Question
Download Solution PDFA car driver travels to a place 150 km away at an average speed of 50 km/hr and returns at 30 km/hr. His average speed for the whole journey in km/hr is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
Distance to the place = 150 km
Speed while going = 50 km/hr
Speed while returning = 30 km/hr
Formula Used:
Average speed = Total distance travelled / Total time
Calculation:
Total distance = Distance to the place + Distance back = 150 + 150 = 300 km
Time taken while going = Distance to the place / Speed while going = 150 / 50
Time taken while going = 3 hours
Time taken while returning = Distance back / Speed while returning = 150 / 30
Time taken while returning = 5 hours
Total time = Time taken while going + Time taken while returning = 3 + 5
Total time = 8 hours
Average speed = Total distance / Total time
Average speed = 300 / 8
⇒ Average speed = 37.5 km/hr
The average speed for the whole journey is 37.5 km/hr.
Last updated on Jul 11, 2025
->The OSSC Junior Stenographer Final Answer Key is out and can be downloaded till 13th July 2025.
-> The OSSC Odisha Junior Stenographer Exam Date 2025 was released for Advt No. 1744/OSSC which will be conducted on 27th July 2025.
->Earlier, OSSC Junior Stenographer 2025 Notification was released for 67 stenographer vacancies.
->The application window was open from 4th April 2025 till 3rd May 2025.
-> Candidates must go through the OSSC Odisha Junior Stenographer Previous Year Papers to understand the trend of the exam.