Mathematics
27.04.2022 09:14
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Orlando invests $1000 at 6% annual interest compounded daily and Bernadette invests

Orlando invests $1000 at 6% annual interest compounded daily and Bernadette invests $1000 at 7%
simple interest. After how many whole years will Orlando's investments be worth more than
Bernadette's investments?
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Geo777
Geo777
4,4(15 marks)

6 Years

Step-by-step explanation:

Orlando invests $1000 at 6% annual interest compounded daily.

Orlando's investment = A=1000(1+\frac{0.06}{365})^{(365\times t)}

Bernadette invests $1000 at 7% simple interest.

Bernadette's investment = A = 1000(1+0.07×t)

By trail and error method we will use t = 5

Bernadette's investment will be after 5 years

1000(1 + 0.07 × 5)

= 1000(1 + 0.35)

= 1000 × 1.35

= $1350

Orlando's investment after 5 years

A=1000(1+\frac{0.06}{365})^{(365\times 5)}

   = 1000(1+0.000164)^{1825}

  = 1000(1.000164)^{1825}

  = 1000(1.349826)

  = 1349.825527 ≈ $1349.83

After 5 years Orlando's investment will not be more than Bernadette's.

Therefore, when we use t = 6

After 6 years Orlando's investment will be = $1433.29

and Bernadette's investment will be = $1420

So, after 6 whole years Orlando's investment will be worth more than Bernadette's investment.

kenken2583
kenken2583
4,8(11 marks)

In 10 years she'll have approximately $21865.4 in her account.

Step-by-step explanation:

When an amount is compounded continuously its value over time is given by the following expression:

v(t) = v(0)*e^{rt}

Applying data from the problem gives us:

v(10) = 12000*e^{(0.06*10)}\\v(10) = 12000*e^{0.6}\\v(10) = 21865.4

In 10 years she'll have approximately $21865.4 in her account.

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