# Text

CHAPTER 5: HOMEWORK EXCERSICES
NAME: CAO THI NGOC ANH
ID: 421 762
CLASS: VISK2010B

SOLUTION

Problem 4 (p.196)
a. We   have a formula :
PV = FVn [PVFk,n]
And substitute what’s known,
\$856 = \$1,122 [PVF7,n]
PVF7,n   = \$856/ \$ 1,122   =   0.7629
n = 4 years
b. First, we calculate the correct k. Because compounding is monthly
k = knom / 12 = 12%/12 = 1%
We   have a formula:
PV = FVn [PVFk,n]
And substitute what’s known,
\$450.00 = \$725.50 [PVF1,n]
PVF1,n = 0.6203
n = 48 months = 4years

c. First, we calculate the correct k. Because compounding is quarterly
k= knom/ 4 = 10% / 4 = 2.5 %
We   have a formula:
PV = FVn [PVFk,n]
And substitute what’s known,
\$5,000 = \$6,724.44 [PVF2.5,n]
PVF2.5,n = 0.7436
n = 12 quarters = 3 years
Problem 7 (p. 196)
a. A formula of the future value factor for k and n is :
FVF k, n = (1 + k) n
(1+k)a+b = (1+k)a (1+k)b
Thus, we have a new formula:
FVFk,a+b = [FVFk,a] [FVFk,b]

And, substitute what’s known on equation
FVF 3, 85 = [FVF 3, 60] [FVF 3,25]
FVF 3, 85 = 5.8916 x 2.0938 = 12.3359
Then, substitute in the formula:   FV = PV [FVF3,85] = \$ 11.50 x 12.3359 = \$141.863   per bond

b. Using equation (a)to calculate :
FVF 7, 85 = [FVF 7, 60] [FVF 7, 25]
FVF 7, 85 = 57.9464 x 5.4274 = 314.49
Then, substitute in the formula:   FV = PV [FVF7, 85] = \$ 11.50 x 314.49 = \$ 3,616.63
Per pond

c.

Problem 9 (p. 197)
a. The amount of note that Charlie will pay in   five years with...