Lesson Loan problems

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Loan problems


Problem 1

You want to buy a car. ​ The loan amount is ​ ​ $25,000.
The company is offering a ​ 5% ​ interest rate for ​ 48 ​ months ​ (4 ​ years).
What will your monthly payments be?

Solution 

Use the formula for the monthly payment for a loan

    M = P%2A%28r%2F%281-%281%2Br%29%5E%28-n%29%29%29


where P is the loan amount; r = 0.05%2F12 is the effective interest rate per month;
n is the number of payments (same as the number of months); M is the monthly payment.


In this problem  P = $25000;  r = 0.05%2F12.


Substitute these values into the formula and get for monthly payment

    M = 25000%2A%28%28%280.05%2F12%29%29%2F%281-%281%2B0.05%2F12%29%5E%28-48%29%29%29 = $575.73.


ANSWER.  The monthly payment is $575.73.


In total, you will pay  4*12*575.73 = 27,635.04 dollars in 4 years.


The difference $27,635.04 - $25,000 = $2,635.04 is the interest you pay to financial company.

Problem 2

You purchased a refrigerator at  Sears for  $2,500  (no sales tax added),
and you decide to pay for it in  4  years.  Sears' interest rate is  10%.
What is the monthly payment that you need to pay?

Solution 

Use the formula for the monthly payment for a loan

    M = P%2A%28r%2F%281-%281%2Br%29%5E%28-n%29%29%29


where P is the loan amount; r = 0.1%2F12 is the effective interest rate per month;
n is the number of payments (same as the number of months in 4 years); M is the monthly payment.


In this problem  P = $2500;  r = 0.1%2F12;  n = 4*12 = 48.


Substitute these values into the formula and get for monthly payment

    M = 2500%2A%28%28%280.1%2F12%29%29%2F%281-%281%2B0.1%2F12%29%5E%28-48%29%29%29 = $63.41  (rounded).


Thus, the monthly payment is $61.41.    ANSWER


In total, you will have to repay  4*12*61.41 = 2,947.68 dollars in 4 years.


The difference  $2,947.68 - $2,500 = $447.68 = $370,681.60  is the interest you pay for this loan.

Problem 3

Jessica borrowed  $5000  from the bank in order to buy a new piano.
She will pay it off by equal payments at the end of  2  years.
The interest rate is  8%  compounded weekly.
Determine the size of the payments and the total interest paid.

Solution 

Use the formula for the periodical payment for a loan

    M = P%2A%28r%2F%281-%281%2Br%29%5E%28-n%29%29%29


where P is the loan amount; 

r = 0.08%2F52 is the effective interest rate per period (per week in this problem);

n is the number of payments (total number of weeks in this problem); 

M is the payment for the period (weekly payment in this problem).



    In this problem  P = $5000;  r = 0.08%2F52;  n = 2*52 = 104.



Substitute these values into the formula and get for monthly payment

    M = 5000%2A%28%28%280.08%2F52%29%29%2F%281-%281%2B0.08%2F52%29%5E%28-104%29%29%29 = $52.07.


Thus, the weekly payment is $52.07.


In total, Jane will have to repay  2*52*52.07 = 5415.28 dollars in 2 years.


The difference $5,415.28 - $5,000 = $415.28  is the interest she will pay for this loan.

Problem 4

You can afford a  $300  per month car payment.  You've found a  5  year loan at  8%  interest.
What the maximum loan could be?

Solution 

Use the formula for the monthly payment for a loan

    M = P%2A%28r%2F%281-%281%2Br%29%5E%28-n%29%29%29


where P is the loan amount; r = 0.03%2F12 is the effective interest rate per month;
n is the number of payments (same as the number of months); M is the monthly payment.


From this formula, the expression for the maximum loan is

    P = M%2F%28%28r%2F%281-%281%2Br%29%5E%28-n%29%29%29%29.


In this problem  M = $300;  r = 0.08%2F12,  n = 5*12 = 60 monthly payments.


Substitute these values into the formula and get for the maximum loan amount

    P = 300%2F%28%28%280.08%2F12%29%2F%281-%281%2B0.08%2F12%29%5E%28-60%29%29%29%29 = $14,795.53.


ANSWER.  The maximum loan amount is $14,795.53.

Problem 5

A bank loaned  $15,000  total,  part of it at  8%  per year,  the rest at  18%   per year.
If the interest the bank received in one year totaled  $2000,  how much was loaned at  8%?

Solution 

Let x be the amount loaned at 8%; then the rest, (15000-x), is the amount loaned at 18%.


Write equation for the total annual interest

    0.08x + 0.18*(15000-x) = 2000.


Simplify and solve for x

    0.08x + 0.18*15000 - 0.18x = 2000

    0.08x - 0.18x = 2000 - 0.18*15000

      -0.1x       =   -700

          x       = -700/(-0.1) = 70000.


ANSWER.  $7000 was  loaned at 8%.

Problem 6

Yankee  Construction agreed to lease payments of  $762.79  on construction equipment to be made
at the end of each month for six years.  Financing is at  15%  compounded monthly.
What is the value of the original lease contract?

