Loan Payment Problems
1.
Name
John borrows 10,000 for 10 years at an annual effective interest rate of 10%. He can repay the
loan using the amortization method with payments of 1627.45 at the end of each year. Instead,
John repays the 10,000 using a sinking fund that pays an annual effective interest rate of 14%.
The deposits to the sinking fund are equal to 1627.45 minus the interest on the loan and are
made at the end of each year for 10 years.
Determine the balance in the sinking fund immediately after repayment of the loan.
A. 2,130
B. 2,180
C. 2,230
D. 2,300
E. 2,370
The annual interest on the loan is 1000. So John pays 1627.45 − 1000 = 627.45 to the sinking
fund each year. Using calculator with N = 10, I = 14, P V = 0, P M T = −627.45, and END,
the accumulated (future) value of the sinking fund at the end of year 10 is 12133.19. Using
10000 to pay off the loan, John has 2133.19 in the sinking fund left which is rounded to 2130.
Ans: A
2.
A 20-year loan of 1000 is repaid with payments at the end of each year. Each of the first ten
payments equals 150% of the amount of interest due. Each of the last ten payments is X. The
lender charges interest at an annual effective rate of 10%. Calculate X.
A. 32
B. 57
C. 70
D. 97
E. 117
For each year in the first 10 years, the payment at the end of year n (0 < n ≤ 10) is 150%
of interest due, so P Rn is 50% of the interest due, that is 12 In . Since In = iOBn−1 , P Rn =
1
2 iOBn−1 . The outstanding balance right after the nth payment is
1
i
OBn = OBn−1 − P Rn = OBn−1 − iOBn−1 = 1 −
OBn−1 .
2
2
Using this formula repeatedly, we get OBn = (1 − i/2)n · 1000. (For example, OB3 = (1 −
i/2)OB2 = (1−i/2)(1−i/2)OB1 = (1−i/2)(1−i/2)(1−i/2)OB0 = (1−i/2)3 ·1000.) Therefore,
OB10 = (1 − i/2)10 · 1000 = (0.95)10 · 1000 = 598.74 which is to be repaid by the next 10 level
annual payments of X each. Using the calculator with N = 10, I = 10, P V = 598.74, F V = 0,
and END, we get P M T = −97.44. So X = 97
Ans: D
3.
A 10-year loan of 2000 is to be repaid with payments at the end of each year. It can be repaid
under the following two options:
(i) Equal annual payments at an annual effective rate of 8.07%.
(ii) Installments of 200 each year plus interest on the unpaid balance at an annual effective
rate of i.
The sum of the payments under option (i) equals the sum of the payments under option (ii).
Determine i.
A. 8.75%
B. 9.00%
C. 9.25%
D. 9.50%
E. 9.75%
For method (i), using N = 10, I = 8.07, P V = 2000, F V = 0 and END, we get P M T = −299.
So the total payment for the 10 years is 2990.
For method (ii), we have
t
0
1
2
3
···
···
8
9
10
OBt
2000
1800
1600
1400
···
···
400
200
0
It
–
2000i
1800i
1600i
···
···
600i
400i
200i
P Rt
–
200
200
200
···
···
200
200
200
Kt
–
200 + 2000i
200 + 1800i
200 + 1600i
···
···
200 + 600i
200 + 400i
200 + 200i
The total payment for the 10 years is
K1 + K2 + · · · + K10 = (200 + 2000i) + (200 + 1800i) + · · · + (200 + 400i) + (200 + 200i)
= 10 × 200 + 200(10 + 9 + · · · + 2 + 1)i
= 2000 + 11000i
2000 + 11000i = 2990 ⇒ i = 0.09 = 9%.
Ans: B
4.
A loan is amortized over five years with monthly payments at a nominal interest rate of 9%
compounded monthly. The first payment is 1000 and is to be paid one month from the date
of the loan. Each succeeding monthly payment will be 2% lower than the prior payment.
Calculate the outstanding loan balance immediately after the 40th payment is made.
A. 6751
B. 6889
C. 6941
D. 7030
E. 7344
Ans: B
5.
A 20-year loan of 20,000 may be repaid under the following two methods:
(i) amortization method with equal payments at an annual effective rate of 6.5%
(ii) sinking fund method in which the lender receives an annual effective rate of 8% and the
sinking fund earns an annual effective rate of j
Both methods require a payment of X to be made at the end of each year for 20 years.
Calculate j.
A. j ≤ 6.5%
D. 10.0% < j ≤ 12.0%
B. 6.5% < j ≤ 8.0%
E. j > 12.0%
C. 8.0% < j ≤ 10.0%
j = 14.179%
Ans: E
6.
Ron is repaying a loan with payments of 1 at the end of each year for n years. The amount of
interest paid in period t plus the amount of principal repaid in period t + 1 equals X.
Calculate X.
A. 1 +
v n−t
i
B. 1 +
v n−t
d
C. 1 + v n−t i
D. 1 + v n−t d
E. 1 + v n−t
Ans: D
7.
Seth borrows X for four years at an annual effective interest rate of 8%, to be repaid with
equal payments at the end of each year. The outstanding loan balance at the end of the third
year is 559.12.
Calculate the principal repaid in the first payment.
A. 444
B. 454
C. 464
D. 474
E. 484
Since OB3 is to be repaid by the last regular payment K at the end of year 4, OB3 = Kv =
K/1.08 ⇒ K = 603.85. P R1 = K · v 4 = 603.85 × (1.08)−4 = 443.85
Ans: A
8.
A loan is being repaid with 25 annual payments of 300 each. With the 10th payment, the
borrower pays an extra 1000, and then repays the balance over 10 years with a revised annual
payment. The effective rate of interest is 8%.
Calculate the amount of the revised annual payment.
A. 157
B. 183
C. 234
D. 257
E. 383
The regular OB10 = 300a15|0.08 = 2567.85 (using calculator). With an extra 1000 payment,
the outstanding balance is 1567.85 which is now to be repaid by 10 revised payments. Using
N = 10, I = 8, P V = 1567.85, F V = 0 and END, we get P M T = −233.66.
Ans: C