A man travels 600 km partly by train and partly by car. If he covers 400 km by train and the rest by car, it takes him 6 hours and 30 minutes. But, if he travels 200 km by train and the rest by car, he takes half an hour longer. Find the speed of the train and that of the car.
Given:
A man travels 600 km partly by train and partly by car. If he covers 400 km by train and the rest by car, it takes him 6 hours and 30 minutes. But, if he travels 200 km by train and the rest by car, he takes half an hour longer.
To do:
We have to find the speed of the train and car respectively.
Solution:
Total distance $=600\ km$.
Let the speed of the train be $x$ km/hr and the speed of the car be $y$ km/hr.
We know that,
Time $=$ Speed $\div$ Distance
In the first case, he takes 6 hours and 30 minutes if he travels 400 km. by train and the rest by car.
Time taken $=\frac{400}{x}+\frac{600-400}{y}$
$\Rightarrow \frac{400}{x}+\frac{200}{y}=6+\frac{30}{60}$
$\Rightarrow \frac{400}{x}+\frac{200}{y}=6+\frac{1}{2}$
$\Rightarrow \frac{400}{x}+\frac{200}{y}=\frac{2\times6+1}{2}$
$\Rightarrow \frac{400}{x}+\frac{200}{y}=\frac{13}{2}$.....(i)
In the second case, he takes 30 minutes more if the travels 200 km by train and the rest by car.
Time taken $=$ 6 hours 30 minutes $+$ 30 minutes
$=7 hours$
Time taken $=\frac{200}{x}+\frac{600-200}{y}$
$\Rightarrow \frac{200}{x}+\frac{400}{y}=7$......(ii)
Multiplying equation (i) by 2, we get,
$2(\frac{400}{x}+\frac{200}{y})=2(\frac{13}{2})$
$\frac{800}{x}+\frac{400}{y}=13$....(iii)
Subtracting equation (ii) from (iii), we get,
$\frac{800}{x}+\frac{400}{y}-\frac{200}{x}-\frac{400}{y}=13-7$
$\frac{800-200}{x}=6$
$\frac{600}{x}=6$
$x=\frac{600}{6}$
$x=100$
Substituting $x=100$ in equation (ii), we get,
$\frac{200}{100}+\frac{400}{y}=7$
$2+\frac{400}{y}=7$
$\frac{400}{y}=7-2=5$
$y=\frac{400}{5}$
$y=80$
Therefore, the speed of the train is $100\ km/hr$ and the speed of the car is $80\ km/hr$ respectively.
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