Math 373
Fall 2013
Homework – Chapter 5
Chapter 5 Section 2
1. (S12HW) Kwaku borrows 100,000 to be repaid with five annual payments. The annual effective
interest rate on the loan is 6%.
Complete an amortization table for this loan.
Solution:
First we need to find the appropriate payments using our calculator:
N  5 I / Y  6 PV  100,000 CPT PMT  23,739.64
Now we can create the Amortization table.
Time
Payment
0
1
2
3
4
5
Interest in Pmt
23,739.64
23,739.64
23,739.64
23,739.64
23,739.64
Principal in
Payment
6,000.00
4,935.62
3,807.38
2,611.44
1,343.75
OLB
17,739.64
18,804.02
19,932.26
21,128.20
22,395.89
100,000.00
82,260.36
63,456.34
43,524.08
22,395.89
0.00
2. (S12HW) Syaza has a loan of 15,000 which is being repaid with ten level annual payments of
2000.
a. Calculate the amount that Syaza will pay in principal over the life of the loan.
Solution: Total Amount of Principal=Total Loan Amount= 15,000
b. Calculate the amount of interest that Syaza will pay.
Solution: Interest Paid=Total Amount Paid-Total Loan Amount
 10*2000 15,000  5,000
October 10, 2013
3. (S08T2) Josh has repaid a loan with 4 annual payments of 950 each. The total interest repaid in
those four payments was 800. Calculate the annual effective interest rate on the loan.
Solution:
Total Amount Paid=Total Amount of Interest Paid+Total Loan Amount
Total Loan Amount  950*4  800  3000
Now we can find the interest rate using our financial calculator.
PV  3000 N  4 PMT  950 CPT I / Y  10.175
4. (S12HW) Cale borrowed money to buy a new car. Payments are made monthly. The loan has an
nominal rate of interest of 12% compounded monthly. Immediately after the 15th payment, Cale
has an outstanding loan balance of 8500.
Calculate the amount of interest in the his 16th payment.
Solution:
 i (12) 
  85000*0.01  85
 12 
Amount of interest in the 16th payment=  OLB15  
5. (S12HW) Daniel took a loan to buy a new couch for his apartment. He is making monthly
payments and the loan has a nominal interest rate of 9% compounded monthly. Immediately after
the 8th payment, Daniel still owes 800 on his loan.
The principal in his 9th payment is 90.
Determine the amount of the 9th payment.
Solution:
i (12)
 0.0075 OLB8  800
12
Princ 9 =90
Payment9=(Principal in Payment 9)+(Interest in Payment 9)= 90  800(0.0075)  96
October 10, 2013
6. (S12HW) Connor has a mortgage that is being repaid with monthly payments. The annual
effective interest rate on his loan is 8%.
The principal in the 120th payment is 890.
Calculate the principal in the 40th payment.
Solution:
To calculate the principle in the kth year use the formula Qvnk 1 .
12
 i (12) 
i (12)
We know that 1 

1.08

 0.00643403

12 
12

Principle in the 120th payment: Qvn121  890
Principle in the 40th payment: Qv n41  x
We need to solve for x using substitution.
First Equation:
Q 1.00643403
Q
 ( n 121)
 890
890
1.00643403
 ( n 121)
 890 1.00643403
Substitute the Q above for the Q in Qv
890 1.00643403
( n 121)
n41
1.00643403
( n 121)
 x.
 ( n  41)
 890 1.00643403
October 10, 2013
80
 532.80
7. (S12HW) Alice is repaying a loan with level annual payments of 1000. The interest rate on the
loan is 7%.
The interest in the 2nd payment is 555.99.
Calculate the interest in the 7th payment.
Solution:
We know Q  1000 and i  0.07 .




We can find n using the formula for the kth payment. Q 1  v n  k 1 .

1000 1  1.07 
 ( n  2 1)
1000  1000 1.07 
1 n
  555.99
 555.99
444.01  1000 1.07 
1 n
1 n 
ln(0.44401)
ln(1.07)
n  13
Now we can find the interest in the 7th payment using Q 1  v n  k 1 .

1000 1  1.07
 (12 7 1)
  377.25
October 10, 2013
8. (S12HW) Taylor has a loan which has an outstanding loan balance of 34,000 immediately after
the 9th payment. The monthly payments of 1000 are level. The interest rate on the loan is a
nominal rate of 12%. Calculate Taylor’s outstanding loan balance immediately after the 11th
payment.
Solution:
i (12)
We know Q  1000 , OLB  34000  1000an 9 , and
 0.01 .
12
First we need to find n.
34000  1000an 9
1  1.01 ( n 9) 
34  

0.01


0.034  1  1.01
9 n
ln(0.66)
ln(1.01)
n  50.75895758
9n 
Now we can find the OLB after the 11th payment.
1  1.01 (50.7589575911) 
OLB  1000 
  32673.40
0.01


October 10, 2013
9. (S08T2) A loan is being repaid with annual payments for 20 years. The principal in the 5th
payment is $4236.99. The principal in the 10 payment is $5670.05. Calculate the amount of the
loan to the nearest dollar.
Solution:
We know n  20 , Qv2051  Qv16  4236.99 , and Qv20101  Qv11  5670.05 .
First, we need to find i.
Q  4236.99v16 .
Using substitution:
4236.99v 16 v11  5670.05
v 5  1.338225958  (1  i) 5
i  0.06
Now we can find Q.
Q  4236.99(1.06)16  10763.44468
However, we are asked for the total amount of the loan, Qa20 .
Qa20
1  1.06 20 
 10763.44468 
  123456
0.06


