Question

2. A person travels 600 km partly by train and partly by car. If he covers 400 km by train and the rest by car he takes 6 hours 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 car and that of the train. 

Answer 1

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Answer 2

The speed of the train is 100 km/hr and speed of the car is 80 km/hr.

Solution:

Let the speed of the train be ‘x’ km/hr and the speed of the car be ‘y’ km/hr.

It is given that he travels 400 km partly by train and the rest i.e. (600-400) = 200 km by car

To travels this distance he takes 6 hours 30 minutes which is equal to $\left(6+\frac{30}{60}\right)=\frac{13}{2} \text { hours }$

Also it is given that he travels 200 km by train and the rest i.e. (600-200) = 400 km by car and the time taken is half an hour longer i.e. $\left(\frac{13}{2}+\frac{1}{2}\right)=7 \text { hours }$

Distance = Speed × Time

Now,

$\frac{400}{x}+\frac{200}{y}=\frac{13}{2}$  → equation 1

$\frac{200}{x}+\frac{400}{y}=7$  → equation 2

Multiplying Equation 2 with 2 we get

$\frac{400}{x}+\frac{800}{y}=14$  → equation 3

Subtracting [Equation 3] from [Equation 2] we get,

$\begin{array}{l}{\frac{600}{y}=14-\frac{13}{2}} \\\\ {\Rightarrow \frac{600}{y}=\frac{28-13}{2}} \\\\ {\Rightarrow \frac{600}{y}=\frac{15}{2}} \\\\ {\Rightarrow y=\frac{600 \times 2}{15}} \\\\ {\Rightarrow y=80 \mathrm{km} / \mathrm{hr}}\end{array}$

Now substituting the value of y in [Equation 2] we get

$\begin{array}{l}{\frac{200}{x}+\frac{400}{80}=7} \\\\ {\Rightarrow \frac{200}{x}+5=7} \\\\ {\Rightarrow \frac{200}{x}=7-5} \\\\ {\Rightarrow \frac{200}{x}=2} \\\\ {\Rightarrow x=\frac{200}{2}} \\\\ {\Rightarrow x=100}\end{array}$

Thus the speed of the train is 100 km/hr and speed of the car is 80 km/hr.