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Updated on: 28 Dec 2025, 04:27
Originally posted by EgmatQuantExpert on 09 May 2018, 08:37.
Last edited by Bunuel on 28 Dec 2025, 04:27, edited 4 times in total.
Last edited by Bunuel on 28 Dec 2025, 04:27, edited 4 times in total.
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
67% (03:40) correct
33%
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wrong
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An aircraft makes a to-and-fro journey every day between the capital cities of two countries M and N, at an average speed of 500 mph. The cities are 2200 miles apart. On a certain day while returning from N to M, the aircraft faced extreme bad weather, and therefore had to make an emergency landing in city P, which is 2000 miles away from city N and 600 miles from city M. Once the weather improved, it started from P and moved to M at normal speed. Find the average speed of the aircraft in the whole journey, starting from M and coming back to the same place, if due to bad weather its speed deceases by 100 mph.
A. 425.5 mph
B. 452.8 mph
C. 465.6 mph
D. 470 mph
E. 475.2 mph
Application of Average Speed in Distance problems - Exercise Question #4
To read the article: Application of Average Speed in Distance problems
A. 425.5 mph
B. 452.8 mph
C. 465.6 mph
D. 470 mph
E. 475.2 mph
Application of Average Speed in Distance problems - Exercise Question #4
To read the article: Application of Average Speed in Distance problems
Updated on: 14 May 2018, 01:46
Originally posted by EgmatQuantExpert on 09 May 2018, 08:38.
Last edited by EgmatQuantExpert on 14 May 2018, 01:46, edited 1 time in total.
Last edited by EgmatQuantExpert on 14 May 2018, 01:46, edited 1 time in total.
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Solution
Given:
• An aircraft makes a to-and-fro journey every day between the capital cities of two countries M and N, at an average speed of 500 mph
• Distance between M and N is 2200 miles
• On a certain day, while returning from N to M, the aircraft faced bad weather and made an emergency landing in P
• City P is 2000 miles away from city N and 600 miles away from city M
• Because of bad weather, the speed of the aircraft becomes 400 mph from N to P
To find:
• The average speed of the aircraft in the whole journey
Approach and Working:
• The whole journey is divided into three separate parts:
- o From M to N – onward journey at normal speed of 500 mph
o From N to P – 1st part of return journey at a speed of 400 mph
o From P to M – 2nd part of the return journey at a speed of 500 mph
This can be demonstrated as follows:
• Therefore, the total journey distance of the aircraft = (2200 + 2000 + 600) miles = 4800 miles
• As we already know the distances and individual speed for every part of the journey, let’s calculate the time taken for the journey
• For the onward journey M to N, the aircraft covered 2200 miles distance at 500 mph
- o Therefore, the time taken in the onward journey = \(\frac{2200}{500}\) hrs = 4.4 hrs
- o Therefore, the time taken in the 1st part of the return journey = \(\frac{2000}{400}\) hrs = 5 hrs
- o Therefore, the time taken in the 2nd part of the return journey = \(\frac{600}{500}\) hrs = 1.2 hrs
• Therefore, the average speed of the whole journey = total distance travelled/total time taken = \(\frac{4800}{10.6}\) mph = 452.8 mph
Hence, the correct answer is option B.
Answer: B
10 May 2018, 03:03
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IMO B.
Average Speed = Distance/Time
Aircraft cruises at average speed of 500mph when weather was fine while moving from M --> N and later from P --> M.
T(a) = (2200 + 600)/500 = 5.6hrs
Its speed reduced by 100mph while moving in bad weather i.e. from N --> P
T(b) = 2000/400 = 5hrs
Average Speed = (2200 + 2000 + 600)/(5 + 5.6) = 452.83
Average Speed = Distance/Time
Aircraft cruises at average speed of 500mph when weather was fine while moving from M --> N and later from P --> M.
T(a) = (2200 + 600)/500 = 5.6hrs
Its speed reduced by 100mph while moving in bad weather i.e. from N --> P
T(b) = 2000/400 = 5hrs
Average Speed = (2200 + 2000 + 600)/(5 + 5.6) = 452.83
