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Q3. Label time t= 0 for March 1, 1998. At time 2 (i.e., March 1, 2000), Robinâ€™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 t1be 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%)t1= 5,000.
Solving the above equation, we obtain t1= 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%)7âˆ’2.317117 = $6,283.43.
On March 1, 2010 (t=12), the balance is therefore $(6,283.43 + 1,000)(1 + 5%)12âˆ’7=
$9,295.71.
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