An aircraft makes a to-and-fro journey every day between the capital cities of two countries M and N, at an average

[BTGmoderatorDC]

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


OA B

Source: e-GMAT

[ravinikhil]

m -> n
500 mph
2000 m
4.4 hrs

n-> p
2000 m
400 mph
5 hrs

p->m
600 m
500 mph
1.2 hr

total - 4800 / 10.6 avg speed
452.8 mph

[swerve]

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


OA B

Source: e-GMAT

Average Speed \(= \dfrac{\text{Distance}}{\text{Time}}\)

Aircraft cruises at average speed of \(500\)mph when weather was fine while moving from \(M \rightarrow N\) and later from \(P \rightarrow M\).

\(T(a) = \dfrac{2200 + 600}{500} = 5.6\)hrs

Its speed reduced by \(100\)mph while moving in bad weather i.e. from \(N \rightarrow P\)

\(T(b) = \dfrac{2000}{400} = 5\)hrs

Average Speed \(= \dfrac{2200 + 2000 + 600}{5 + 5.6} = 452.83\)mph

[Scott@TargetTestPrep]

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


OA B

Solution:

The plane traveled a total of 2200 + 2000 + 600 = 4800 miles.

It took 2200/500 = 4.4 hours to travel from M to N.
It took 2000/400 = 5 hours to travel from N to P.
It took 600/500 = 1.2 hours to travel from P to M.

We use the formula: average speed = total distance / total time. Therefore, the average speed for the round trip was 4800 / (4.4 + 5 + 1.2) = 4800/10.6 ≈ 452.8 mph.

Answer: B