The problem involves calculating hire purchase details for a computer. The cash price is N$10500. A 10% deposit is required, followed by monthly installments of 0.5% of the remaining amount plus N$250. We need to calculate: (a) The deposit amount. (b) The monthly installment (at least for the first month). (c) The duration of the installments if the total hire purchase price is N$10849.63. (d) The percentage difference between the hire purchase price and the cash price.

1. Problem Description

The problem involves calculating hire purchase details for a computer. The cash price is N$

1

0

5

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0. A 10% deposit is required, followed by monthly installments of 0.5% of the remaining amount plus N$

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0. We need to calculate:

(a) The deposit amount.

(b) The monthly installment (at least for the first month).

(c) The duration of the installments if the total hire purchase price is N$10849.

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3. (d) The percentage difference between the hire purchase price and the cash price.

2. Solution Steps

(a) Calculating the deposit:

The deposit is 10% of the cash price, which is N$

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0. $Deposit = \frac{10}{100} \times 10500$

$Deposit = 0.10 \times 10500$

$Deposit = 1050$

(b) Calculating the monthly installment (first month):

After the deposit, the remaining amount is:

$Remaining = 10500 - 1050 = 9450$

The first monthly installment is 0.5% of the remaining amount plus N$

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0. $Installment = (0.005 \times 9450) + 250$

$Installment = 47.25 + 250$

$Installment = 297.25$

(c) Calculating the duration of installments:

Total hire purchase price = N$10849.63

Deposit = N$1050

Total installments = N$10849.63 - N$1050 = N$9799.63

Let $n$ be the number of months. The installment is $0.005 \times Remaining + 250$ each month. This implies solving a recurrence relation, which is beyond the scope of a simple calculation. Let's approximate the solution by assuming the monthly installment to be constant, which is not strictly correct but it is a reasonable first approximation.

The average installment paid:

$\approx$ $\frac{9799.63}{n} \approx 297.25$ from (b).

So $n \approx \frac{9799.63}{297.25} \approx 32.96 \approx 33$ months.

Let's see what happens if we simplify things and take the *average* remaining principal across all months:

We know the total repayment is $10849.

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3. The interest paid is 10849.63 - 10500 = 349.

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3. The deposit is

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0. Thus the sum of the installments is $10849.63-1050 = 9799.

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3. Let $R$ be the remaining amount (principal) each month.

Then the monthly payment is $0.005 R + 250$.

So the total amount paid on installments is the sum of the monthly payment:

$\sum_1^n (0.005 R_i + 250) = 9799.63$ where $R_i$ is the remaining balance at month $i$.

$0.005 \sum_1^n R_i + 250n = 9799.63$.

Let the average value of $R_i$ be $R_{avg} = \frac{1}{n} \sum R_i$.

Thus $\sum R_i = n R_{avg}$.

$0.005 n R_{avg} + 250n = 9799.63$.

Assume the reduction is linear:

$R_{avg} \approx \frac{9450+0}{2} = 4725$.

$0.005 n (4725) + 250n = 9799.63$.

$23.625 n + 250n = 9799.63$.

$273.625 n = 9799.63$.

$n = \frac{9799.63}{273.625} \approx 35.81$ months.

$35$ months is $2$ years and $11$ months.

$0.81$ months $\times 4$ weeks per month is about $3$ weeks.

Therefore, it is approximately 2 years, 11 months, and 3 weeks.

(d) Calculating the percentage difference:

Difference = Total paid - Cash price = N$10849.63 - N$10500 = N$349.63

Percentage difference = $\frac{Difference}{Cash \ price} \times 100$

Percentage difference = $\frac{349.63}{10500} \times 100$

Percentage difference = $0.033298 \times 100$

Percentage difference = $3.33$%

3. Final Answer

(a) The deposit: N$1050.00

(b) First monthly installment: N$297.25

(c) Duration of installments: 2 years, 11 months, 3 weeks

(d) Percentage difference: 3.33%