Suggested time to complete: 20 minutes

1. A laptop is advertised for $\scriptsize \text{R}469$ per month over a period of three years on hire purchase. If the rate of simple interest charged is $\scriptsize 22.95\%$, calculate the original cash price of the laptop. Calculate the amount saved by buying for cash rather than on hire purchase.
2. After a period of $\scriptsize 20$ years, an investment of $\scriptsize \text{R}12\text{ 750}$ has a total value of $\scriptsize \text{R}23\text{ 205}$. Calculate the rate of simple interest that was paid on this investment.
3. A new fridge is advertised for $\scriptsize \text{R}262$ per month over a period of two years. If the simple interest charged is $\scriptsize 45.3\%$ per annum, calculate the cash price of the fridge.
4. Calculate the time it will take $\scriptsize \text{R}8\text{ 000}$ to amount to $\scriptsize \text{R}8\text{ 960}$ at $\scriptsize 4\%$ simple interest per annum.

The full solutions are at the end of the unit.

Exercise 1.1

1. .
   $\scriptsize P=\text{R}3\text{ 750}$
   $\scriptsize i=3.2\%$
   $\scriptsize n=4$
   $\scriptsize A=?$
   .
   $\scriptsize \begin{align*}A&=P(1+i.n)\\&=3750\left[ {1+\displaystyle \frac{{3.2}}{{100}}(4)} \right]\\&=3750[1+(0.032)(4)]\\&=3750[1+0.128]\\&=3750[1.128]\\&=4230\end{align*}$
   The total value of the investment after the period of four years is $\scriptsize \text{R}4\text{ 230}$.
2. .
   $\scriptsize P=30\text{ }000$
   $\scriptsize n=2$
   $\scriptsize A=30\text{ }000+9\text{ 000=39 }000$
   $\scriptsize i=?$
   .
   $\scriptsize \begin{align*}A&=P(1+i.n)\\ \displaystyle \frac{A}{P}&=1+i.n\\ \displaystyle \frac{A}{P}-1&=i.n\\ \displaystyle \frac{1}{n}\left( {\displaystyle \frac{A}{P}-1} \right)&=i\\ i&=\displaystyle \frac{1}{2}\left( {\displaystyle \frac{{39000}}{{30000}}-1} \right)\\ &=\displaystyle \frac{1}{2}(1.3-1)\\ &=\displaystyle \frac{1}{2}(0.3)\\ &=0.15\text{ or } 15\% \text{ } \end{align*}$
   The rate of simple interest charged was $\scriptsize 15\%$ .
3. .
   $\scriptsize \begin{align*}n&=3\text{ years = 36 months}\\i&=35\%\\A&=299\times 36=10764\end{align*}$
   .
   $\scriptsize \begin{align*}A&=P(1+i.n)\\\displaystyle \frac{A}{{(1+i.n)}}&=P\\P&=\displaystyle \frac{{10\text{ }764}}{{1+\left( {\displaystyle \frac{{35}}{{100}}} \right)(3)}}\\&=\displaystyle \frac{{10\text{ }764}}{{1+(0.35)(3)}}\\&=\displaystyle \frac{{10\text{ }764}}{{1+1.05}}\\&=\displaystyle \frac{{10\text{ }764}}{{2.05}}\\&=5\text{ }250.732\end{align*}$
   The original cash price of the laptop is $\scriptsize \text{R}5\text{ 250}\text{.73}$.
   Buying for cash rather than hire purchase would save $\scriptsize 10\text{ }764-5\text{ }250.73=\text{R}5\text{ }513.27$.
4. .
   $\scriptsize \begin{align*}P&=5\text{ }000\\A&=1\text{ }050+5\text{ }000=6\text{ }050\\I&=3.5\%=\displaystyle \frac{{3.5}}{{100}}=0.035\end{align*}$
   .
   $\scriptsize \begin{align*}A&=P(1+i.n)\\\displaystyle \frac{A}{P}&=1+i.n\\\displaystyle \frac{A}{P}-1&=i.n\\\displaystyle \frac{1}{i}\left( {\displaystyle \frac{A}{P}-1} \right)&=n\\n&=\displaystyle \frac{1}{{0.035}}\left( {\displaystyle \frac{{6\text{ }050}}{{5\text{ }000}}-1} \right)\\&=\displaystyle \frac{1}{{0.035}}(1.21-1)\\7&=\displaystyle \frac{1}{{0.035}}(0.21)\\&=6\end{align*}$
   It would take six years.

