Charles invests $3000{\text{ USD}}$ in a bank that offers compound interest at a rate of $3.5\%$ per annum, compounded half-yearly.

Calculate the number of years that it takes for Charles’s money to double.

$6000 = 3000{\left( {1 + \frac{{3.5}}{{200}}} \right)^{2n}}$     (M1)(A1)

Note: (M1) for substituting values into a compound interest formula, (A1) for correct values with a variable for the power.

$n = 20{\text{ years}}$     (A1)     (C3)

Note: If $n$ used in formula instead of $2n$, can allow as long as final answer is halved to get $20$.

[3 marks]

Part (a) on simple interest was answered well - common errors being $0.04$ in the numerator as well as $100$ in denominator, and using $6000$ as the interest. Part (b) was not well done. Candidates struggled with interest that was not compounded yearly although such questions have been asked on previous papers.