Answer:
- $1,094,748.09
- $8600.28
- 11 years, 10 months
Step-by-step explanation:
a) The compound interest formula can be used:
A = P(1 +r/n)^(nt)
where P is the principal invested at annual rate r compounded n times per year for t years.
A = $750,000(1 +.038/4)^(4·10) ≈ $1,094,748.09
Tamsyn's account will have a balance of $1,094,748.09 when she retires.
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b1) The amortization formula is good for this.
A = P(r/n)/(1 -(1 +r/n)^(-nt))
where P is the amount earning interest at annual rate r compounded n times per year for t years.
A = $1,094,748.09(0.049/12)/(1 - (1 +0.049/12)^(-12·15)) ≈ $8600.28
Tamsyn can withdraw $8600.28 per month for 15 years.
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b2) The account balance after n months will be ...
B = P(1 +r/12)^n -A((1+r/12)^n -1)/(r/12)
Filling in the known values and solving for n, we have ...
300,000 = 1,094,748.09(1.1.00408333^n) -8600.28(1.00408333^n -1)/.000408333
300,000 = 1,094,748.09(1.1.00408333^n) -2,106,191.02(1.00408333^n -1)
-1,806,191.02 = -1,011,442.93(1.00408333^n)
1.785757 = 1.00408333^n
n = log(1.785757)/log(1.00408333) = 142.3
After about 11 years 10 months, the account balance will be $300,000.
