# PSTAT 171 Loans Homework With solutions

1. A loan of 50,000 is repaid with level payments at the end of each year for 10 years at a rate of 9%. Find: a) the balance of the loan at the end of 3 years and b) the principal and interest paid in year 3.

__Solution__

First find the amount of the annual payment: 50,000/ *a* 10|
at 9% = 50,000/6.418 = 7,791.00

Balance of the loan at the end of year 3 = 7,791.00 |
a 7| |
= 39,211.74. |
---|

To find capital/interest in year 3, you need the balance of the loan at the

end of year 2: 7,791.00 |
a 8| |
= 43,121.78. |
---|

Interest paid in year 3 = 0.09 * 43,121.78 = 3,880.96

Therefore capital repaid = 7,791.00 â€“ 3,880.96 = 3,910.04

This is a prospective solution to the problem; as we have seen in class you can also solve these problems retrospectively. You should get the same result.

2. A borrower borrows 50,000 for 5 years at 7.5%; he pays 6,000 for each of the first 2 years and then catches up in the remaining 3 years. What is the payment in the last 3 years?

__Solution__

This problem has to be done retrospectively. The accumulation of the $6,000 payments at 7.5% is 6000(1.075) + 6000 = 12,450

(Note as we covered in class, when you have only one or two payments it is easier to value them directly rather than by formula.)

2

At the end of year 2, the 50,000 initial loan has accumulated with
interest and is now 57,781.25. The balance of the loan at that time is
57,781.25 less 12,450 = 45,331.25. We need to amortize this remaining
loan over 3 years, and the annual payment is 45,331.25/ *a* =
17,431.57. ~~3|~~

Question: is this reasonable? First reasonability check: the interest on the original 50,000 loan at 7.5% is 3,750 per year. So if the borrower only pays 6,000 the capital repayment in the first two years is about 2,250/year, or 4,500 in total. So at the end of the first 2 years we would expect the loan balance to be about 45,500, which it is. Now we have to amortize this loan over 3 years; without interest the payment would be 15,110; we calculated 17,431 with interest, which seems reasonable.

3. A loan made at an annual rate of 6% has 10 remaining payments of 1000. What is the loan balance?

__Solution__

Simple problem: we know that the balance on a loan is the present
value of the outstanding payments, or 1000 *a* = 7,360.09.
~~10|~~

4. A 30-year, monthly payment mortgage of 500,000 is offered at a nominal rate of 8.4% convertible monthly. Find:

a. Monthly payment;

b. The total principal and interest paid over the life of the loan; c.
The balance after 5 years;

d. Principal and interest paid over the first 5 years.

__Solution__

The simplest way to approach a monthly payment mortgage is to convert it into a 360-pay loan at 0.7% (8.4%/12).

a. Monthly payment = 500,000/131.2616 = 3,809.19.

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b. Total principal/interest over the loan life = 360* 3,809.19 = 1,371,307.79.

c. Balance after 5 years: there are 300 payments remaining, so

balance = (3,809.19) |
a 300| |
= 477,043.59. |
---|

d. The principal repaid in the first 5 years is the difference between 500,000 and 477,043.59 = 22,956.41. The total payments made in the 5 years is 3,809.19*60 = 228,551.40, so interest paid in the first 5 years is 205,594.99.

5. An annual payment loan of 65,000 is made for a term of 10 years at
7.3%. The lender only wants interest payments until the end of year 10
when a balloon payment is made. The borrower will make annual year-end
deposits into a sinking fund earning 4.8%. Find:

a. the level sinking fund deposit;

b. the balance in the sinking fund at time 5; and

c. The total payment and the principal in the 6th payment.

__Solution__

a. The borrower needs to pay sufficient annual deposits to accumulate
to 65,000 over 10 years at 7.3%; X ~~10~~| = 65,000.

X = 65,000/12.461 = 5,216.23.

b. Balance in the sinking fund at the end of year 5: (5,216.23)
~~5|~~ *s* =

28,708.03.

c. Total payment and principal in the 6th payment: i. Interest: 0.073* 65000 = 4,745.

ii. Total payment = 4,745+ 5,216.23 = 9,961.23. iii. Principal: (1.048)5 5,216.23 = 6,594.22.

6. John borrows 10,000 for 10 years at an effective interest rate of 10%. He can repay the loan using the amortization method with payments of 1,627.45 at the end of each year. Instead John repays the loan using a sinking fund that pays an annual effective rate of 14%. Deposits to the

4

sinking fund are equal to 1,627.45 less interest on the loan, and are made at the end of each year for 10 years. What is the balance in the Sinking Fund immediately after repayment of the loan?

__Solution__

The interest payment is 10% of 10,000 = 1,000. So the balance (payment into the SF) is 627.45. This will accumulate to 627.45 10| after 10 years at

14% = (627.45)(19.3373) = 12,133.19.

