Hi Fly 5K748 Flight Status Today

International flight Hi Fly 5K748 takes off from Mauritius (MRU) Mauritius to Mumbai (BOM) India. It's operated by Hi Fly. The plane leaves Sir Seewoosagur Ramgoolam International Airport at 01:00 Indian/Mauritius. The flight is expected to land at Chhatrapati Shivaji International Airport at None Asia/Kolkata. The flight will last about 5 hours 36 minutes.

5K748 - HFY748 Hi Fly (5K)
MRU Mauritius Mauritius
Flight MRU to BOM
BOM Mumbai India
Contact Airlines It may not operate on the date requested
Departure Sir Seewoosagur Ramgoolam International Airport
IATA: MRUICAO: FIMP
Scheduled 01:00
None None
Arrival Chhatrapati Shivaji International Airport
IATA: BOMICAO: VABB
Scheduled None
None None
5K748 Detail
  • Type: International Flight
  • Flight Duration: 5 hours 36 minutes
  • Flight Distance: 4675 kms / 2905 miles
Airline
  • Hi Fly
  • IATA: 5K
  • ICAO: HFY
  • Operating Days: Monday, Tuesday, Wednesday, Friday, Saturday, Sunday
Passenger Services
  • Service type: None
  • Seats: None
  • Freight capacity: None
  • Passenger classes: None
More Detail
  • Aircraft: Airbus A340-312
  • Callsign: MAU748
  • Departure Timezone: Indian/Mauritius
  • Arrival Timezone: Asia/Kolkata
  • Current Time in Mauritius: Wednesday 2026-01-21 14:33 PM
  • Current Time in Mumbai: Wednesday 2026-01-21 16:03 PM

Flight routes similar to Hi Fly 5K748

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Air India AI6748
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Air Mauritius MK110
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Air India AI6759
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None

16:40

Hi Fly Flight 5K748 FAQs

What is the scheduled flight duration for Hi Fly 5K748 flight?

On average, nonstop flight takes 5 hour(s) 36 minutes, with the flight distance of 4675 km (2905 miles).

What type of aircraft is used for the Hi Fly 5K748 flight?

All Hi Fly 5K748 flights are operated using Airbus A340-312 aircraft.

How many Hi Fly 5K748 flights are operated a week?

6 flights per week. The Flight Hi Fly 5K748 is operated on Monday, Tuesday, Wednesday, Friday, Saturday, Sunday.