Solution 

In this problem, the value of the original lease contract is the amount of a loan, which requires 
monthly payments of $762.79 at the end of each month for six years at 15% compounded monthly.


Use the standard formula for monthly payments of a loan

    M = P%2A%28r%2F%281-%281%2Br%29%5E%28-n%29%29%29,    (1)


where P is the loan amount; r = 0.15%2F12 is the effective interest rate per month;
n is the number of payments (same as the number of months, n = 6*12 = 72);
M is the monthly payment.


In this problem  M = $762.79;  r = 0.15%2F12.


Substitute these values into the formula and get this equation

    762.79 = P%2A%28%28%280.15%2F12%29%29%2F%281-%281%2B0.15%2F12%29%5E%28-72%29%29%29.    (2)


In equation (2), calculate separately the coefficient (the multiplier, or the factor)

    %28%280.15%2F12%29%29%2F%281-%281%2B0.15%2F12%29%5E%28-72%29%29 = 0.021145013  (rounded).


Now from equation (2), find the ANSWER

    P = 762.79%2F0.021145013 = 36074.23.


ANSWER.  The value of the original lease contract is  $36074.23.

Problem 7

If a loan of  $300,000  is to be settled by  $3,000 monthly payments for  15  years,
what interest rate compounded monthly is charged on the loan?

Solution 

Write a loan equation

    M = %28L%2Ar%29%2F%281-%281%2Br%29%5E%28-n%29%29,


where 

M is the monthly payment;

L is the loaned amount;

r is the monthly effective rate as a decimal;

n is the number of payments (= the number of months).


In our case this equation takes the form

    3000 = %28300000%2Ar%29%2F%281-%281%2Br%29%5E%28-15%2A12%29%29,

or

    3000%2F300000 = r%2F%281-%281%2Br%29%5E%28-180%29%29.

    0.01 = r%2F%281-%281%2Br%29%5E%28-180%29%29.


In this equation, r is the unknown.


This equation is unsolvable algebraically, so use the numerical methods.

You may use any of numerous online calculators.

I used online calculator DESMOS of common use at web-site www.desmos.com/calculator


It gave me  r = 0.00729945, approximately.


    Here is the link to the DESMOS solution.  

    https://www.desmos.com/calculator/ey97hhvirs

    Click on the intersection point to get its coordinates.



This value  r = 0.00729945  is the monthly effective rate - so, the annual nominal rate is 12 times this value

    r%5Bannual%5D = 12*0.00729945 = 0.0875934,  or about 0.0876,

which corresponds to 8.76%.


ANSWER. In this problem, the monthly compounded interest rate is about 8.76 %.


To check my solution, I substituted this value r= 0.00729945 

into the loan function  f(r) = r%2F%281-%281%2Br%29%5E%28-180%29%29 = 0.00729945%2F%281-1.00729945%5E%28-180%29%29.


I got the value 0.0100000030, which is quite close to  0.01, so the solution is confirmed.

Problem 8

A  $38,000  loan bears interest at  10%  compounded semi-annually and is to be repaid
in semi-annual payments of  $2,000 each.
How many semi-annual payments must be the debtor make?

Solution 

Use the standard formula for the semi-annual payment for a loan

    P = L%2A%28r%2F%281-%281%2Br%29%5E%28-n%29%29%29%29,


where L is the loan amount; r = 0.1%2F2 is the effective semi-annual compounding interest rate;
n is the number of payments; P is the semi-annual payment.


In this problem  P = $2000;  r = 0.1%2F2 = 0.05.


Substitute these values into the formula and get for monthly payment

    2000 = 38000%2A%280.05%2F%281-1.05%5E%28-n%29%29%29%29.


In this equation, n is the unknown: we should find n from this equation.


Simplify step by step

    2000%2F38000 = %280.05%2F%281-1.05%5E%28-n%29%29%29,

    0.052631579 = %280.05%2F%281-1.05%5E%28-n%29%29%29,

    0.052631579%2F0.05 = %281%2F%281-1.05%5E%28-n%29%29%29,

    1.05263158 = %281%2F%281-1.05%5E%28-n%29%29%29,

    1%2F1.05263158 = 1-1.05%5E%28-n%29,

    0.95 = 1-1.05%5E%28-n%29,

    1.05%5E%28-n%29 = 1 - 0.95,

    1.05%5E%28-n%29 = 0.05,

    1%2F1.05%5En = 0.05,

    1.05^n = 1/0.05,

    1.05^n = 20,

    n*log(1.05) = log(20),

    n = log%28%2820%29%29%2Flog%28%281.05%29%29 = 61.4.


So, 61 full semi-annual payments should be made of 2,000 each,
and then the last,62-th payment, should be made of the lesser amount.


ANSWER.  61 full semi-annual payments should be made of 2,000 each,
         and then the last, 62-th payment, should be made of the lesser amount.

         The total number of semi-annual payments is 62.

Problem 9

How many quarterly payments of  $15,000  will be necessary to pay off a debt
of  $175,000  if the interest rate charged is  8%  compounded quarterly?

