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.