The Theory of Interest - Solutions Manual
23
Chapter 3
1.
The equation of value using a comparison date at time
20
t
=
is
20
10
50,000
1000
at
7%.
s
Xs
=
+
Thus,
20
10
50,000
1000
50,000
40,995.49
$651.72.
13.81645
s
X
s
−
−
=
=
=
2. The down payment (
D
) plus the amount of the loan (
L
) must equal the total price paid
for the automobile. The monthly rate of interest is
.18/12
.015
j
=
=
and the amount
of the loan (
L
) is the present value of the payments, i.e.
(
)
48 .015
250
250 34.04255
8510.64.
L
a
=
=
=
Thus, the down payment needed will be
10,000
8510.64
$1489.36.
D
=
−
=
3.
The monthly interest rate on the first loan (
L
1
) is
1
.06/12
.005
j
=
=
and
(
)(
)
1
48 .005
500
500
42.58032
21,290.16.
L
a
=
=
=
The monthly interest rate on the second loan (
L
2
) is
2
.075/12
.00625
j
=
=
and
2
1
25,000
25,000
21,290.16
3709.84.
L
L
=
−
=
−
=
The payment on the second loan (
R
) can be determined from
12 .00625
3709.84
Ra
=
giving
3709.84
$321.86.
11.52639
R
=
=
4. A’s loan:
8 .085
20,000
Ra
=
20,000
3546.61
5.639183
R
=
=
so that the total interest would be
( )(
)
8
3546.61
20,000
8372.88.
−
=
B’s loan: The annual interest is
(
)
(
)
.085
20,000
1700.00
=
so that the total interest would be
( )(
)
8
1700.00
13,600.00.
=
Thus, the difference is
13,600.00
8372.88
$5227.12.
−
=

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