October 10, 2013
10. (S09T2) A 30 year mortgage is being repaid with level monthly payments. The principal in the
30th payment is 90.43. The principal in the 60th payment is 106.61.
Calculate the interest in the 90th payment.
Solution:
We know, n  30(12)  360 , Qv360301  Qv331  90.43 , and Qv360601  Qv301  106.61.
First, we need to find i.
Q  90.43v 331
90.43v 331v301  106.61
v 30  1.178922924  (1  i)30
i  0.005501788
Now we can find Q.
Q(1.005501788)331  90.43
Q  555.9489515
We are asked to calculate the interest in the 90th payment:
555.9489515 1  (1.005501788)  (360901)   430.26
October 10, 2013
Chapter 5, Section 3
11. (S12HW) Chen Corporation borrows 100,000. The loan will be repaid with annual payments for
ten years using the sinking fund method. The loan has an annual effective interest rate of 9%.
The sinking fund earns an annual effective interest rate of 6%. The payments to the sinking fund
will result in the sinking fund being exactly equal to the loan at the end of ten years.
a. Calculate the amount of interest that will be paid to the bank each year.
Solution:
I  iL  0.09(100000)  9000
b. Calculate the sinking fund deposit each year.
Solution:




 L 
100000 

D

 7586.80
 s    1.0610  1  
 10  


  0.06  
c. Calculate the amount that will be in the sinking fund immediately after the 4th payment.
 1.064  1 
SF4  Ds4  7586.80 
  33,189.82
 .06 
October 10, 2013
12. (S12HW) Cui Corporation wants to borrow 200,000. Cui has the choice of the following two
loans:
Bolle Bank offers a sinking fund loan with 10 annual payments. The annual effective interest rate
on the loan is 10%. The annual effective interest rate to be earned by the sinking fund will be
7.5%. The amount in the sinking fund at the end of 10 years will exactly repay the loan.
Fang Finance Company offers an amortization loan with 10 level annual payments at an annual
effective interest rate of i .
The total annual payments are the same under either loan.
Calculate i .
Solution:
First, find the payment:


 200, 000 
L
D  I  iL   0.1(200, 000)  
  34,137.18549
10
sn
 1.075  1 


 .075 
Then use your calculator to find i.
PV  200000 N  10 PMT  34,137.1859 CPT I / Y  11.12
October 10, 2013
13. (S08T2) Wozny-Wiggins Corporation wants to borrow 500,000 to be repaid with annual
payments over ten years.
Dummitt Bank offers a loan using the sinking fund method. The interest rate on the loan is i and
the sinking fund will earn 5%. Each year, Wozny-Wiggins must pay the interest on the loan and
make a payment into the sinking fund. The payments into the sinking fund are such that the
amount in the sinking fund after 10 years will exactly repay the loan.
Lumley Bank offers a loan based on the amortization method and an annual effective interest rate
of 6.5%. The amount of the payment under this loan is exactly equal to the sum of the interest
payment and sinking fund deposit on the loan from Dummitt Bank.
Calculate i, the annual effective interest rate on the loan from Dummitt Bank?
Solution:
Payment  I  D  i (500, 000) 
500, 000
 i (500, 000)  39,572.287
 1.0510  1 


 .05 
Payment * a10  500, 000
Payment 
500, 000
 69,552.345
  1 10 
 1 
 
  1.065  


.065




500, 000i  39,352.287  69,552.345
i  0.0596
October 10, 2013
Chapter 5, Section 4
14. (S08T2) A four year loan is being repaid with annual payments. The first payment is 3000. The
second payment is 4000. The third payment is 5000. The final payment is 6000. The annual
effective interest rate is 10%. Create an amortization table for this loan.
Solution:
Loan  3000v  4000v2  5000v3  6000v4  13887.71
Time
0
1
2
3
4
Payment
Interest
Principal
3000
4000
5000
6000
1388.77
1227.64
950.41
545.45
1611.23
2772.35
4049.58
5454.54
OLB
13887.71
12276.48
9504.128
5454.54
0
15. (S09T2) A loan of 50,000 is being repaid with 30 annual payments. The annual effective interest
rate on the loan is 6%. The first payment on the loan is 30P. Each subsequent payment decreases
by P. Therefore, the second payment is 29P. The third payment is 28P, etc.
Calculate the principal in the 29th payment.
Solution:
First, we need to find P using the P and Q formula where
P  30 P
.
Q  P
Note that there are two different “P’s” which is pretty confusing. But we only need to know what
we are going to plug into the formula.
 P 
30
30 Pa30  
  a30  30v   50,000
 0.06 

 1 
30 
P 30a30  
  a30  30v    50,000
 0.06 


P  184.78
Now we can find the OLB after the 28th payment by finding the present value of future payments.
The future payments are 2P at time 29 and P at time 30.
OLB28  2Pv  Pv2  2(184.78)(1.06)1  184.78(1.06)2  513.10
The interest in the 29th payment is (OLB28 )(i)  513.10(0.06)  30.79 .
Then the principal in the 29th payment is the 29th payment less the interest in the 29th payment:
2 P  Interest  2(184.78)  30.79  338.78
October 10, 2013
Answers
1. Not Given. The table is self checking as the last OLB should be zero.
2.
a. 15,000
b. 5,000
3. 10.175%
4. 85
5. 96
6. 532.80
7. 377.25
8. 32,673.40
9. 123,456
10. 430.26 *
11.
a. 9000.00
b. 7586.80
c. 33,189.34
12. 11.124%
13. 5.96%
14. Not Given. The table is self checking as the last OLB should be zero.
15. 338.78 *
*Note: You may get slightly different answers due to rounding. This may also be true of other problems.
October 10, 2013