Mathematics Standard • Year 11 • Module 2 • Lesson 20
Timetables and Elapsed Time — Past-Paper Style
HSC Mathematics Standard 2-style writing: short-answer time conversions and elapsed-time calculations plus one extended response with explicit marking criteria.
1. Short-answer questions
1.1 A train departs at 9:48 pm and arrives at 1:23 am the next morning. Calculate the elapsed time of the journey in hours and minutes. Show all steps. 3 marks Band 3
1.2 A bus timetable shows the following stops for one service.
Wynyard 0742; Town Hall 0748; Central 0755; Redfern 0801; Newtown 0810.
(a) How long does the bus take from Wynyard to Newtown?
(b) A passenger needs to arrive at Newtown by 0815. Is this service suitable? Explain in one sentence. 3 marks Band 3–4
1.3 A pizza delivery driver is paid for two shifts in one day: 11:15 am to 2:48 pm, and again from 5:30 pm to 11:12 pm.
(a) Calculate the length (h, min) of each shift.
(b) Calculate the total hours worked across both shifts.
(c) The driver is paid $24.80 per hour. Calculate her total pay for the day. 4 marks Band 4
2. Extended response
2.1 A traveller is planning a trip from Newcastle to a regional town. The schedule is:
Leg 1: Train from Newcastle, departing at 6:42 pm, taking 2 h 38 min to Sydney Central.
Wait at Central: 47 min before catching the next service.
Leg 2: Overnight bus from Central, taking 7 h 15 min to the regional town.
The traveller has been told they must be at the destination by 8:00 am the next morning for a connecting work meeting.
(a) At what 24-hour time does the train arrive at Sydney Central?
(b) At what 24-hour time does the overnight bus depart Central?
(c) At what 24-hour time does the traveller arrive at the regional town?
(d) State whether the traveller arrives in time for the 8:00 am meeting, and by how many minutes they have to spare (or are late). Justify your answer in a sentence. 7 marks Band 5–6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — converts 6:42 pm correctly to 1842.
• 1 mark — adds 2 h 38 min correctly to give 2120.
Part (b) — 1 mark
• 1 mark — adds 47 min to 2120 to give 2207.
Part (c) — 2 marks
• 1 mark — recognises midnight crossing and counts up via 0000 (or adds 2400 then subtracts).
• 1 mark — correct arrival 0522 next morning.
Part (d) — 2 marks
• 1 mark — correct difference between 0522 and 0800 in hours/minutes.
• 1 mark — clear sentence stating arrives in time, with a numerical buffer.
Your response:
Stuck on (c)? Bus departs 2207, takes 7 h 15. 2207 + 7 h = 2907 → 0507 next day; + 15 min = 0522 next day.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Elapsed time crossing midnight (3 marks)
Sample response.
Depart 9:48 pm = 2148; arrive 1:23 am = 0123.
Crosses midnight (0123 < 2148).
Count up: 2148 → 2200 = 12 min; 2200 → 0000 = 2 h; so 2148 → 0000 = 2 h 12 min.
Then 0000 → 0123 = 1 h 23 min.
Total elapsed = 2 h 12 min + 1 h 23 min = 3 h 35 min.
Marking notes. 1 mark — correct conversion of both times to 24-hour. 1 mark — correct midnight-crossing handling (count to midnight + count past midnight). 1 mark — correct final answer with h and min units. Common error: doing 0123 − 2148 directly gives a negative — must add 2400 first.
1.2 — Bus timetable (3 marks)
(a) Sample response. Newtown 0810 − Wynyard 0742 = 28 min.
(b) Sample response. 0810 is earlier than 0815, so yes — this service arrives 5 minutes before the deadline, with time to spare.
Marking notes. (a) 1 mark — correct elapsed minutes. (b) 1 mark — correct yes/no. 1 mark — sentence with a numerical buffer (5 min). A bare "yes" with no justification scores 1/2.
1.3 — Two shifts plus pay (4 marks)
(a) Sample response.
Shift 1: 11:15 am to 2:48 pm → 1115 to 1448. Min: 48 − 15 = 33; hr: 14 − 11 = 3 → 3 h 33 min.
Shift 2: 5:30 pm to 11:12 pm → 1730 to 2312. Min: 12 − 30 → borrow 1 h: 72 − 30 = 42; hr: 23 − 1 − 17 = 5 → 5 h 42 min.
(b) Sample response. Total min = 33 + 42 = 75 = 1 h 15 min. Total hr = 3 + 5 + 1 = 9 h 15 min. Total = 9 h 15 min.
(c) Sample response. Decimal: 9 + 15/60 = 9.25 h. Pay = 9.25 × $24.80 = $229.40.
Marking notes. (a) 1 mark — both shift lengths correct. (b) 1 mark — correct sum with carry of 60-minute overflow. (c) 1 mark — decimal conversion. 1 mark — correct dollar pay.
2.1 — Newcastle → regional town journey, sample Band-6 (7 marks)
Sample Band-6 response.
(a) Train arrival at Central.
6:42 pm = 1842. [1 mark.]
Add 2 h 38 min: 1842 + 2 h = 2042; + 38 min: 42 + 38 = 80 → 80 min = 1 h 20 min → 2042 + 1 h 20 min = 2200 + 20 min = 2120 (9:20 pm). [1 mark.]
(b) Bus departure from Central.
2120 + 47 min: 20 + 47 = 67 = 1 h 7 → 2120 + 1 h 7 min = 2207 (10:07 pm). [1 mark.]
(c) Bus arrival at destination.
2207 + 7 h 15 min. 2207 + 7 h = 2907; subtract 2400 (crosses midnight) → 0507 next day. [1 mark — midnight handled.]
Add 15 min: 0507 + 15 min = 0522 next morning. [1 mark.]
(d) Arrival vs deadline.
Deadline = 0800; arrival = 0522. Buffer = 0800 − 0522. Min: 0 − 22 → borrow 1 h: 60 − 22 = 38; hr: 7 − 1 − 5 = 2. Buffer = 2 h 38 min. [1 mark — correct difference.]
Conclusion: the traveller arrives at 0522, which is 2 h 38 min before the 8:00 am meeting — they have time to spare. [1 mark — clear conclusion with numerical buffer and direction.]
Total: 7/7.
Band descriptors for marker.
Band 3: Correct conversion of times to 24-hour and correct (a); struggles with the 38 min carry in (a) or omits (d) entirely. ≈ 3 marks.
Band 4: Parts (a)–(c) correct but slips on the midnight crossing in (c) (gives 2907 as the answer instead of 0522); (d) attempted but loses the conclusion mark. ≈ 4–5 marks.
Band 5: All numerical answers correct; (d) has the correct buffer but no conclusion sentence (bare "2 h 38 min"). ≈ 6 marks.
Band 6: Complete and correct, with all times in 24-hour, midnight handled cleanly, and a clear conclusion sentence in (d) that includes the buffer and the direction (early/late). 7/7.