Question

A small aeroplane can travel at 320 km/hr in still air. The wind is blowing at a constant speed of 40 km/hr. The total time for a journey against the wind is 135 minutes. What will be the time in minutes for the return journey with the wind? (Ignore take off and landing times for the aeroplane)
A. 94.5
B. 105
C. 108.125
D. 120

Answer 1

Hint: Distance travelled by aeroplane is the same, in both the cases aeroplane travelling against the wing and aeroplane travelling with the wind. Wind blowing against the aeroplane will resist the aeroplane and hence the speed of aeroplane is reduced. On the other hand, wind blowing with the aeroplane in return will help aeroplane travel fast. So we will find the distance using \[\text{Distance}=\text{speed}\times \text{time}\] for the first case and then use it in the second case to get the time.

Complete step-by-step answer:
We have been given that the speed of the aeroplane in still air = \[320\] \[km/hr\]
Speed of the wind blowing = \[40\] \[km/hr\]
Total time for journey against the wind = \[135\] minutes = \[\dfrac{135}{60}\] hours
Let us take the distance travelled in the journey as \[x\] \[km\].
Now, we know that speed of the aeroplane when travelling against the wind will be = \[(320-40)=280\] \[km/hr\]
Also, we know that the speed of the aeroplane travelling with the wind will be = \[(320+40)=360\] \[km/hr\]
Considering case when the aeroplane travels against the wind, applying speed, distance and time formula we have
\[\text{Distance}=\text{speed}\times \text{time}\]
\[x=280\times \dfrac{135}{60}\]
\[x=630\]\[km\]
Distance of the journey by aeroplane is calculated to be 630 km which will also be same in case of the return journey of the aeroplane.
Let us take the case when the aeroplane travels with the wind, we will apply the speed distance formula for the return journey.
In return, the aeroplane will cover 630 km with a speed of $ 360 $ $ km/hr $ and let the time of travel be t. So, using formula we have
\[\text{Distance}=\text{speed}\times \text{time}\]
\[\begin{align}
  & 630=360\times t \\
 & t=\dfrac{630}{360} \\
\end{align}\]
Converting to minutes, we have
\[t=\dfrac{630}{360}\times 60\] minutes
\[t=105\] minutes
Hence, the required time of travel of the aeroplane in return journey with the wind is 105 minutes.
Option B is the correct answer.

Note: Keep in mind when aeroplane travel with wind in that case speed of the wind is added to aeroplane’s speed and in the other case when the aeroplane travel against the speed of wind is subtracted from the aeroplane’s speed. If the student gets this concept wrong and applies it, then the solution will not be valid. Here, we converted time into hours as the speed was given in km/hr. But since we required the final answer in minutes, we could have converted speed in terms of km/min also. This might confuse some students, so it is not advised unless concepts are clear.