Higher June 2023 Paper 3 Q13
13 Charlie is driving 293 miles home.
He
- leaves at 9.00 am
- travels the first 176 miles at an average speed of 48 mph
- drives the rest of the way at an average speed of 65 mph
Will he be home by 2.30 pm?
You must show your working. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(176 \div 48\) or 3.66… or 3.67 or \(\dfrac{11}{3}\) or 3 h 40 mins | M1 | oe eg 220 mins implied by 12 40 pm |
| \((293 - 176) \div 65\) or \(117 \div 65\) or 1.8 or \(\dfrac{9}{5}\) or 1 h 48 mins | M1 | oe eg 108 mins |
| their 3.66… \(+\) their 1.8 or \(\dfrac{82}{15}\) or [5.46, 5.47] or 5 h 28 mins or [2 27 (pm), 2 28.2 (pm)] | M1dep | oe eg 328 mins dep on M2 implied by adding times eg \(9 + 3\ 40 + 1\ 48\) |
| 5.5 and [5.46, 5.47] and Yes or 5 h 30 mins and 5 h 28 mins and Yes or 330 mins and 328 mins and Yes or [2 27 (pm), 2 28.2 (pm)] and Yes | A1 | oe arrival time must be in a comparable time format |
Additional guidance
| Up to M3 may be awarded for correct work seen in multiple attempts even if not subsequently used | |
| Accept use of 24 hour clock throughout | |
| Do not accept 2 28 am as a correct arrival time | |
| \(\dfrac{176}{48} = 3.6\), \(\dfrac{117}{65} = 1.8\), \(3.6 + 1.8 = 5.4\), 2 24 pm and Yes | M1M1M1A0 |
| \(\dfrac{176}{48} = 3.7\), \(\dfrac{117}{65} = 1.8\), \(3.7 + 1.8 = 5.5\), 2 30 pm and Arrives on time | M1M1M1A0 |
| \(3.6 + 1.8 = 5.4\), 2 24 pm and Yes | M0M1M0A0 |
| \(3.7 + 1.8 = 5.5\), 2 30 pm and Arrives on time | M0M1M0A0 |