Back to Exercise 1.1

Unit 1: Assessment

1. .
   $\scriptsize \begin{align*}n&=3\text{ years}=36\text{ months}\\A&=469\times \text{36=16 884}\\i&=22.95\%\\P&=?\end{align*}$
   .
   $\scriptsize \begin{align*}A&=P(1+i.n)\\\displaystyle \frac{A}{{1+i.n}}&=P\\P&=\displaystyle \frac{{16\text{ }884}}{{1+\left( {\displaystyle \frac{{22.95}}{{100}}} \right)(3)}}\\&=\displaystyle \frac{{16\text{ }884}}{{1+(0.2295)(3)}}\\&=\displaystyle \frac{{16\text{ }884}}{{1+0.6885}}\\&=\displaystyle \frac{{16\text{ }884}}{{1.6885}}\\&=9\text{ }999.408\end{align*}$
   The cash price of the laptop is $\scriptsize \text{R}9\ 999.41$.
   The amount saved by buying for cash rather than on hire purchase would be $\scriptsize 16884-9999.41=R6884.59$
2. .
   $\scriptsize \begin{align*}A&=23\text{ }205\\P&=12\text{ }750\\n&=20\\i&=?\end{align*}$
   .
   $\scriptsize \begin{align*}A&=P(1+i.n)\\\displaystyle \frac{A}{P}&=1+i.n\\\displaystyle \frac{A}{P}-1&=i.n\\\displaystyle \frac{1}{n}\left( {\displaystyle \frac{A}{P}-1} \right)&=i\\i&=\displaystyle \frac{1}{{20}}\left( {\displaystyle \frac{{23\text{ }205}}{{12\text{ }750}}-1} \right)\\&=\displaystyle \frac{1}{{20}}(1.82-1)\\&=\displaystyle \frac{1}{{20}}(0.82)\\&=0.041\text{ or 4}\text{.1 } \% \text{ p}\text{.a}\text{.}\end{align*}$
   The rate of simple interest paid was $\scriptsize 4.1\%$ per annum.
3. .
   $\scriptsize \begin{align*}i&=45.3\%\\n&=2\text{ years = 24 months}\\A&=262\times 24=6\ 288\end{align*}$
   .
   $\scriptsize \begin{align*}A&=P(1+i.n)\\\displaystyle \frac{A}{{(1+i.n)}}&=P\\P&=\displaystyle \frac{{6\text{ }288}}{{1+\left( {\displaystyle \frac{{45.3}}{{100}}} \right)(2)}}\\&=\displaystyle \frac{{6\text{ }288}}{{1+(0.453)(2)}}\\&=\displaystyle \frac{{6\text{ }288}}{{1+0.906}}\\&=\displaystyle \frac{{6\text{ }288}}{{1.906}}\\&=3\text{ }299.056\end{align*}$
   The cash price of the fridge is $\scriptsize \text{R}3\text{ }299.06$
4. .
   $\scriptsize \begin{align*}P&=8\text{ }000\\A&=8\text{ }960\\i&=4\%=\displaystyle \frac{4}{{100}}=0.04\end{align*}$
   .
   $\scriptsize \begin{align*}A&=P(1+i.n)\\\displaystyle \frac{A}{P}&=1+i.n\\\displaystyle \frac{A}{P}-1&=i.n\\\displaystyle \frac{1}{i}\left( {\displaystyle \frac{A}{P}-1} \right)&=n\\n&=\displaystyle \frac{1}{{0.04}}\left( {\displaystyle \frac{{8\text{ }960}}{{8\text{ }000}}-1} \right)\\&=\displaystyle \frac{1}{{0.04}}(1.12-1)\\&=\displaystyle \frac{1}{{0.04}}(0.12)\\&=3\end{align*}$
   It will take three years.

Back to Unit 1: Assessment