Answer:

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 }(6+

60

30

)=

2

13

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 }(

2

13

+

2

1

)=7 hours

Distance = Speed × Time

Now,

\frac{400}{x}+\frac{200}{y}=\frac{13}{2}

x

400

+

y

200

=

2

13

→ equation 1

\frac{200}{x}+\frac{400}{y}=7

x

200

+

y

400

=7 → equation 2

Multiplying Equation 2 with 2 we get

\frac{400}{x}+\frac{800}{y}=14

x

400

+

y

800

=14 → equation 3

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

\begin{gathered}\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}\end{gathered}

y

600

=14−

2

13

⇒

y

600

=

2

28−13

⇒

y

600

=

2

15

⇒y=

15

600×2

⇒y=80km/hr

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

\begin{gathered}\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}\end{gathered}

x

200

+

80

400

=7

⇒

x

200

+5=7

⇒

x

200

=7−5

⇒

x

200

=2

⇒x=

2

200

⇒x=100

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