assgn1-soln__updated_

# assgn1-soln__updated_ - re-t Afro =/4'eD l'ozzVzJ v l lo =...

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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 \$[6,945.75-2,000]=\$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.
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assgn1-soln__updated_ - re-t Afro =/4'eD l'ozzVzJ v l lo =...

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