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The Theory of Interest  Solutions Manual
1
Chapter 1
1.
(
a
)
Applying formula (1.1)
( )
( )
2
2
3
and
0
3
At
t
t
A
=++
=
so that
()
( )
( )
2
1
23
.
03
at
t
t
kA
==
=
+
+
(
b
)
The three properties are listed on p. 2.
(1)
1
1
.
3
a
(2)
1
2
2
0
for
0,
3
t
t
′
=+
>
≥
s
o
t
h
a
t
( )
is an increasing function.
(3)
( )
is a polynomial and thus is continuous.
(
c
)
Applying formula (1.2)
(
)
[ ]
()()
2
2
22
12
31
2
1
3
21
3
.
n
IA
nA
n
n
n
n
n
nn
n
n
⎡
⎤
=−
−
=
+
+
−
−
+
−
+
⎣
⎦
−+−
−+
−
2.
(
a
)
Appling formula (1.2)
( )
[ ] ( ) ( )
[ ] (
)(
)
[ ]
10
1
0.
n
II
I
A
A
A
A
A
n
An
A
+++=
−
+
−
++
−
−
…"
(
b
)
The LHS is the increment in the fund over the
n
periods, which is entirely
attributable to the interest earned. The RHS is the sum of the interest earned
during each of the
n
periods.
3.
Using ratio and proportion
5000
12,153.96 11,575.20
$260
11,130
.
−=
4.
We have
( )
2
,
a
t
b
so that
01
3
9
1.72.
ab
aa
b
=+=
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Chapter 1
2
Solving two equations in two unknowns
.08
and
1.
ab
=
=
Thus,
( )
(
)
2
5
5
.
0
8
1
3
a
=
+=
(
)
(
)
2
10
10 .08
1 9
a
=
.
and the answer is
(
)
()
10
9
100
100
300.
53
a
a
==
5.
(
a
)
From formula (1.4
b
) and
( )
100 5
A
tt
=
+
( )
5
5
4
125 120
5
1
.
4
120
120
24
AA
i
A
−−
=
=
(
b
)
(
)
( )
10
10
9
150 145
5
1
.
9
145
145
29
i
A
=
=
6.
(
a
)
( )
()()
54
5
4
100 1.1
and
5
4
100 1.1
1.1
1.1 1 .1.
4
100 1.1
t
At
i
A
=
⎡⎤
⎣⎦
=
−
=
(
b
)
() (
)
() ()
10
9
10
9
10
9
100 1.1
1.1
1.1 1 .1.
9
100 1.1
i
A
=
−
=
7.
From formula (1.4
b
)
(
)
(
)
1
1
n
An
i
−
−
=
−
so that
(
)
(
)
(
)
11
n
iAn
−
−=
−
and
(
)
( )
(
)
.
n
i An
=
+−
8.
We have
567
.05,
.06,
.07,
iii
===
and using the result derived in Exercise 7
( )( )( )
74
1
1
1
1000 1.05 1.06 1.07
$1190.91.
=+
+
+
9.
(
a
)
Applying formula (1.5)
(
)
615
500 1 2.5
500 1250
ii
=
+
so that
The Theory of Interest  Solutions Manual
Chapter 1
3
1250
115 and
115/1250 .092,
or
9.2%.
ii
==
=
(
b
)
Similarly,
(
)
630
500 1 .078
500 39
tt
=+=
+
so that
1
3
39
130
and
130/39 10/3 3
years.
=
=
10.
We have
(
)
1110 1000 1
1000 1000
it
it
=+
=
+
1000
110
and
.11
it
it
=
=
so that
()
( )
[]
3
500 1
2
500 1 1.5
4
500 1
1.5 .11
$582.50.
it
i
t
⎡⎤
⎛⎞
+=
+
⎜⎟
⎢⎥
⎝⎠
⎣⎦
=
11.
Applying formula (1.6)
.04
and
.025
11
1
.
0
4
1
n
i
i
in
n
+−
+
−
so that
.025 .001
1
.04,
.001
.016,
and
16
.
nn
n
=
=
=
12.
We have
12 34 5
.01
.02
.03
.04
.05
i
=====
and adapting formula (1.5)
( )
(
)
12345
1000 1
1000 1.15
$1150.
iiiii
++
+
+
+
=
=
13.
Applying formula (1.8)
2
600 1
600
264
864
i
+
=
which gives
2
1
864/600 1.44,
1
1.2,
and
.2
i
=
+
=
=
so that
(
)
33
2000 1
2000 1.2
$3456.
i
=
14.
We have
1
and
1
1
1
n
n
n
rr
j
j
+
+
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Chapter 1
4
so that
(
)
( )
11
1
1.
1
ij
ii
j
r
j
jj
+−+
+−
=−
=
=
++
+
This type of analysis will be important in Sections 4.7 and 9.4.
15.
From the information given:
()
12
21
3
1
3
/2
31
15
1
5
61
10
1
5/3
.
aa
bb
cc
nn
+=
By inspection
52
1
5.
33
2
=⋅⋅
Since exponents are addictive with multiplication, we
have
.
ncab
=−−
16.
For one unit invested the amount of interest earned in each quarter is:
Quarter:
Simple:
Compound:
23
2
4
3
1
2
3
4
.03
.03
.03
.03
1.03 1
1.03
1.03
1.03
1.03
1.03
1.03
−−
−
−
Thus, we have
43
32
4
1.03
1.03
.03
1.523.
3
1.03
1.03
.03
D
D
⎡⎤
⎣⎦
==
17.
Applying formula (1.12)
18
19
20
21
:
10,000 1.06
1.06
6808.57
10,000 1.06
1.06
6059.60
Difference
$748.97.
A
B
=
18.
We have
2
1
vv
+
=
and multiplying by
2
1
n
i
+
2
1
=
+
or
()()
2
1
0
+
−
=
which is a quadratic.
The Theory of Interest  Solutions Manual
Chapter 1
5
Solving the quadratic
()
11
4
15
1
22
n
i
±+
+
+=
=
rejecting the negative root.
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This note was uploaded on 11/18/2010 for the course MATH 320 taught by Professor Dr.k during the Spring '10 term at Nevada.
 Spring '10
 Dr.K

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