Problem 51
Question
Some lending institutions calculate the monthly payment \(M\) on a loan of \(L\) dollars at an interest rate \(r\) (expressed as a decimal) by using the formula $$ M=\frac{L r k}{12(k-1)} $$ where \(k=[1+(r / 12)]^{12 t}\) and \(t\) is the number of years that the loan is in effect. Car loan An automobile dealer offers customers no-downpayment 3 -year loans at an interest rate of \(10 \%\). If a customer can afford to pay \(\$ 500\) per month, find the price of the most expensive car that can be purchased.
Step-by-Step Solution
Verified Answer
The maximum car price is about $15,516.78.
1Step 1: Identifying Variables and Given Values
We need to determine the maximum loan amount (or car price 'L') the customer can afford. We know the monthly payment, \(M = 500\), the interest rate, \(r = 0.10\) (since 10% is expressed as a decimal), and the loan term \(t = 3\) years.
2Step 2: Calculate k
The formula for \(k\) is \(k = [1 + (r / 12)]^{12t}\). Substituting the values: \(r = 0.10\), and \(t = 3\), the expression becomes:\[k = \left[1 + \frac{0.10}{12}\right]^{12 \times 3} = [1 + 0.0083]^{36} \approx 1.34885.\]
3Step 3: Substitute Values into the Monthly Payment Formula
We rearrange the formula to solve for \(L\):\[L = \frac{M \times 12 \times (k-1)}{r \times k}.\]Substituting \(M = 500\), \(r = 0.10\), and the calculated \(k \approx 1.34885\):\[L = \frac{500 \times 12 \times (1.34885 - 1)}{0.10 \times 1.34885}.\]
4Step 4: Perform the Calculation for L
Calculate the numerator:\[500 \times 12 \times (1.34885 - 1) = 500 \times 12 \times 0.34885 \approx 2093.1.\]Calculate the denominator:\[0.10 \times 1.34885 \approx 0.134885.\]Finally, divide the results to find \(L\):\[L \approx \frac{2093.1}{0.134885} \approx 15516.78.\]
5Step 5: Conclusion
The maximum price of the car that the customer can afford is approximately $15,516.78.
Key Concepts
Interest RateMonthly PaymentLoan TermMathematical Formula
Interest Rate
The interest rate is a key component in loan payment calculations. It represents the cost of borrowing money, expressed as a percentage of the loan amount. For example, a 10% interest rate means the borrower must pay back an additional 10% of the original loan value over the loan term. When expressed as a decimal for calculation purposes, 10% becomes 0.10.
Understanding the interest rate allows you to determine the total cost of a loan over time. In calculations, this is vital for determining the monthly payment amount or the total repayment.
Understanding the interest rate allows you to determine the total cost of a loan over time. In calculations, this is vital for determining the monthly payment amount or the total repayment.
- Higher interest rates increase monthly payments and total loan costs.
- Lower interest rates make loans more affordable by reducing these payments.
Monthly Payment
The monthly payment is the amount the borrower is required to pay each month to repay the loan over the agreed-upon time frame. This amount includes both the principal and interest.
In our car loan example, the customer can make a monthly payment of $500. Knowing the maximum you can afford to pay monthly helps you determine the loan's total size and whether the loan fits your budget.
In our car loan example, the customer can make a monthly payment of $500. Knowing the maximum you can afford to pay monthly helps you determine the loan's total size and whether the loan fits your budget.
- Monthly payments are calculated using the loan amount, interest rate, and loan term.
- A higher monthly payment can reduce the total interest paid over the life of the loan.
Loan Term
The loan term is the period over which the borrower agrees to repay the loan. Loan terms are typically expressed in years.
In our exercise, the loan term is 3 years, meaning the total debt is to be paid off in 36 monthly installments. The choice of a loan term can significantly influence the monthly payment and total interest paid.
In our exercise, the loan term is 3 years, meaning the total debt is to be paid off in 36 monthly installments. The choice of a loan term can significantly influence the monthly payment and total interest paid.
- Longer loan terms generally result in smaller monthly payments but can increase cumulative interest costs.
- Shorter loan terms typically mean higher monthly payments but less total interest over the loan's lifespan.
Mathematical Formula
Calculating loan payments involves specific mathematical formulas. For instance, the formula to determine the monthly payment in our example incorporates variables like the loan amount, interest rate, and term.
The formula used is \(M=\frac{L r k}{12(k-1)}\) where \(k=[1+(r / 12)]^{12 t}\). This captures the compounding nature of interest, making it suitable for calculating monthly payments.
The formula used is \(M=\frac{L r k}{12(k-1)}\) where \(k=[1+(r / 12)]^{12 t}\). This captures the compounding nature of interest, making it suitable for calculating monthly payments.
- Understanding these formulas helps decipher how changes in parameters affect payments.
- Practice with examples enhances familiarity with these calculations.
Other exercises in this chapter
Problem 51
Use the compound interest formula to determine how long it will take for a sum of money to double if it is invested at a rate of \(6 \%\) per year compounded mo
View solution Problem 51
Exer. 51-52: Approximate \(x\) to three significant figures. (a) \(\log x=3.6274\) (b) \(\log x=0.9469\) (c) \(\log x=-1.6253\) (d) \(\ln x=2.3\) (e) \(\ln x=0.
View solution Problem 52
Solve the compound interest formula $$ A=P\left(1+\frac{r}{n}\right)^{n t} $$ for \(t\) by using natural logarithms.
View solution Problem 52
Exer. 51-52: Approximate \(x\) to three significant figures. (a) \(\log x=1.8965\) (b) \(\log x=4.9680\) (c) \(\log x=-2.2118\) (d) \(\ln x=3.7\) (e) \(\ln x=0.
View solution