Question

1.Find the amount in an account where $200 is invested at 3.5% compounded quarterly for a period of 3 years.

Answer 1

Part 1 :

P = $200

r = 3.5% or 0.035

t = 3

n = 4

Compound interest formula = $P(1+\frac{r}{n})^{nt}$

= $200(1+\frac{0.035}{4})^{3*4}$

= $200(1.00875)^{12}$

= $200\times1.1102$

= 222.04

So, money in the bank will be = 200+222.04 = $422.04

Part 2 :

P = $200

r = 3.5% or 0.035

t = 3

Continuous compound interest formula is =

A = $Pe^{rt}$

Putting e =  2.71828 we get.

A = $200(2.71828)^{0.035*3}$

= $200(2.71828)^{0.105}$

= 200*1.11071 = 222.14

Hence, amount in account after 3 years will be= 200+222.14 = $422.14