Question

A train after travelling $150\text{km},$ meets with an accident and then proceeds at $\frac{3}{5}$ of its former speed and arrives at its destination $8$ hours late. Had the accident occurred $360\text{km}$ further, it would have reached the destination $4$ hours late. What is the total distance travelled by the train?

• A. $870\text{km}$
• B. $510\text{km}$
• C. $1740\text{km}$
• D. $600\text{km}$

Answer 1

The correct option is A $870\text{km}$

Let the original speed of the train be $x\text{km/hr}.$
Then, $\frac{360}{\left(\frac{3}{5}x\right)}-\frac{360}{x}=4$
$\Rightarrow \frac{600}{x}-\frac{360}{x}=4\Rightarrow 4x=240$
$\Rightarrow x=60$

Let the total distance travelled by the train be $y\text{km}.$
Then, $[\frac{150}{60}+\frac{(y-150)}{(\frac{3}{5}\times 60)}]-\frac{y}{60}=8$
$\Rightarrow \frac{5}{2}+\frac{(y-150)}{36}-\frac{y}{60}=8$
$\Rightarrow 2y=1740$
$\Rightarrow y=870$
Hence, required distance $=870\text{km}$