Higher November 2019 Paper 6 Q5
5 Kay invests £1500 in an account paying 3% compound interest per year.
Neil invests £1500 in an account paying \(r\)% simple interest per year.
At the end of the 5th year, Kay and Neil’s accounts both contain the same amount of money.
Calculate \(r\).
Give your answer correct to 1 decimal place. [6]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 3.2 nfww | 6 | M3 for \(1500 \times 1.03^5\) or M2 for \(1500 \times 1.03^k\) where \(k\) is 2, 3 or 4 or M1 for 1.03 soi perhaps by 1545 AND M2 for \(\frac{\textit{their}\ 1738.91 - 1500}{5 \times 1500}\) [× 100] oe or M1 for (their 1738.91 – 1500) ÷ 5 or for (their 1738.91 – 1500) ÷ 1500 Alternative (not using a base amount) M5 for [\(r\) =] \((1.03^5 - 1) \div 5\) or M4 for \(1.03^5 - 1\) or M3 for \(1.03^5\) or M2 for \(1.03^k\) (where \(k\) is 2, 3 or 4) or M1 for 1.03 | Condone 3.2% as final answer soi by 1738 to 1739 soi by [2 yr =] 1591[.35], [3 yr =] 1639[.09..] or [4 yr =] 1688.[26…] their 1738.91 must come from a valid attempt to find compound interest for at least 2 years M2 soi by 0.0317 to 0.032 or soi by 3.17 to 3.19 M1 soi by 47.6[0] to 47.8[0] or soi by [0].1586 to 0.1594 |