Question

A man travels 600km apart by train and partly by car. It takes 8 hours and 40 minutes if he travels 320 km by train and rest by car. It would take 30 minutes more if he travels 200 km by train and the rest by the car/. Find the speed of the train and by car separately.
I am back
sorry it took me a lot of time _/\_ !​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

Speed of the train be x km per hour

Speed of the car be y km per hour

Case :- 1

A man travels 600km apart by train and partly by car. It takes 8 hours and 40 minutes if he travels 320 km by train and rest by car.

Distance travelled by train = 320 km

So,

Distance travelled by car = 600 - 320 = 280 km

Time taken = 8 hr 40 min

So,

$\rm \: \dfrac{320}{x} + \dfrac{280}{y} = 8\dfrac{40}{60}$

$\rm \: \dfrac{320}{x} + \dfrac{280}{y} = 8\dfrac{2}{3}$

$\rm \: \dfrac{320}{x} + \dfrac{280}{y} = \dfrac{26}{3} - - - (1)$

Case :- 2

It would take 30 minutes more if he travels 200 km by train and the rest by the car.

Distance travelled by train = 200 km

So,

Distance travelled by car = 600 - 200 = 400 km

Time taken = 8 hr 40 min + 30 min = 9 hr 10 min

So,

$\rm \: \dfrac{200}{x} + \dfrac{400}{y} = 9\dfrac{10}{60}$

$\rm \: \dfrac{200}{x} + \dfrac{400}{y} = 9\dfrac{1}{6}$

$\rm \: \dfrac{200}{x} + \dfrac{400}{y} = \dfrac{55}{6} - - - (2)$

Let assume that

$\rm \: \dfrac{1}{x} \: = \: u - - - (3)$

and

$\rm \: \dfrac{1}{y} \: = \: v - - - (4)$

So, equation (1) and (2), can be rewritten as

$\rm \: 320u + 280v = \dfrac{26}{3} - - - (5)$

and

$\rm \: 200u + 400v = \dfrac{55}{6} - - - (6)$

On multiply equation (5) by 5 and (6) by 8, we get

$\rm \: 1600u + 3200v = \dfrac{220}{3} - - - (7)$

and

$\rm \: 1600u + 1400v = \dfrac{130}{3} - - - (8)$

On Subtracting equation (8) from (7), we get

$\rm \: 1800v = \dfrac{90}{3}$

$\rm \: v = \dfrac{1}{60}$

On substituting the value of v, we get

$\rm \: \dfrac{1}{y} = \dfrac{1}{60}$

$\rm\implies \:y \: = \: 60 \: km \: per \: hour$

On substituting the value of v in equation (5), we get

$\rm \: 320u + 280 \times \dfrac{1}{60} = \dfrac{26}{3}$

$\rm \: 320u + \dfrac{14}{3} = \dfrac{26}{3}$

$\rm \: 320u = \dfrac{26}{3} - \dfrac{14}{3}$

$\rm \: 320u = \dfrac{26 - 14}{3}$

$\rm \: 320u = \dfrac{12}{3}$

$\rm \: u = \dfrac{1}{80}$

On substituting the value of y from equation (3), we get

$\rm \: \dfrac{1}{x} = \dfrac{1}{80}$

$\rm\implies \:x \: = \: 80 \: km \: per \: hour$

Hence,

Speed of the train is 80 km per hour

Speed of the car is 60 km per hour

Answer 2

Step-by-step explanation:

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 Find

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,

$\bf \: Time = Speed \div Distance$

Case 1

$\tt \: Time taken =\frac{400}{x}+\frac{600-400}{y} \\ \tt\Rightarrow \frac{400}{x}+\frac{200}{y}=6+\frac{30}{60} \\ \tt\Rightarrow \frac{400}{x}+\frac{200}{y}=6+\frac{1}{2} \\ \tt\Rightarrow \frac{400}{x}+\frac{200}{y}=\frac{2\times6+1}{2} \\ \tt\Rightarrow \frac{400}{x}+\frac{200}{y}=\frac{13}{2}...(i)$

Case 2

Time taken = 6 hours 30 minutes + 30 minutes

=7 hours

$\tt \: Time taken =\frac{200}{x}+\frac{600-200}{y} \\ \tt\Rightarrow \frac{200}{x}+\frac{400}{y}=7...(ii) \\ \tt \: Multiplying \: equation \: (i) \: by 2 \: , we \: get, \\\tt2(\frac{400}{x}+\frac{200}{y})=2(\frac{13}{2}) \\ \tt\frac{800}{x}+\frac{400}{y}=13...(iii)$

Now ,

Subtracting equation (ii) from (iii), we get,

$\tt\frac{800}{x}+\frac{400}{y}-\frac{200}{x}-\frac{400}{y}=13-7 \\ \tt\frac{800-200}{x}=6 \\ \tt\frac{600}{x}=6 \\ \tt \: x=\frac{600}{6} \\ \tt \: x=60$

Now ,

Substituting equation (ii), we get,

$\tt\frac{200}{100}+\frac{400}{y}=7 \\ \tt \: 2+\frac{400}{y}=7 \\ \tt\frac{400}{y}=7-2=5 \\ \tt \: y=\frac{400}{5} \\ \tt \: y=80$

$\boxed{ \bf \red{Therefore \: the \: speed \: of \: the \: train \: is \: 10km/hr \: and \: the \: speed \: of \: the \: car \: is \: 80 km/hr \: respectively.}}$