Question

A man travels 600 km partly by train and partly by car. It takes 8 hours and 40 minutes if he travels 320 km by train and the rest by car. It would take 30 minutes more if he travels 200km by train and rests by car. Find the speed of the train.

Answer 1

Hint:For solving this problem, let the speed of the train be x and the speed of car be y. By using the fact that speed is the ratio of distance and time, we form two equations for two different times as given in the problem. By using this methodology, we can easily solve the problem.

Complete step-by-step answer:
According to the problem, a man travels 600 km partly by train and partly by car. Let the speed of the train be x and the speed of the car be y in km/hr.
As we know that speed is defined as the ratio of distance and time. Therefore, time can be related as:
$\begin{aligned} & \text{speed}=\dfrac{\text{distance}}{\text{time}} \\\ & \text{time = }\dfrac{\text{distance}}{\text{speed}} \\\ \end{aligned}$
A person travels 320km by train and rests by car, then time taken is 8 hours and 40 minutes.
For the train, the time taken is $\dfrac{320}{x}$. For a car, the time taken is $\dfrac{600-320}{y}$.
So, after equating the total time with train and car time,we get
$\begin{aligned} & \dfrac{320}{x}+\dfrac{600-320}{y}=8+\dfrac{40}{60} \\\ & \dfrac{320}{x}+\dfrac{280}{y}=8+\dfrac{2}{3} \\\ & \dfrac{320}{x}+\dfrac{280}{y}=\dfrac{26}{3} \\\ & \dfrac{8}{x}+\dfrac{7}{y}=\dfrac{26}{3\times 40} \\\ & \dfrac{8}{x}+\dfrac{7}{y}=\dfrac{13}{60}\ldots (1) \\\ \end{aligned}$
Again, if the person travels 200 km by train and rests by car, then the time taken is 30 minutes more than the previous time. Similarly we get,
$\begin{aligned} & \dfrac{200}{x}+\dfrac{600-200}{y}=8+\dfrac{40}{60}+\dfrac{30}{60} \\\ & \dfrac{200}{x}+\dfrac{400}{y}=8+\dfrac{2}{3}+\dfrac{1}{2} \\\ & \dfrac{200}{x}+\dfrac{400}{y}=\dfrac{55}{6} \\\ & \dfrac{1}{x}+\dfrac{2}{y}=\dfrac{55}{200\times 6} \\\ & \dfrac{1}{x}+\dfrac{2}{y}=\dfrac{11}{240}\ldots (2) \\\ \end{aligned}$
Multiplying equation (2) with 8 and subtracting it with equation (1), we get
$\begin{aligned} & 8\left( \dfrac{1}{x}+\dfrac{2}{y} \right)-\left( \dfrac{8}{x}+\dfrac{7}{y} \right)=\dfrac{8\times 11}{240}-\dfrac{13}{60} \\\ & \left( \dfrac{8}{x}+\dfrac{16}{y} \right)-\left( \dfrac{8}{x}+\dfrac{7}{y} \right)=\dfrac{88-13\times 4}{240} \\\ & \dfrac{16}{y}-\dfrac{7}{y}=\dfrac{36}{240} \\\ & \dfrac{9}{y}=\dfrac{3}{20} \\\ & y=\dfrac{20\times 9}{3} \\\ & y=60km/hr \\\ \end{aligned}$
Putting in equation (2), we get

& \dfrac{1}{x}+\dfrac{2}{60}=\dfrac{11}{240} \\\ & \dfrac{1}{x}=\dfrac{11}{240}-\dfrac{2}{60} \\\ & \dfrac{1}{x}=\dfrac{11-8}{240} \\\ & \dfrac{1}{x}=\dfrac{3}{240} \\\ & x=\dfrac{240}{3} \\\ & x=80km/hr \\\ \end{aligned}$$ Therefore, the speed of the train is 80 km/hr. Note: This problem could be alternatively solved by letting $u=\dfrac{1}{x}\text{ and }v=\dfrac{1}{y}$. In this way, we obtain two linear equations in simplified form $\left( ax+by+c=0 \right)$. Now by applying simple substitution, we can obtain the answer with less complexity and reduction in calculation.