Homework #7 Selected Solutions
Math 105: Fall 2003, Instructor: Erin McNicholas
Assigned: Consumer Math
6.1/ 2, 4, 5, 10, 21, 23, 24, 29, 38, 47, 50
6.2/ 2, 6, 12, 13, 17, 24, 26, 30, 33, 42
6.3/ 2, 6, 12, 18, 19, 24, 28, 32, 36, 38
Relevant Formulas
Simple Interest:
Interest: I = Prt where P = principal, r = annual interest rate, t = time in years Future
Value: F = P(1+rt)
Compounded Interest:
mt
r

Future Value: F  P1  
 m
where m = # of times per year the interest is compounded.
mt
 r 
Effective Annual Rate: EAR  1  c   1
 m
where rc = the compounded interest rate
Amortized Loan:
12t
r 
 r 
P 1  
12
12
Monthly Payment: mpmt    12t  APR = (1.8)(Add-On Interest Rate)


r 
1    1
 12 

APR = (1.8)(Annual simple interest associated with Rent to Own cost)
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6.1/ 2, 4, 5, 10, 21, 23, 24, 29, 38, 47, 50
2)
a. I = $525*0.05*2 =
b. I = $300*0.03*5 =
c. I = $7934*0.0415*8 = $
4)
a. I = $525*0.05*(48/12) =
b. I = $300*0.03*(40/12) =
c. I = $7934*0.0415*(33/12) = $
10)
a. 1/96 through 3/99 = 3*12 + 3 = 39 mo
b. 3/96 through 2/99 = 10 + 2*12 + 2 = 36 mo
I = $700*0.05*(39/12) = $
I = $385*0.085*(36/12) = $
( 365 / 5 )
24)

0.07 

EAR= 1 
 (365 / 5) 
38)
 0.05 
Future Value = $50001 

4 

 1  7.25%
4*20
 $13507.42
50)
 .06 
4 P  P 1 

12 

12*t
4  1.005
12t
With trial and error we find that it will take between 23 and 24 years.
(1.005) 12*23  3.96
(1.005)12*24  4.21
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6.2/ 2, 6, 12, 13, 17, 24, 26, 30, 33, 42
2)
a. Finance Charge = I = $183.65*0.169*(31/365) = $2.64
b. Finance Charge = $194.85*0.165*(30/365) = $2.64
6)
Finance Charge = $325.5(0.149)(31/365) = $4.12
New Balance = $485.88 + $4.12 = $490
12)
Dates
9/11-14
9/15-21
9/22-10/1
10/2-10
Balance
$385.56
$185.56
$228.41
$421.34
Number of Days
4
7
10
9
a. Av. Daily Balance = (4($385.56) + 7($185.56) + 10($228.41) + 9($421.34))/30
= $297.24
Finance Charge = $297.24(0.149)(30/365) = $3.64
b. New Balance = $421.34 + $3.64 = $424.98
24)
Total paid = 12 x $34.53 = $414.36
Using the future value formula to solve for the associated principal we find:
$414.36 = P(1+(0.105)(1yr)) = P(1+0.105)=P(1.105)
P = $414.36/(1.105) = $374.99
26)
Use Table 6.2
a. 21.5%
c. 18.2%
b. 10.8%
d. 14.1%
30)
a. Total Paid = 2 x 12 x $48.18 = $1,156.32
Using the future value formula to solve for the interest rate we find:
$1,156.32 = $925 (1 + 2r)
r = ($1156.32-$925)/(2 x $925) = 0.125 = 12.5%
b. Total Paid = 3 x 12 x $58.17 = $2094.12
r = ($2094.12 - $1500)/(3 x $1500) = 0.132 = 13.2%
42)
Total Paid = $42.50 x 24 = $1020
Interest paid over 2 years = $1020 - $796 = $224
Interest paid per year = $112, which is $112/$796 = 14.07% of the principal.
Thus, the associated annual simple interest rate is 14.07%
APR~(1.8)(14.07%)=25.3%
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6.3/ 2, 6, 12, 18, 19, 24, 28, 32, 36, 38
2)
After 1 month, accrued interest = $5000(0.08)(1/12) = $33.33
Thus, $33.33 of the $101.40 payment will go towards paying off the interest
and the remaining $68.07 will go towards paying off the balance.
The new balance = $5000 - $68.07 = $4931.93
6)
We did a similar example in class on 10/16
12)
Using table 6.4 we find that the monthly payment on a $1000 loan for 15 yrs at
10% is $10.75.
Thus, the monthly payment on the $28,000 loan = 28 x $10.75 = $301
125
18)
.1 
 .1 
$5000 1` 
 12  12 
monthly payment 
125


.1 
1    1
 12 

24)
 .1175  .1175 
$10,800
1`

12 
12 

monthly payment 
 .1175 124 
  1
1 
12 


28)
 $106.24
124
 $283.08
Principal = $85,000 – down payment = $85,000 – (0.1)($85,000) = $76,500
 .08  .08 
$76500
1`

 12  12 
monthly payment 
 .08 1230 
 1

1 
 12 

1230
 $561.33
32)
Using Table 6.5 we find that the monthly payment for a 15 year loan of $1000 at
12% is $12.001681
$300 / $12.001681 = 24.996
Thus, we can afford to borrow 24.996 loans, each for $1000. In other words, we
can afford to borrow $24996.
36)
In a method similar to that used in problem #32 we use Table 6.5 to find the
associated monthly payment for $1000, which is $12.667577
$300/$12.667577 = 23.6825
Thus we can afford to borrow $23, 682.50
38)
A monthly payment of $231.55 on a 25-year, $30,000 loan, corresponds to a
monthly payment of $231.55/30 = $7.72 on a 25-year, $1000 loan at the same
interest. Using table 6.5 we find this monthly payment corresponds to an 8%
interest rate.