
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