Foundation November 2023 Paper 3 Q18
18 Nina invests £540 at a simple interest rate of 2% per year.
Kareem invests £540 at a compound interest rate of 2% per year.
Work out the difference in value between the two investments at the end of 5 years.
You must show your working. [6]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [£]2.2[0] with correct working | 6 | B5 for answer 2 or 2.203 to 2.204 with correct working OR M2 for [simple] [£] [540 +] \(\dfrac{540 \times 2 \times 5}{100}\) oe soi 594 or M1 for \(\dfrac{540 \times 2}{100}\) oe soi 10.8[0] and M2 for [compound] [£] \(540 \times 1.02^5\) oe soi 596.2[0] or M1 for \(540 \times 1.02^k\) oe (\(k\) positive integer) If 0 or 1 awarded, instead award SC3 for answer 2.2[0] or \(-2.2\)[0] OR SC2 for an answer that rounds to 2.2[0] or to \(-2.2\)[0] | Correct working requires evidence of at least M1 and M1 With correct working, accept \(-2.2\)[0] for 6 marks and \(-2.203\) to \(-2.204\) for B5 May be implied by 54 nfww See appendix for non-calculator methods with values not 54, 594 or 10.8 May be \(540 \times 1.02^5 - 540\) soi 56.2… Implied by 561.8… or 573.05… or 584.5… etc with no working or insufficient working |
Appendix
Non Calculator methods for percentages.
Labels only
This is when labels such as 10% = are used. If only labels are used the final answer scores full marks if it is correct.
Condone a numerical slip if the answer is correct.
If there is an error in the values and so the final answer is incorrect this cannot score method marks
e.g. Find 65% of 80
| Method scoring M1A1 | 10% = 8 5% = 4 50% = 40 65% = 52 ✓ M1A1 | 10% = 8 5% = 5 ✖ condone this slip as answer correct 50% = 40 65% = 52 ✓ M1A1 |
| Method scoring M0A0 | 10% = 8 5% = 6 ✖ Do not condone this slip as answer incorrect 50% = 40 65% = 54 ✖ M0 |
Build up method
This is where the candidate finds the percentages to build up to the required value but shows the operations used.
e.g. Find 65% of 80
10% = \(80 \div 10 = x\)
5% = \(x \div 2 = y\)
50% = \(x \times 5 = z\)
65% = \(x + z + y\)
Because the operations have been shown and they are correct, if there is an error in one of \(x\), \(y\) or \(z\), method marks can still be earned