Higher June 2022 Paper 2R Q18
18 Kaidan and Sonja went on two different car journeys.
For Kaidan’s journey
distance = 80 km correct to the nearest 5 km
time = 2.7 hours correct to 1 decimal place
For Sonja’s journey
distance = 33 km correct to 2 significant figures
time = 1 hour correct to the nearest 0.1 hour
Kaidan says,
“My average speed could have been greater than Sonja’s average speed.”
By considering bounds, show that Kaidan is correct.
Show your working clearly.
(4)
| Scheme | Marks |
|---|---|
| 77.5 or 82.5 or 2.65 or 2.75 or 32.5 or 33.5 or 0.95 or 1.05 or 77500 or 82500 or 159 or 165 or 32500 or 33500 or 57 or 63 | B1 |
| eg 82.5 ÷ 2.65 (= 31.13…) or 82500 ÷ 159 (= 518.867…) or km/min or m/h | M1 |
| eg 32.5 ÷ 1.05 (= 30.95…) or 32500 ÷ 63 (= 515.873……) or km/min or m/h | M1 |
| UB K = 31132…..m/h LB S = 30952……m/h UB K = 0.51886….km/min LB S = 0.51587…km/min Answer: Shown | A1 |
| (4) | |
| (4 marks) |
Notes
B1: For a UB or LB for one of the distances or times in hours or in minutes
M1: for a method to find the upper bound of Kaidan’s average speed
eg \(UB_K \div LB_K\) where \(80 \lt UB_K \leqslant 82.5\) and \(2.65 \leqslant LB_K \lt 2.7\)
or
use of m/min to find upper bound for Kaidan’s average speed
eg \(UB_K \div LB_K\) where \(80000 \lt UB_K \leqslant 82500\) and \(159 \leqslant LB_K \lt 162\)
can use km/min or m/h
M1: indep for a method to find the lower bound of Sonja’s average speed
eg \(LB_S \div UB_S\) where \(32.5 \leqslant LB_S \lt 33\) and \(1 \lt UB_S \leqslant 1.05\)
or
use of m/min to find lower bound for Sonja’s average speed
\(LB_S \div UB_S\) where \(32500 \leqslant LB_S \lt 33000\) and \(60 \lt UB_S \leqslant 63\)
can use km/min or m/h
A1: shown with accurate figures in the same units – sufficient figures for comparison (can be truncated) but must be from correct working and UB for Kaidan and LB for Sonja selected
eg UB Kaidan = 31.13… (km/h) and LB Sonja = 30.95…( km/h)
or
UB Kaidan = 518.867…(m/min) and LB Sonja = 515.873…( m/min)
(dep on correct method)