(6.9) Doris buys a hi - fi sets for R13 700 . She pays $$ R 2200 $$ in cash and the balance is paid by a hire purchase loan agreement with an interest rate of $$ 15 \% $$ p a. The loan is repaid over three years. Calculate the following: 6.9.1 The monthly instalment 6.9.2 The total amount of money that Doris paid for the hi-fi set. 6.9.3 How much Doris would have saved had she paid in cash?

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To solve this problem, we will break it down into the required parts step by step.

### Known Conditions:
- Cost of the hi-fi set: $$ C = 13700 $$
- Cash payment: $$ P = 2200 $$
- Loan amount: $$ L = C - P = 13700 - 2200 = 11500 $$
- Interest rate: $$ r = 15\% $$ per annum
- Loan term: $$ t = 3 $$ years
- Monthly interest rate: $$ r_m = \frac{r}{12} = \frac{15\%}{12} = 1.25\% = 0.0125 $$
- Total number of payments: $$ n = t \times 12 = 3 \times 12 = 36 $$

### 6.9.1 Calculate the Monthly Instalment
The formula for the monthly instalment $$ M $$ for a loan is given by:

$$
M = L \cdot \frac{r_m(1 + r_m)^n}{(1 + r_m)^n - 1}
$$

Now, we will calculate $$ M $$.

### 6.9.2 Calculate the Total Amount Paid
The total amount paid $$ T $$ is the sum of the cash payment and the total of the monthly instalments:

$$
T = P + (M \cdot n)
$$

### 6.9.3 Calculate the Savings
The savings $$ S $$ if paid in cash is the difference between the total amount paid and the cost of the hi-fi set:

$$
S = T - C
$$

Now, let's perform the calculations for $$ M $$, $$ T $$, and $$ S $$.

First, we will calculate $$ M $$.

$$
M = 11500 \cdot \frac{0.0125(1 + 0.0125)^{36}}{(1 + 0.0125)^{36} - 1}
$$

Let's calculate this value.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{11500\left(0.0125\left(1+0.0125\right)^{36}\right)}{\left(\left(1+0.0125\right)^{36}-1\right)}$$
- step1: Remove the parentheses:
  $$\frac{11500\times 0.0125\left(1+0.0125\right)^{36}}{\left(1+0.0125\right)^{36}-1}$$
- step2: Add the numbers:
  $$\frac{11500\times 0.0125\times 1.0125^{36}}{\left(1+0.0125\right)^{36}-1}$$
- step3: Add the numbers:
  $$\frac{11500\times 0.0125\times 1.0125^{36}}{1.0125^{36}-1}$$
- step4: Convert the expressions:
  $$\frac{11500\times 0.0125\left(\frac{81}{80}\right)^{36}}{1.0125^{36}-1}$$
- step5: Convert the expressions:
  $$\frac{11500\times 0.0125\left(\frac{81}{80}\right)^{36}}{\left(\frac{81}{80}\right)^{36}-1}$$
- step6: Multiply:
  $$\frac{\frac{23\times 81^{36}}{2^{146}\times 5^{34}}}{\left(\frac{81}{80}\right)^{36}-1}$$
- step7: Subtract the numbers:
  $$\frac{\frac{23\times 81^{36}}{2^{146}\times 5^{34}}}{\frac{81^{36}-80^{36}}{80^{36}}}$$
- step8: Multiply by the reciprocal:
  $$\frac{23\times 81^{36}}{2^{146}\times 5^{34}}\times \frac{80^{36}}{81^{36}-80^{36}}$$
- step9: Rewrite the expression:
  $$\frac{23\times 81^{36}}{2^{146}\times 5^{34}}\times \frac{16^{36}\times 5^{36}}{81^{36}-80^{36}}$$
- step10: Rewrite the expression:
  $$\frac{23\times 81^{36}}{2^{146}\times 5^{34}}\times \frac{2^{144}\times 5^{36}}{81^{36}-80^{36}}$$
- step11: Reduce the numbers:
  $$\frac{23\times 81^{36}}{2^{2}}\times \frac{5^{2}}{81^{36}-80^{36}}$$
- step12: Multiply the fractions:
  $$\frac{23\times 81^{36}\times 5^{2}}{2^{2}\left(81^{36}-80^{36}\right)}$$
- step13: Multiply:
  $$\frac{575\times 81^{36}}{4\times 81^{36}-4\times 80^{36}}$$
Calculate or simplify the expression $$ 2200 + (398.651278 * 36) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$2200+\left(398.651278\times 36\right)$$
- step1: Multiply the numbers:
  $$2200+14351.446008$$
- step2: Add the numbers:
  $$16551.446008$$
Calculate or simplify the expression $$ 2200 + (398.651278 * 36) - 13700 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$2200+\left(398.651278\times 36\right)-13700$$
- step1: Multiply the numbers:
  $$2200+14351.446008-13700$$
- step2: Calculate:
  $$2851.446008$$
### Calculations:

1. **Monthly Instalment $$ M $$**:
   $$
   M \approx 398.65
   $$

2. **Total Amount Paid $$ T $$**:
   $$
   T = 2200 + (398.65 \times 36) \approx 16551.45
   $$

3. **Savings $$ S $$**:
   $$
   S = T - C = 16551.45 - 13700 \approx 2851.45
   $$

### Summary of Results:
- **6.9.1** The monthly instalment is approximately $$ R 398.65 $$.
- **6.9.2** The total amount of money that Doris paid for the hi-fi set is approximately $$ R 16551.45 $$.
- **6.9.3** Doris would have saved approximately $$ R 2851.45 $$ had she paid in cash.
- **6.9.1** Monthly instalment: $$ R 398.65 $$ - **6.9.2** Total amount paid: $$ R 16,551.45 $$ - **6.9.3** Savings if paid in cash: $$ R 2,851.45 $$

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