| Date | May 2014 | Marks available | 2 | Reference code | 14M.1.sl.TZ1.12 |
| Level | SL only | Paper | 1 | Time zone | TZ1 |
| Command term | Calculate | Question number | 12 | Adapted from | N/A |
Question
Ludmila takes a loan of 320 000 Brazilian Real (BRL) from a bank for two years at a nominal annual interest rate of 10%, compounded half yearly.
Write down the number of times interest is added to the loan in the two years.
Calculate the exact amount of money that Ludmila must repay at the end of the two years.
Ludmila estimates that she will have to repay \({\text{360}}\,{\text{000}}\) BRL at the end of the two years.
Calculate the percentage error in her estimate.
Markscheme
4 (A1) (C1)
[1 mark]
\(320\,000{\left( {1 + \frac{{10}}{{2 \times 100}}} \right)^{2 \times 2}}\) (M1)(A1)
Note: Award (M1) for substituted compound interest formula, (A1) for correct substitutions.
OR
\({\text{N}} = 2\)
\({\text{I}}\% = 10\)
\({\text{PV}} = - 320000\)
\({\text{P }}/{\text{ Y}} = 1\)
\({\text{C }}/{\text{ Y}} = 2\) (A1)(M1)
Note: Award (A1) for \({\text{C }}/{\text{ Y}} = 2\) seen, (M1) for correctly substituted values from the question into the finance application.
OR
\({\text{N}} = 4\)
\({\text{I}}\% = 10\)
\({\text{PV}} = - 320000\)
\({\text{P }}/{\text{ Y}} = 2\)
\({\text{C }}/{\text{ Y}} = 2\) (A1)(M1)
Note: Award (A1) for \({\text{C }}/{\text{ Y}} = 2\) seen, (M1) for correctly substituted values from the question into the finance application.
amount to repay \( = 388962\) (A1) (C3)
Note: Award (C2) for final answer \(389000\) if \(388962\) not seen previously.
[3 marks]
\(\left| {\frac{{360\,000 - 388\,962}}{{388\,962}}} \right| \times 100\) (M1)
Note: Award (M1) for correctly substituted percentage error formula.
\( = 7.45{\text{ (% ) }}(7.44597 \ldots )\) (A1)(ft) (C2)
Notes: Follow through from their answer to part (b).
[2 marks]
