Higher November 2017 Paper 2 Q15
15 Mirek invests £6000 at a compound interest rate of 1.5% per year.
He wants to earn more than £1000 interest.
Work out the least time, in whole years, that this will take. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| 1.015 | M1 | oe eg 101.5% or \(1 + \dfrac{1.5}{100}\) Implied by 6090 |
| \(6000 \times 1.015^n\) for any positive integer \(n \gt 1\) | M1dep | oe Implied by 6181.(…) |
| 11 | A1 | If showing trials for 10 and/or 11 years, must have \(6000 \times 1.015^{10} = 6963.(\ldots)\) and/or \(6000 \times 1.015^{11} = 7067.(\ldots)\) or 7068 If showing totals from year on year for 10 and/or 11 years, must have (Y10) [6963.21, 6963.30] and/or (Y11) [7067.65, 7067.75] |
| Alternative method 2 | ||
| 1.015 | M1 | oe eg 101.5% or \(1 + \dfrac{1.5}{100}\) Implied by 6090 |
| Evaluates \(1.015^n\) for any positive integer \(n \gt 1\) and \(7000 \div 6000\) or 1.166… or 1.167 or 1.17 | M1dep | |
| 11 | A1 | If showing trials for \(n = 10\) and/or 11 must have \(1.015^{10} = [1.160, 1.161]\) and/or \(1.015^{11} = [1.177, 1.178]\) |
Additional guidance
| Values for working year on year Y1 \(6000 \times 1.015 = 6090\) Y2 \(6090 \times 1.015 = 6181.35\) Y3 \(6181.35 \times 1.015 = [6274.07, 6274.08]\) Y4 \([6274.07, 6274.08] \times 1.015 = [6368.18, 6368.20]\) Y5 \([6368.18, 6368.20] \times 1.015 = [6463.70, 6463.73]\) Y6 \([6463.70, 6463.73] \times 1.015 = [6560.65, 6560.69]\) Y7 \([6560.65, 6560.69] \times 1.015 = [6659.05, 6659.11]\) Y8 \([6659.05, 6659.11] \times 1.015 = [6758.93, 6759.00]\) Y9 \([6758.93, 6759.00] \times 1.015 = [6860.31, 6860.39]\) Y10 \([6860.31, 6860.39] \times 1.015 = [6963.21, 6963.30]\) Y11 \([6963.21, 6963.30] \times 1.015 = [7067.65, 7067.75]\) | |
| Answer 11 with no working | M2A1 |
| \(1000 \div 90 = 11.1\) Answer 11 | Zero |