Question

A bus travels one third of the distance with a speed of $12km{h}^{-1}$ and the remaining distance with a speed of $20km{h}^{-1}$. The average speed of the bus is ______ $km{h}^{-1}$

• A. $14$
• B. $20$
• C. $12$
• D. $16.4$

Answer 1

The correct option is D

$16.4$

Step 1: Given data

A bus travels one-third of the distance at a speed $=12km{h}^{-1}$.

It travels the remaining distance at a speed $=20km{h}^{-1}$

Step 2: Determine the time taken by the bus to travel one-third of the distance

Let the total distance traveled by bus be $dkm$.

As the bus travels one-third of the distance with a speed of $12km{h}^{-1}$, the distance covered will be $\frac{d}{3}km$.

Let the time taken by the bus for covering one-third of the distance with a speed of $12km{h}^{-1}$ be $t_1$.

As we know the formula for calculating time is $\mathit{t}\mathit{i}\mathit{m}\mathit{e}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{d}\mathbf{i}\mathbf{s}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{c}\mathbf{e}\mathbf{\left(}\mathbf{d}\mathbf{\right)}}{\mathbf{s}\mathbf{p}\mathbf{e}\mathbf{e}\mathbf{d}\mathbf{\left(}\mathbf{s}\mathbf{\right)}}$,

$\Rightarrow t_1=\frac{\left({\displaystyle \frac{d}{3}}\right)}{12}$

$\Rightarrow t_1=\frac{d}{36}h$

Step 3: Determine the time taken by the bus to travel the remaining distance

The remaining distance $=d-\frac{d}{3}=\frac{2d}{3}km$.

Let the time taken by the bus for covering the remaining distance with a speed of $20km{h}^{-1}$ be $t_{2}h$.

Similarly, the time taken by the bus to travel the remaining distance with a speed of $20km{h}^{-1}$ is,

$\Rightarrow t_2=\frac{\left({\displaystyle \frac{2d}{3}}\right)}{20}$

$\Rightarrow t_2=\frac{d}{30}h$

Step 4: Determine the average speed of the bus

The formula to calculate the average speed of the body is $\mathit{A}\mathit{v}\mathit{e}\mathit{r}\mathit{a}\mathit{g}\mathit{e}\mathbf{}\mathit{s}\mathit{p}\mathit{e}\mathit{e}\mathit{d}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{T}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}\mathbf{}\mathbf{d}\mathbf{i}\mathbf{s}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{c}\mathbf{e}}{\mathbf{T}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}\mathbf{}\mathbf{t}\mathbf{i}\mathbf{m}\mathbf{e}}$.

On applying the required values in the above formula we get,

$$\begin{gathered} Averagespeed=\frac{d}{t_1+t_2} \\ =\frac{d}{\left({\displaystyle \frac{d}{36}}\right)+\left({\displaystyle \frac{d}{30}}\right)} \\ =\frac{d}{\left({\displaystyle \frac{5d+6d}{180}}\right)} \\ =\frac{d}{\left({\displaystyle \frac{11d}{180}}\right)} \\ =\frac{180}{11} \\ =16.4km{h}^{-1} \end{gathered}$$

Hence, option D is the correct answer.