Q3
. Label time
t
= 0 for March 1, 1998. At time 2 (i.e., March 1, 2000), Robin’s balance
is $6
,
300(1 + 5%)
2
= $6
,
945
.
75. After the withdrawal of $2,000 on March 1, 2000, the
balance becomes $[6,945.752,000]=$4,945.75.
Let
t
1
be the time length needed for $4,945,75 to grow to $5,000 at an annual interest
rate of 3.5%. Then, we have the equation
4
.
945
.
75(1 + 3
.
5%)
t
1
= 5
,
000
.
Solving the above equation, we obtain
t
1
= 0
.
317117. This implies that the balance
after 2.317117 years (since t=0), is $5,000. Therefore, on March 1, 2005 (i.e., t=7),
right before the deposit of $1,000, the balance will be
$(5
,
000)(1 + 5%)
7

2
.
317117
= $6
,
283
.
43
.
On March 1, 2010 (t=12), the balance is therefore $(6
,
283
.
43 + 1
,
000)(1 + 5%)
12

7
=
$9
,
295
.
71.