Question: A savings account gathers compound interest at a rate of 5% per annum. The amount of money in the account over the first two years is shown below. The expression below shows how to find the amount of money in the account after $n$ years. Which numbers should replace $A$ and $B$?
Start £600.00
After 1 year £630.00
After 2 years £661.50
£ $A \times B^n$
1. **State the problem:** We want to find the values of $A$ and $B$ in the expression $$A \times B^n$$ that models the amount of money in a savings account after $n$ years with compound interest.
2. **Given data:**
- Initial amount (start): £600.00
- After 1 year: £630.00
- After 2 years: £661.50
3. **Formula for compound interest:**
$$\text{Amount after } n \text{ years} = P \times (1 + r)^n$$
where $P$ is the principal (initial amount) and $r$ is the annual interest rate as a decimal.
4. **Identify $A$ and $B$:**
- $A$ corresponds to the initial amount $P$.
- $B$ corresponds to the growth factor $(1 + r)$.
5. **Calculate $B$:**
Given the interest rate is 5% per annum,
$$r = 0.05$$
So,
$$B = 1 + r = 1 + 0.05 = 1.05$$
6. **Calculate $A$:**
The initial amount is £600.00, so
$$A = 600$$
7. **Verification:**
- After 1 year:
$$A \times B^1 = 600 \times 1.05 = 630$$
- After 2 years:
$$A \times B^2 = 600 \times (1.05)^2 = 600 \times 1.1025 = 661.5$$
This matches the given data.
**Final answer:**
$$A = 600, \quad B = 1.05$$