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El día 6 en un lugar de Longitud= 140º18'E son HcL= 3h 42m 10s. En ese mismo momento, calcular la HcL y Hz que tendrá un lugar de Longitud= 086º24'W.On the 6th at longitude 140º 18' E the local civil time (HcL) is 3h 42m 10s. At that same instant, calculate the HcL and zone time (Hz) at longitude 086º 24' W.
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Correct answer: d) HcL = 12h 35m 22s (5th) and Hz = 12h 20m 58s (5th).
Why it is correct
140º 18′ E is 9h 21m 12s, so HcG = 03h 42m 10s − 09h 21m 12s = 18h 20m 58s on the 5th. At 086º 24′ W (5h 45m 36s W, zone +6), HcL = 18h 20m 58s − 5h 45m 36s = 12h 35m 22s on the 5th and Hz = 18h 20m 58s − 6h = 12h 20m 58s on the 5th.
Why the others are not
- a) Subtracts the longitudes instead of adding them for opposite names.
- b) Inverts the conversion signs at both meridians.
- c) Gives the Greenwich civil time (18h 20m 58s) as the second place's HcL.
How to work it out in the exam
- Convert first to Greenwich (HcG = 18h 20m 58s on the 5th) and from there to the second place.
Where this question comes from
- Ministerio de Transportes (Madrid) · 2024 · Convocatoria de abril · CY Test 1 · question 14 Official paper · Official answer key
- Ministerio de Transportes (Madrid) · 2024 · Convocatoria de abril · CY Test 2 · question 18 Official paper · Official answer key
Pending instructor review. Machine translation, not reviewed.