/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 Maximum loan amount. Desmond pla... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Maximum loan amount. Desmond plans to purchase a new car. He qualifies for a loan at an annual interest rate of \(5.8 \%,\) compounded monthly for 6 yr. He is willing to pay up to \(\$ 300\) per month. What is the largest loan he can afford?

Short Answer

Expert verified
The largest loan Desmond can afford is approximately \(\$ 18,015.57\).

Step by step solution

01

Understanding the Formula

To determine the maximum loan amount Desmond can afford, we will use the formula for the monthly payment of an amortizing loan, which is calculated as \[ M = P \frac{r(1+r)^n}{(1+r)^n -1} \]where \( M \) is the monthly payment, \( P \) is the principal or loan amount, \( r \) is the monthly interest rate, and \( n \) is the total number of payments.
02

Calculating Monthly Interest Rate

The annual interest rate is \(5.8\%\). We convert this to a monthly rate by dividing by 12 (the number of months in a year): \[ r = \frac{0.058}{12} = 0.0048333 \]
03

Calculating Total Number of Payments

The loan term is 6 years. Since the payments are monthly, we calculate the total number of payments as:\[ n = 6 \times 12 = 72 \]
04

Substituting Known Values

Now using the known values, set \( M = 300 \) (Desmond's maximum monthly payment), \( r = 0.0048333 \), and \( n = 72 \) in the formula:\[ 300 = P \frac{0.0048333(1+0.0048333)^{72}}{(1+0.0048333)^{72} -1} \]
05

Solving for Loan Amount \( P \)

First, compute the numerator and the denominator separately:- Compute \( (1+0.0048333)^{72} \).- Compute the entire expression for \( P \):\[ P = \frac{300 \times ((1+0.0048333)^{72} -1)}{0.0048333(1+0.0048333)^{72}} \]Calculate the expression using a calculator to determine \( P \).
06

Final Calculation

After performing the calculations, we find the largest loan Desmond can afford to be approximately \( P = \$ 18,015.57 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Monthly Payment Formula
To find out how much Desmond can borrow for his car, we use a special formula. It's called the monthly payment formula for amortizing loans. This formula helps us find out what the monthly payments would be if you know the amount of the loan, the interest rate, and how long you will be paying the loan back.

The formula is: \[ M = P \frac{r(1+r)^n}{(1+r)^n -1} \]
  • \( M \) is the monthly payment. In Desmond's case, he wants this to be \( \\(300 \) or less.
  • \( P \) is the principal, or the total loan amount we're solving for.
  • \( r \) is the monthly interest rate, calculated from the annual rate.
  • \( n \) is the total number of monthly payments Desmond will make.
This formula lets us plug in different numbers to see what monthly payment we can handle, or in Desmond's situation, what loan principal he can afford with a \( \\)300 \) payment.
Loan Interest Rate
The loan interest rate is a percentage that affects how much you end up paying for a loan. For Desmond, the interest rate on his car loan is set at \( 5.8\% \) per year. To make our calculations easier, we convert this annual interest rate to a monthly rate.

This is done by dividing the annual interest rate by 12, as there are 12 months in a year. So, for Desmond:\[ r = \frac{0.058}{12} = 0.0048333 \]
  • This monthly rate \( 0.0048333 \) is what we use in the formula to find the maximum loan amount.
  • A higher interest rate means more cost in paying back the loan.
  • Adjusting this rate can significantly affect how much you can borrow.
Keeping an eye on the interest rate helps you understand the overall cost of a loan.
Loan Term Calculation
The loan term refers to how long you will be paying off the loan. Desmond's car loan term is planned to be 6 years. When we calculate for monthly loan payments, we need to change years into months. This gives us the total number of monthly payments that need to be made.

Desmond's loan term calculation goes as follows:\[ n = 6 \times 12 = 72 \]
  • So, \( n = 72 \) represents the total number of payments Desmond will make.
  • A longer loan term means smaller monthly payments, but usually more interest paid over time.
  • Conversely, a shorter term involves larger monthly payments but often less interest overall.
By knowing your loan term, you can make smart decisions about how much you can afford to borrow and repay.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A monk weighing 170 lb begins a fast to protest a war. His weight after \(t\) days is given by $$W=170 e^{-0.008 t}$$ a) When the war ends 20 days later, how much does the monk weigh? b) At what rate is the monk losing weight after 20 days (before any food is consumed)?

The tortoise population, \(P(t),\) in a square mile of the Mojave Desert after \(t\) years can be approximated by the logistic equation $$ P(t)=\frac{3000}{20+130 e^{-0.214 t}} $$ a) Find the tortoise population after \(0 \mathrm{yr}, 5 \mathrm{yr}, 15 \mathrm{yr},\) and 25 yr. b) Find the rate of change in the population, \(P^{\prime}(t)\). c) Find the rate of change in the population after \(0 \mathrm{yr}\) \(5 \mathrm{yr}, 15, \mathrm{yr},\) and \(25 \mathrm{yr}\) d) What is the limiting value (see Exercise 42 ) for the population of tortoises in a square mile of the Mojave Desert?

The Rule of 70. The relationship between doubling time \(T\) and growth rate \(k\) is the basis of the Rule of \(70 .\) Since $$ T=\frac{\ln 2}{k}=\frac{0.693147}{k}=\frac{69.3147}{100 k} \approx \frac{70}{100 k} $$ we can estimate the length of time needed for a quantity to double by dividing the growth rate \(k\) (expressed as a percentage) into \(70 .\) Estimate the time needed for an amount of money to double, if the interest rate is \(7 \%,\) compounded.

A student made the following error on a test: \(\frac{d}{d x} e^{x}=x e^{x-1}\) Identify the error and explain how to correct it.

The demand, \(D(x)\), and supply, \(S(x)\), functions for a multipurpose printer are as follows: $$D(x)=q=480 e^{-0.003 x}$$ and $$S(x)=q=150 e^{0.004 x}$$ a) Find the equilibrium point. Assume that \(x\) is the price in dollars, b) Find the elasticity of demand when \(x=\$ 100\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.