Question

A train travels at a certain average speed for a distance of 54 km and then travels a distance of 63 km at an average of 6 km/hr more than the first speed. If it takes 3 hours to complete the total journey, what is its first speed? [CBSE 2015]

Answer 1

Let the first speed of the train be x km/h.

∴ Time taken to cover 54 km = $\frac{54}{x}$ h $\left(\mathrm{Time}=\frac{\mathrm{Distance}}{\mathrm{Speed}}\right)$

New speed of the train = (x + 6) km/h

∴ Time taken to cover 63 km = $\frac{63}{x+6}$ h

According to the given condition,

Time taken to cover 54 km + Time taken to cover 63 km = 3 h
$$\begin{gathered} \therefore \frac{54}{x}+\frac{63}{x+6}=3 \\ \Rightarrow \frac{54x+324+63x}{x\left(x+6\right)}=3 \\ \Rightarrow 117x+324=3\left(x^2+6x\right) \\ \Rightarrow 117x+324=3x^2+18x \end{gathered}$$
$$\begin{gathered} \Rightarrow 3x^2-99x-324=0 \\ \Rightarrow x^2-33x-108=0 \\ \Rightarrow x^2-36x+3x-108=0 \\ \Rightarrow x\left(x-36\right)+3\left(x-36\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow \left(x-36\right)\left(x+3\right)=0 \\ \Rightarrow x-36=0\mathrm{or}x+3=0 \\ \Rightarrow x=36\mathrm{or}x=-3 \end{gathered}$$
∴ x = 36 (Speed cannot be negative)

Hence, the first speed of the train is 36 km/h.