Question

How much more would $8,000 earn in four years compounded daily at 5% than compounded annually at 5%?

Answer 1

Answer:

Given the statement:

8,000 earn in four years compounded daily at 5%

To find the amount we use formula:

$A = P(1+\frac{r}{n})^{nt}$

where P is the principal , A is the amount , n is number of times compounded per year and t is the time in year.

Here, Principal(P) = $8000, r = 5% and n = 365

Substitute these given values we get;

$A_1= 8000(1+\frac{5}{365})^{365 \cdot 4}$

$A_1= 8000 \cdot 1.000137^{1460}$

$A_1= 8000 \cdot 1.22141$

Simplify:

$A_1= \ 9771.28$

To find the Interest we use formula:

$I_1= A_1-P$

$I_1= 9771.28 -8000 = \ 1771.28$

It is also given that:

8,000 earn in four years compounded annually at 5%.

Here, P = $8000, r = 5% , t =4 year and n = 1

Using the same formula to calculate the amount:

$A_2 = 8000(1+\frac{5}{1})^{1 \cdot 4}$

$A_2= 8000(1.05)^4$

Simplify:

$A_2= \ 9724.05$

To find the Interest :

$I_2= A_2 - P$

$I_2= 9724.05 - 8000= \ 1724.05$

Then;

$I_1-I_2 = 1771.28-1724.05 = \ 47.23$

Therefore, $47.23 more would  $8,000 earn in four years compounded daily at 5% than compounded annually at 5%