Question

A bus maintains an average speed of \[60{\rm{km/hr}}\] while going from P to Q and maintains an average speed of \[90km/hr\] while coming back from Q to P. The average speed of the bus is
A) \[75km/hr\]
B) \[72km/hr\]
C) \[70km/hr\]
D) \[30km/hr\]

Answer 1

Hint:
Here we will use the basic formula of the speed to get the average speed of the bus. Firstly we will assume the distance between the points P to Q. Then we will find the time taken by the bus while going and time taken by bus while coming back. We will then divide the total distance travelled by the bus by the total time taken by the bus to get the average speed of the bus.

Complete step by step solution:
Let us assume the distance between the points P and Q be \[x\,km\]
Speed of the bus while going is \[60km/hr\].
Speed of the bus while coming back is \[90km/hr\].
Now we will calculate the time taken by the bus while going is
So,
Time taken by the bus while going \[ = \dfrac{{{\rm{distance}}}}{{{\rm{speed}}}}\]
\[ \Rightarrow \]Time taken by the bus while going \[ = \dfrac{x}{{60}}hr\]
Similarly, now we will calculate the time taken by the bus while coming back.
Time taken by the bus while coming back \[ = \dfrac{x}{{90}}hr\]
We know that the average speed is the ratio of the total distance to the total time take to travel that distance. Therefore, we get
Average speed of the bus \[ = \dfrac{{2x}}{{\dfrac{x}{{60}} + \dfrac{x}{{90}}}}\]
Simplifying the expression, we get
\[ \Rightarrow \]Average speed of the bus \[ = \dfrac{{2x}}{{\dfrac{{3x + 2x}}{{180}}}} = \dfrac{{2x}}{{\dfrac{{5x}}{{180}}}}\]
Further simplifying the terms, we get
\[ \Rightarrow \]Average speed of the bus \[ = \dfrac{{180 \times 2}}{5} = 72km/hr\]
Hence, \[72km/hr\] is the average speed of the bus.

So, option B is the correct option.

Note:
We should know that the distance is equal to the product of the speed and time. Another important thing is that the units of the distance should be the same. Distance is measured as the speed with which the object is traveling in some particular amount of time. Speed is a scalar quantity and we must not confuse speed with the velocity. Velocity is a vector quantity as velocity is the ratio of the displacement to the time taken to travel that distance of displacement.
\[\begin{array}{l}{\rm{speed}} = \dfrac{{{\rm{distance}}}}{{{\rm{time}}}}\\{\rm{velocity}} = \dfrac{{{\rm{displacement}}}}{{{\rm{time}}}}\end{array}\]