Question

Q1 a) Kamahi invests some money which earns compound interest every year. His investment

Answer 1

Without any more information, it sounds like interest is compounded continuously, in which case the value of the investment A is given by

$A=Pe^{rt}$

where P is the principal investment, r is the interest rate, and t is the number of years.

At the end of the second year (t = 2), the value is A = 60,500, and after the fourth year (t = 4), the value is A = 73,205. So solve the system

$\begin{cases}60,500=Pe^{2r}\\73,205=Pe^{4r}\end{cases}$

for r. You can eliminate P by dividing

$\dfrac{Pe^{4r}}{Pe^{2r}}=\dfrac{73,205}{60,500}\implies e^{2r}=\dfrac{121}{100}$

Take the logarithm (log here means natural log) of both sides to get

2r = log(121/100)

r = 1/2 log(121/100)

r ≈ 0.0953

So the interest rate is about 9.53%.

Answer 2

Answer: 2=2

Step-by-step explanation: