Question

A man travels 370 km, partly by train and partly by car. If he covers 250 km by train and the rest by car, it takes him 4 hours. But, if he travels 130 km by train and the rest by car, he takes 18 minutes longer. Find the speed of the train and that of the car.

Answer 1

Let the speed of the train be x km/h and the speed of the car be y km/h.
Then, we have:
Time taken to cover 250 km by train = $\left(\frac{250}{x}\right)$ hrs
Time taken to cover 120 km by car = $\left(\frac{120}{y}\right)$ hrs (∵ Total distance = 370 km)
Total time taken = 4 hrs
∴ $\frac{250}{x}+\frac{120}{y}=4$
⇒ $\frac{125}{x}+\frac{60}{y}=2$
⇒ 125u + 60v = 2 ...(i) where $\frac{1}{x}=u\mathrm{and}\frac{1}{y}=v$
Again, we have:
Time taken to cover 130 km by train = $\left(\frac{130}{x}\right)$ hrs
Time taken to cover 240 km by car = $\left(\frac{240}{y}\right)$ hrs (∵ Total distance = 370 km)
Total time taken = 4 hours 18 minutes = $\left(4+\frac{18}{60}\right)$ hrs = $\left(4+\frac{3}{10}\right)$ hrs = $\left(\frac{43}{10}\right)$ hrs
∴ $\frac{130}{x}+\frac{240}{y}=\frac{43}{10}$
⇒ $\left(\frac{1300}{x}+\frac{2400}{y}\right)=43$
⇒ 1300u + 2400v = 43 ...(ii) Here, $\frac{1}{x}=u\mathrm{and}\frac{1}{y}=v$
On multiplying (i) by 40, we get:
5000u + 2400v = 80 ...(iii)
On subtracting (ii) from (iii), we get:
3700u = 37
⇒ u = $\frac{37}{3700}=\frac{1}{100}$
On substituting $u=\frac{1}{100}$ in (i), we get:
$125\times \frac{1}{100}+60v=2$
⇒ $\frac{5}{4}+60v=2$
⇒ $60v=\left(2-\frac{5}{4}\right)=\frac{3}{4}$
⇒ v = $\left(\frac{3}{4\times 60}\right)=\frac{1}{80}$
∴ $x=\frac{1}{u}=\frac{1}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$100$}\right.}}=100$
$y=\frac{1}{v}=\frac{1}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$80$}\right.}}=80$

∴ Speed of the train = 100 km/h
And, speed of the car = 80 km/h