Theory of Interest – Assignment 1 Answer key
1. Questions from ch. 1
A(5)  A(4) 125  120
5
i5 


1
5a) A(t )  100  5t
24
A(4)
120
120
150  145
5

1
b) i10 
29
145
145
6a)
b)
A(t )  100(1.1) t i5 
i10 
A(10)  A(9) 100(1.1)10  100(1.1) 9

 .1
A(9)
100(1.1) 9
11. 1000(1  it )  1110
14.
A(5)  A(4) 100(1.1) 5  100(1.1) 4

 .1
A(4)
100(1.1) 4
it  .11 500(1  (.75i)( 2t ))  500(1  1.5it )  500(1.165)  582.50
600(1  i) 2  864 (1  i) 2  1.44
19. v  v
n
2n
1
(1  i) n 
Set x  v
n
(1  i )  1.2
Then x  x  1  0
2
2000(1  i) 3  $3456
x
1 5
(positive root)
2
2
1 5
4
4
2
3 5 3 5
(1  i ) 2 n 
(optional ) 

x

 2.618
2
62 5 3 5 3 5
(1  5) 2
33.
i ( m ) d ( m ) i ( m )  d ( m ) .1844144  .1802608 .1844144  .1802608




m
m
m
m2
m2
.1844144  .1802608
m
8
.1844144  .1802608
45.
2
2
.01tdt
a(2)  e 0
 e .01t / 2 0  e .02  (1  i) 2
2
1  i  e .01
i  e .01  1
50.

20
exp  (.01t  .1)dt  exp .01t 2 / 2  .1t
0

20
0
 e4
(1  i ) 20  e 4
(1  i )1.5  e 4 x (1.5 / 20)  e .3
2. Compound i  .06
r  .06
Simple
( 2)
Compound i
3. i
(12)
 (1.06

d ( 4) 
1 

4 

1
6
I 3  A(3)  A(2)  5000(1.06) 3  5000(1.06) 2  337.08
I 3  A(3)  A(2)  5000(1  .06(3))  5000(1  .06(2))  300.00
 .06 I 3  A(3)  A(2)  5000(1.03) 6  5000(1.03) 4  342.72
 1)12  .11710553
or 11.71%
4
 1.06 2
d ( 4)  .0114856551
or 11.49%
m
 .17 
f ( m )  1 
  1.18
m 

f (3)  1.1798
f (4)  1.1811 Therefore, m=4
4. Want smallest integer m, such that
f (2)  1.1772
m
5. Want smallest integer m, such that
 .16 
f ( m )  1 
  1.18 , but first check to make sure that there
m 

m
 .16 
.16
is an answer lim 1 
  e  1.1735 The max effective rate that can be reached is 17.35%
m 
m 

Question 4 would not work with a nominal rate of 16% (Students must show this logic in order to get credit
for this question)
6.
1  i
i
(12)
(12)
/ 12

108
 12 .2691
3
7.
Chapter 2

6a) 10000(.06) 62
365
  101.92
b) 10000(.06)(60 / 360)  100.00
c)
10000(.06)(62 / 360)  103.33
9. 200v  500v
5
v 5  .40188
10
 400.94v 5
500v 10  200.94v 5
1  i 5  2.48831
v 10  .40188v 5
1002.483312  1202.48331  917.76
10. d
( 4)
 4 / 41
d  1 / 41 per quarter
100(1.025)1 (1  1 / 41) 24  100(1.025) 3 (1  1 / 41)16  100(1.025) 5 (1  1 / 41) 8  185.39  159.87  137.85  483.11
8. 30000v 17  30000v 15  30000v 12  29478.73
where
i  .08