Higher November 2019 Paper 2 Q20
20 \(d = \dfrac{1}{8}c^3\)
\(c = 10.9\) correct to 3 significant figures.
By considering bounds, work out the value of \(d\) to a suitable degree of accuracy.
Give a reason for your answer. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| 160 (supported) | B1 | stating bound of 10.85 or 10.95 |
| M1 | using both UB and LB to work out value of \(d\) eg [UB of \(c\)]\(^3 \div 8\) and [LB of \(c\)]\(^3 \div 8\) or gives a bound of 159.66… from correct working or gives a bound of 164.11… from correct working | |
| A1 | for 159.66… and 164.11… from correct working | |
| C1 | for 160 from 159.66… and 164.11… with a supporting reason eg “since both UB and LB round to 160” |
Additional guidance
Accept 10.949 or 10.9499… for 10.95
\(10.9 \lt \text{UB} \leqslant 10.98\)
\(10.85 \leqslant \text{LB} \lt 10.9\)
Accept bounds rounded or truncated to at least 4 sig fig