Solution 

Use the standard formula for the quarterly payment for a loan

    P = L%2A%28r%2F%281-%281%2Br%29%5E%28-n%29%29%29%29,


where L is the loan amount; r = 0.08%2F4 = 0.02 is the effective quarterly compounding interest rate;
n is the number of payments; P is the quarterly payment.


In this problem  P = $15000;  r = 0.08%2F4 = 0.02.


Substitute these values into the formula and get for quarterly payment

    15000 = 175000%2A%280.02%2F%281-1.02%5E%28-n%29%29%29%29.


In this equation, n is the unknown: we should find n from this equation.


Simplify step by step

    15000%2F175000 = %280.02%2F%281-1.02%5E%28-n%29%29%29,

    0.085714286 = %280.02%2F%281-1.02%5E%28-n%29%29%29,

    0.085714286%2F0.02 = %281%2F%281-1.02%5E%28-n%29%29%29,

    4.2857143 = %281%2F%281-1.02%5E%28-n%29%29%29,

    1%2F4.2857143 = 1-1.02%5E%28-n%29,

    0.233333333 = 1-1.02%5E%28-n%29,

    1.02%5E%28-n%29 = 1 - 0.233333333,

    1.02%5E%28-n%29 = 0.766666667,

    1%2F1.02%5En = 0.766666667,

    1.02^n = 1/0.766666667,

    1.02^n = 1.304347826
,

    n*log(1.02) = log(1.304347826),

    n = log%28%281.304347826%29%29%2Flog%28%281.02%29%29 = 13.4.


So, 13 full semi-annual payments should be made of 2,000 each,
and then the last,14-th payment, should be made of the lesser amount.


ANSWER.  13 full semi-annual payments should be made of 2,000 each,
         and then the last, 14-th payment, should be made of the lesser amount.

         The total number of semi-annual payments is 14.

Problem 10

A company  "Warehouse, Inc"  is going to loan  $16,000 to  construct a warehouse.
The warehouse will provide the storage space-value at  $3,600 per year.  It will have the annual
maintenance and operating cost of  $360  per year.  The loan interest is  10%  compounded yearly.
How long the warehouse must function to recoup the cost of the loan and the operating cost?
    (a)  8 years.   (b)  9 years.   (c)  7 years.   (d)  6 years.

Solution 

Let PMT be the annual payment for the loan (which is some constant value over the years).


Then an inequality for the project to be profitable is

    PMT + 360 <= 3600,           (1)

which implies

    PMT <= 3600 - 360 = 3240.    (2)


Now, let's calculate PMT for 6, 7, 8 and 9 years with the interest rate r = 10% = 0.1.
Use the standard formula

    for n = 6 years  PMT = L%2A%28r%2F%281-%281%2Br%29%5E%28-n%29%29%29 = 16000%2A%280.1%2F%281-1.1%5E%28-6%29%29%29 = 3673.72;

    for n = 7 years  PMT = L%2A%28r%2F%281-%281%2Br%29%5E%28-n%29%29%29 = 16000%2A%280.1%2F%281-1.1%5E%28-7%29%29%29 = 3286.49;

    for n = 8 years  PMT = L%2A%28r%2F%281-%281%2Br%29%5E%28-n%29%29%29 = 16000%2A%280.1%2F%281-1.1%5E%28-8%29%29%29 = 2999.10;

    for n = 9 years  PMT = L%2A%28r%2F%281-%281%2Br%29%5E%28-n%29%29%29 = 16000%2A%280.1%2F%281-1.1%5E%28-9%29%29%29 = 2778.25.


Comparing these numbers with 3240 in the right side of (2), you see that to be profitable, 
the warehouse should function at least 8 years.    <<<---===  ANSWER


My other lessons on Finance problems in this site are
    - Problems on simple interest accounts
    - Problems on discretely compounded accounts
    - Problems on continuously compounded accounts
    - Find future value of an Ordinary Annuity
    - Find regular deposits for an Ordinary Annuity
    - How long will it take for an ordinary annuity to get an assigned value?
    - Find future value for an Annuity Due saving plan
    - Regular withdrawals from an annuity account
    - Ordinary annuity account with non-zero initial deposit as a combined total of two accounts
    - Annual depositing and semi-annual compounding in ordinary annuity saving plan
    - Variable withdrawals from a compounded account (sinking fund)
    - Present value of an ordinary annuity cumulative saving plan
    - Problems on sinking funds
    - Find the compounding rate of an ordinary annuity
    - Accumulate money using ordinary annuity; then spend money via sinking fund
    - Calculating a retirement plan
    - Accumulating money via ordinary annuity and spending simultaneously via sinking fund
    - Mortgage problems
    - Amortizing a debt on a credit card
    - One level more complicated non-standard problems on ordinary annuity plans
    - One level more complicated problems on sinking funds
    - One level more complicated non-standard problems on loans
    - Using Excel to find the principal part of a certain loan payment
    - Using Excel to find the interest part of a certain loan payment
    - Tricky problems on present values of annuities
    - OVERVIEW of my lessons on Finance section in this site

Use this file/link  ALGEBRA-I - YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-I.

Use this file/link  ALGEBRA-II - YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-II.


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