Mathematics Standard • Year 11 • Module 2 • Lesson 22
Latitude, Longitude, and Location — Past-Paper Style
HSC Mathematics Standard 2-style writing on global coordinates and the 15°-rule: short answers and one extended response with explicit marking criteria.
1. Short-answer questions
1.1 A coastal town is at longitude 105°E. Using the 15°-per-hour rule, calculate its theoretical UTC offset, and find the local time at this town when it is 0700 UTC on a Monday. 3 marks Band 3
1.2 Two cities have coordinates A (40°N, 90°E) and B (40°N, 75°W).
(a) State the hemispheres of each.
(b) Calculate the total longitude difference and the time difference in hours using only the 15°-rule. 3 marks Band 3–4
1.3 Sydney (≈ 33.9°S, 151.2°E) and Santiago, Chile (≈ 33.5°S, 70.7°W) are often called "sister cities" because they share a similar latitude.
(a) Explain in one sentence why latitude similarity gives them similar climates.
(b) Calculate the total longitude difference between them.
(c) Use the 15°-rule to calculate the corresponding time difference, to the nearest hour. 4 marks Band 4
2. Extended response
2.1 A live science webinar is scheduled to start at 7:00 pm Friday local time at a research base in Hawaii, longitude approximately 158°W (the base is observed using only its longitude — assume no daylight saving).
(a) Use the 15°-per-hour rule to calculate the base's theoretical UTC offset, in hours and minutes (to the nearest minute).
(b) Find the UTC time and day when the webinar starts.
(c) A viewer in Sydney (longitude ≈ 151°E) wishes to watch. Use the 15°-rule (no official AEST or AEDT) to calculate Sydney's theoretical UTC offset, to the nearest minute.
(d) Find the Sydney local time and day when the webinar starts. State whether this is a convenient time for the viewer (e.g. within 7 am – 9 pm). 7 marks Band 5–6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — 158 ÷ 15 ≈ 10.53 h (or equivalent).
• 1 mark — expresses as h:min (10 h 32 min) and applies negative sign (UTC−10:32).
Part (b) — 1 mark
• 1 mark — UTC = 1900 + 10 h 32 min = 0532 Saturday (next day).
Part (c) — 1 mark
• 1 mark — 151 ÷ 15 ≈ 10 h 4 min → UTC+10:04.
Part (d) — 3 marks
• 1 mark — adds Sydney offset to UTC arrival: 0532 + 10 h 4 min = 1536 Saturday.
• 1 mark — correct local time and day stated.
• 1 mark — clear conclusion sentence with convenience comment (e.g. "≈ 3:36 pm Saturday, a convenient afternoon time").
Your response:
Stuck? Always work via UTC: base local → UTC → Sydney local.How did this worksheet feel?
What I'll revisit before next class:
1.1 — 105°E offset and local time (3 marks)
Sample response.
Offset = 105 ÷ 15 = 7 h east → UTC+7.
Local time = 0700 UTC + 7 h = 1400 Monday (2:00 pm).
Marking notes. 1 mark — correct division to get 7 h. 1 mark — east → positive sign. 1 mark — correct local time and day with units.
1.2 — Two cities at 40°N (3 marks)
(a) Sample response. A (40°N, 90°E) → Northern, Eastern; B (40°N, 75°W) → Northern, Western.
(b) Sample response. Total longitude difference = 90 + 75 = 165°. Time difference = 165 ÷ 15 = 11 hours (A is east of B → A is 11 h ahead).
Marking notes. (a) 1 mark — both hemispheres correctly stated. (b) 1 mark — correct Δλ. 1 mark — correct Δt with direction.
1.3 — Sydney and Santiago sister cities (4 marks)
(a) Sample response. Both cities are about 34° south of the equator, so they receive similar amounts of solar energy through the year, giving them comparable temperature ranges and seasonal patterns.
(b) Sample response. Δλ = 151.2 + 70.7 = 221.9°.
(c) Sample response. Δt = 221.9 ÷ 15 ≈ 14.79 ≈ 15 hours (to the nearest hour). Sydney (east) is ahead of Santiago.
Marking notes. (a) 1 mark — links latitude to climate. (b) 1 mark — correct Δλ. (c) 1 mark — correct division. 1 mark — correctly rounded to the nearest hour with direction stated.
2.1 — Hawaii webinar viewed in Sydney, sample Band-6 (7 marks)
Sample Band-6 response.
(a) Hawaii UTC offset from longitude.
158 ÷ 15 = 10.5333… h = 10 h + 0.5333 × 60 min ≈ 10 h 32 min. [1 mark — division correct.]
West → negative: UTC−10:32. [1 mark — sign and h:min form.]
(b) UTC time when webinar starts.
Hawaii local 7:00 pm Friday = 1900. UTC = 1900 + 10 h 32 min = 2932 → subtract 2400 → 0532 next day = 0532 UTC Saturday. [1 mark.]
(c) Sydney offset from longitude.
151 ÷ 15 = 10.0666… h = 10 h + 0.0666 × 60 ≈ 10 h 4 min → UTC+10:04. [1 mark.]
(d) Sydney local time and day.
Sydney = 0532 UTC + 10 h 4 min = 0532 + 10 h = 1532; + 4 min = 1536 Saturday (≈ 3:36 pm Saturday). [1 mark — offset applied. 1 mark — local time + day stated.]
Conclusion: a Sydney viewer can watch the webinar at approximately 3:36 pm on Saturday — a convenient mid-afternoon time, well within the 7 am – 9 pm window. [1 mark — clear convenience sentence.]
Total: 7/7.
Band descriptors for marker.
Band 3: Correctly converts 158 ÷ 15 ≈ 10.5 h and applies the negative sign but does not express in h:min form; (b) and (c) attempted but with rounding inconsistencies. ≈ 3 marks.
Band 4: Correct offsets in both directions; correctly handles the midnight crossing in (b); (d) attempted but does not state the day or omits the convenience comment. ≈ 4–5 marks.
Band 5: Full numerical solution including day; conclusion sentence in (d) is a bare time without convenience comment. ≈ 6 marks.
Band 6: Complete and correct throughout, with hours-and-minutes for both offsets, midnight rollover handled cleanly, and a clear conclusion sentence that includes a convenience judgement. 7/7.