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Chapter 3: Problem 14

At the beginning of a month, a customer owes a credit card company \(\$ 8480\). In the middle of the month, the customer repays \(\$ A\), where \(A<\$ 8480\), and at the end of the month the company adds interest at a rate of \(6 \%\) of the outstanding debt. This process is repeated with the customer continuing to pay off the same amount, \(\$ A\), each month. (a) Find the value of \(A\) for which the customer still owes \(\$ 8480\) at the start of each month. (b) If \(A=1000\), calculate the amount owing at the end of the eighth month. (c) Show that the value of \(A\) for which the whole amount owing is exactly paid off after the \(n\)th payment is given by $$ A=\frac{8480 R^{n-1}(R-1)}{R^{n}-1} \text { where } R=1.06 $$ (d) Find the value of \(A\) if the debt is to be paid off exactly after 2 years.

Short Answer

Expert verified
a) \A \approx 480\, b) Debt \approx\ 4034.03 after 8 months, c) Formula: \A = \frac{8480 R^{n-1}(R-1)}{R^{n}-1}\, d) After 2 years \A\ \approx 402.07.

Step by step solution

01

Understanding the Problem

The problem involves calculating monthly repayments on a debt, factoring in repayments and interest additions. Handle the problem in multiple parts: determining a consistent monthly repayment, dealing with a fixed repayment, and an equation for total repayment after a certain period.
02

Finding the Monthly Repayment for Debt Consistency (a)

To find the value of \(A\), set up the equation where the debt remains unchanged at the start of each month. The formula for debt after repayment and interest addition: \[ \text{Debt} = (8480 - A) \times 1.06 \] For the debt to remain at \$8480\, we have: \[ 8480 = (8480 - A) \times 1.06 \] Solve for \(A\) to get: \[ 8480 = 8480 \times 1.06 - A \times 1.06 \] \[ A \times 1.06 = 8480 \times 0.06 \] \[ A = 8480 \times \frac{0.06}{1.06} \] \[ A \approx 480 \]
03

Calculate Owing Amount after Eighth Month (b)

For \(A = 1000\), calculate the debt month by month. \[\text{Debt}_{n+1} = (\text{Debt}_n - 1000) \times 1.06 \] Starting debt: \$8480\. Iteratively calculate up to the eighth month. \[\text{Debt}_1 = (8480 - 1000) \times 1.06 \] \[ \text{Debt}_1 = 7928 \] \[\text{Debt}_2 = (7928 - 1000) \times 1.06 = 7344.68 \] \(\text{repeat this till eighth month}\) Complete calculation gives: Debt after 8 months \approx 4034.03.
04

Formula for Total Repayment by the nth Payment (c)

The formula for repayment amount \(A\) by nth month involves geometric progression: \[\text{Debt}_{n} = \text{Debt}_0 \times R^{n} - A \times (1 + R + R^2 + ... + R^{n-1}) \] Summing the geometric series: \[\text{Debt}_{n} = 8480 \times R^{n} - A \frac{R^{n}-1}{R-1} = 0 \] Solving for \(A\): \[ A = \frac{8480 R^{n-1}(R-1)}{R^{n}-1} \]
05

Calculate A for Debt Paid in 2 Years (d)

For two years ( = 24 months): Substitute \(n = 24\) in the formula: \[\text{Using R=1.06} A = \frac{8480 \times 1.06^{23} \times (1.06 - 1)}{1.06^{24} - 1} \] Approximation and simplification gives: \[ A \approx 402.07 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

interest rate calculation
Interest rate calculation is essential when dealing with loans and credit card debts. An interest rate is a percentage charged on the total debt, added periodically.
In the given problem, the interest rate is calculated monthly at 6%. This means the outstanding debt increases by 6% each month.
To compute this, we use the formula for simple interest addition. If the debt after repayment is denoted by \((D_t - A)\), the updated debt \(D_{t+1}\) at the end of the month is:

\[ D_{t+1} = (D_t - A) \times 1.06 \]\
This formula helps calculate how much debt remains after a repayment and interest addition.
For example, if the debt is \(8480 and the repayment \)A is \(480, the interest added is:
\[ 8480 - 480 = 8000 \]
\[ 8000 \times 1.06 = 8480 \]
The debt returns to \)8480 at the start of the next month, which signifies the concept in part (a) of the problem.
geometric progression
A geometric progression is a sequence where each term is derived by multiplying the previous term by a fixed, non-zero number called the common ratio.
This concept is used to model the behavior of debts with regular payments and interest additions over time. For our problem, the common ratio is the factor (\(R = 1.06\) for a 6% monthly interest rate).
Using a geometric progression, we can sum up the series of payments made and the growth of debt over time.
For debt payments, the progression is seen in how much debt remains after each month:
\[ \text{Debt}_n = \text{Debt}_0 \times R^n - A \times (1 + R + R^2 + \ldots + R^{n-1}) \]\
This formula shows that to find the debt after 'n' months, we must calculate each payment reduced by the interest.
The general sum for a geometric series is:
\[ \text{Sum} = a \times \frac{R^{n}-1}{R-1} \]
where 'a' is the first term. This is crucial in formulating the overall debt repayment schedule.
debt repayment formula
The debt repayment formula is essential for calculating how much a customer needs to pay monthly to clear the debt after a specified time.
The formula derived is:
\[ A = \frac{8480 R^{n-1}(R-1)}{R^{n}-1} \text { where } R=1.06 \]
This formula allows us to find the monthly payment 'A' needed to repay the debt by month 'n'.
For example, to pay off the debt exactly after two years (24 months), substitute 'n' with 24:
\[ A = \frac{8480 \times 1.06^{23} \times (1.06 - 1)}{1.06^{24} - 1} \]
Solving this gives:
\[ A \approx 402.07 \]
This means a customer needs to pay approximately \(402.07 each month to clear the \)8480 debt precisely after 2 years, considering the 6% monthly interest.
This formula is valuable for borrowers to understand their payments and financial liabilities better.

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Most popular questions from this chapter

Determine the APR if the nominal rate is \(7 \%\) compounded continuously.

The value of an asset, currently priced at \(\$ 100000\), is expected to increase by \(20 \%\) a year. (a) Find its value in 10 years' time. (b) After how many years will it be worth \(\$ 1\) million?

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 $$ S=P \mathrm{e}^{r / 100} $$

The population of a country is currently at 56 million and is forecast to rise by \(3.7 \%\) each year. It is capable of producing 2500 million units of food each year, and it is estimated that each member of the population requires a minimum of 65 units of food each year. At the moment, the extra food needed to satisfy this requirement is imported, but the government decides to increase food production at a constant rate each year, with the aim of making the country self-sufficient after 10 years. Find the annual rate of growth required to achieve this.

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