john invested n$ 20 000 on 1/06/2009 at 9% p. a. interest compounded monthly

a) what is the value of the investment by 1/06/2019?

b) if he only takes the interest earned on this investment on this date and invests this at 10.5% p. a. ( compounded continuously), how much can he expect back by 1/06/2025

  a) $49,027.14

  b) $54,501.67

Step-by-step explanation:

a) The account balance multiplier for interest rate r compounded n times per year for t years is ...

  multiplier = (1 +r/n)^(nt)

For your numbers, r=.09, n=12, t=10, this is 2.45135708, so the account balance after 10 years will be ...

  $20,000 × 2.45135708 ≈ $49,027.14

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b) All but $20,000 of the above balance is interest, so the amount invested for the second period is $29,027.14.

The multiplier for a continuously compounded interest rate r for t years is ...

  multiplier = e^(rt)

For your numbers, r=.105 and t=6, so the multiplier is e^.63 ≈ 1.8776106 and the account balance for an investment of $29,027.14 will be ...

  $29,027.14 × 1.8776106 ≈ $54,501.67

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step-by-step explanation:

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