Find the value of the investment after 3 years.

Find the difference in the final value of the investment if the interest was compounded quarterly at the same nominal rate.

\(FV = 5000{\left( {1 + \frac{{6.25}}{{1200}}} \right)^{3 \times 12}}\)   (M1)(A1)

Note: Award (M1) for substituted compound interest formula, (A1) for correct substitutions.

OR

\(N = 3\)
\(I\%  = 6.25\)
\(PV =  - 5000\)
\(P/Y = 1\)
\(C/Y = 12\)     (M1)(A1)

Note: Award (A1) for \(C/Y = 12\) seen, (M1) for other correct entries.

OR

\(N = 36\)
\(I\%  = 6.25\)
\(PV =  - 5000\)
\(P/Y = 12\)
\(C/Y = 12\)     (M1)(A1)

Note: Award (A1) for \(C/Y = 12\) seen, (M1) for other correct entries.

\( = 6028.22\)     (A1)     (C3)

Note: The answer should be given correct to two decimal places or the final (A1) is not awarded.

\(FV = 5000{\left( {1 + \frac{{6.25}}{{400}}} \right)^{3 \times 4}}\)     (M1)

Note: Award (M1) for correctly substituted compound interest formula.

OR

\(N = 3\)
\(I\%  = 6.25\)
\(PV = - 5000\)
\(P/Y = 1\)
\(C/Y = 4\)     (M1)

Note: Award (M1) for all correct entries seen.

OR

\(N = 12\)
\(I\%  = 6.25\)
\(PV = - 5000\)
\(P/Y = 4\)
\(C/Y = 4\)     (M1)

Note: Award (M1) for all correct entries seen.

\(FV = 6022.41\)     (A1)
\({\text{Difference}} = 5.80\)     (A1)(ft)     (C3)

Notes: Accept \(5.81\). This answer should be given correct to two decimal places or the final (A1) is not awarded unless this has already been penalized in part (a). Follow through from part (a).
Notes: Illustrating use of GDC notation acceptable in this case only. However on P2 an answer given with no working would receive G2.