Higher November 2018 Paper 3 Q21
21 Priya and Joe travel the same 16.8 km route.
Priya starts at 9.00 am and walks at a constant speed of 6 km/h
Joe starts at 9.30 am and runs at a constant speed.
Joe overtakes Priya at 10.20 am
At what time does Joe finish the route? [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| 1(h) 20 (min) and 50 (min) or \(1\dfrac{20}{60}\) (h) or \(1\dfrac{1}{3}\) (h) or 1.33…(h) or \(\dfrac{50}{60}\) (h) or \(\dfrac{5}{6}\) (h) or 0.83…(h) | B1 | oe Journey time(s) at 10.20 am |
| \(6 \times \text{their } 1\dfrac{1}{3}\) or 8 | M1 | oe Priya’s distance at 10.20 am |
| \(\text{their } 8 \div \text{their } \dfrac{50}{60}\) or 9.6 or \(16.8 \div 8\) or 2.1 | M1dep | oe Joe’s speed in km/h Multiplier for distance comparison |
| \(16.8 \div \text{their } 9.6\) or 1.75(h) or 1(h) 45 (min) or 105 (min) or \(16.8 \div 8 \times 50\ (\div 60)\) or \(\dfrac{16.8 - \text{their } 8}{\text{their } 9.6}\) or \(\dfrac{8.8}{\text{their } 9.6}\) or 0.91(6…)(h) or 0.917(h) or 0.92(h) or 55(min) | M1dep | oe Joe’s total journey time Joe’s journey time after overtaking Priya |
| 11.15 (am) | A1 | oe eg quarter past 11 (in the morning) |
Additional guidance
| If 11.15 comes from correct method but with premature rounding eg \(8 \div 0.83 = 9.64\) \(16.8 \div 9.64 = 1.743\) h \(1.743 \times 60 = 104.58\) minutes ie 11 : 14 : 58 so 11 : 15 | B1M3A0 |
| 8 km implies | B1M1 |
| \(16.8 \div 6\) or 2.8 with no further valid working | B0M0 |