The loan is repaid at this time (10,000) leaving 2,133.19.

7. A 10 year loan of 2,000 is repaid with payments the end of each year. There are 2 options:

a. Equal annual payments at an annual effective rate of 8.07%;

b. Installment of 200 each year plus interest on the outstanding balance
at an effective rate of i %.

The sum of the payments under each option is the same. What is i?

__Solution__

We first find the payments under each option:

a.X 10| *a* = 2000 => X = 2000/6.689 => X = 299.00. Total
payments is

2,990.

b. Under option b, there is a flat payment of 200 per year plus a decreasing payment of i[2000 + 1800+ 1600â€¦+200] = 200i[10+9+..1] 200i[55] = 11,000i. Principal payments are (10)(200).

Total payments = 2000 + 11,000 I = 2,990.

Hence i = 0.09.

5

8. A loan is repaid with level annual payments based on an annual effective interest rate of 7%. The 8th payment consists of 789 interest and 211

principal.

Calculate the amount of interest in the 18th payment.

__Solution__

Annual payment is X *a* . The 8th payment is 1000. The
original loan was ~~|~~*n*

therefore 1000 *a* . The question then is what is the
outstanding balance at ~~|~~*n*

the end of year 17?

At the end of year 7, the balance of the loan is 1000 *a*
~~âˆ’~~ = 1000(1- vn-7)/i. The interest paid in the 8th year is
equal to this loan balance times i.

ï¿½ 1000(1- vn-7) = 789

ï¿½ (vn-7) = 0.211 => -(n-7)ln (1.07) = ln (0.211) -(n-7)(0.06766) =
-1.5559 ï¿½

ï¿½ (0.06766) n = 1.5559+0.47362 = 2.02952 ï¿½ n = 30

We can now calculate the balance at the end of the 17th year:

1000 13| = 835.76.

Interest payment in year 18 is then 585.04.

There are other ways to solve this problem. Ask at Section if you tried a

different method.

9. A 20-year loan of 20,000 may be repaid under one of two methods:

a. Amortization method with equal annual payments at an effective

rate of 6.5%

b. Sinking fund method in which the lender receives an effective rate of

8% and the sinking fund earns an annual effective rate of j.

6

Both methods require a payment of X at the end of the each year for 20 years.

Calculate j.

__Solution__

a. Payment = 20,000 |
a 20| |
= 20000/11.0185 = 1,815.13 |
---|

b. The interest payment under the SF method is 0.08*20,000 = 1,600. Hence the payment into the Sinking Fund is 215.13.

215.13 20| | = 20,000 at rate j => 20| | = 92.967 |
---|

((1+j)20 -1)/j = 92.967 => j â‰ˆ 0.1415.

10. A loan at 6.5% annually has an initial payment of 300 and 9 further payments. The payment amount increases by 20 each year. Find the loan balance immediately after the 6th payment.

__Solution__

There are two payment streams. We construct a time-line:

|_____|_____|_____|_____|_____|_____|_____|____|___|____|

Time | 0 1 2 3 4 5 6 7 8 9 /th> | ) | 10 | ||||||
---|---|---|---|---|---|---|---|---|---|

Paytâ€™s | 300 | 320 | 340 | 360 â€¦â€¦ |
480 | ||||

Original Loan = 300 10| | + 20 [ v2 + 2v3 â€¦9v10] = 2,156.65 + 20v2 | ( | Ia |
. |

The increasing annuity has value:

9| |
âˆ’ | 9 | v |
9 |
= (7.089 â€“ 5.106)/.065 | ||||||||||||||||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|

i |

= 30.505; so the increasing annuity value is 537.90.

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Original loan was therefore 2,694.55. (you donâ€™t actually have to calculate the original loan.)

At the end of 6 years:

400 4| *a* + 20 [ v+ 2v2 + 3v3â€¦4v4] = 1370.32 + (8.295)(20) =
1,536.22 .

11. A loan at 7% annually has an initial payment of 250 and 9
further

payments. The payment amount increases by 3% annually. Find the loan
balance immediately after the 7th payment.

__Solution__

Payments are 250, 250 (1.03) etc.

After the 7th payment, there are 3 further payments and present value =

250 [(__1.03__)7 + (__1.03__)8 + (__1.03__)9] = 830.24

1.07 | 1.072 | 1.073 |
---|

12. You have a loan of 20,000 at 7.2% annually for 8 years. You agree to pay off the principal in installments of 2,500 per year, and to pay interest on the outstanding balance each year. Find the interest due in the 6th payment.

__Solution__

At the end of year 5, there have been 5 payments. This is an
installment loan, so interest and annual payments vary. The total amount
of capital reduction is 5 * 2,500 = 12,500. At the beginning of year 6
the

outstanding balance is 20,000 â€“ 12,500 = 7,500.

Interest is 7.2% of 7500 or $540.

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