A man interest # 15,000 for two yrs at compound interest after one year his many amount to $166000. Find a) Rate of interest b) Interest for 2nd year.

The formula for the amount after $n$ years with compound interest is $A = P(1 + r)^n$. For the first year, $n=1$.

$$A_1 = P(1 + r)$$

$$\text{N}16,000 = \text{N}15,000(1 + r)$$

Divide both sides by $\text{N}15,000$:

$$\frac{16,000}{15,000} = 1 + r$$ $$\frac{16}{15} = 1 + r$$

Subtract 1 from both sides:

$$r = \frac{16}{15} - 1$$ $$r = \frac{16 - 15}{15}$$ $$r = \frac{1}{15}$$

To express this as a percentage:

$$r = \frac{1}{15} \times 100\%$$ $$r = \frac{100}{15}\%$$ $$r = \frac{20}{3}\%$$ $$r \approx 6.67\%$$

The rate of interest is $\boxed{\frac{1}{15} \text{ or } 6.67\%}$.

The amount after 1 year ($A_1$) becomes the new principal for the second year.

$$A_2 = A_1(1 + r)$$

Substitute $A_1 = \text{N}16,000$ and $r = \frac{1}{15}$:

$$A_2 = \text{N}16,000 \left(1 + \frac{1}{15}\right)$$ $$A_2 = \text{N}16,000 \left(\frac{15 + 1}{15}\right)$$ $$A_2 = \text{N}16,000 \left(\frac{16}{15}\right)$$ $$A_2 = \frac{\text{N}256,000}{15}$$ $$A_2 \approx \text{N}17,066.67$$

The interest for the 2nd year is the difference between the amount after 2 years and the amount after 1 year.

$$\text{Interest for 2nd year} = A_2 - A_1$$ $$\text{Interest for 2nd year} = \text{N}17,066.67 - \text{N}16,000$$ $$\text{Interest for 2nd year} = \text{N}1,066.67$$

The interest for the 2nd year is $\boxed{\text{N}1,066.67}$.