Question

A man travelled 300 km by train and 200 km by taxi, and he completed this journey in 5 hours and 30 minutes. However, if he travels 260 km by train and 240 km by taxi, it will take 6minutes more to complete the journey. The speed of the train is:

Answer 1

Let the speed of train is x km/h and speed of taxi is y km/h.

a/c to question,

A man travelled 300 km by train and 200 km by taxi, and he completed this journey in 5 hours and 30 minutes.

that means, time taken by train + time taken by bus = (5 + 1/2) hrs

or, 300/x + 200/y = 11/2 .....(1)

again, if he travels 260 km by train and 240 km by taxi, it will take 6minutes more to complete the journey.

that means, time taken by train + time taken by bus = 5 hrs + 30minutes + 6 minutes

or, 260/x + 240/y = (5 + 36/60) hrs

or, 260/x + 240/y = 28/5 .....(2)

multiplying 6 with equation (1) and 5 with equation (2) and then subtracting equation (2) from (1) ,

that means, 6(300/x + 200/y) - 5(260/x + 240/y) = 6 × 11/2 - 5 × 28/5

or, 1800/x - 1300/x = 33 - 28

or, 500/x = 5

or, x = 100

hence, speed of train is 100km/h

Answer 2

Answer:

Speed of train is 100km/hr

Step-by-step explanation:

Let the speed of train = x km/hr

Let the speed of taxi = ykm /hr

()

5hours 30minutes = 55/10hr

According to the question ;

Journey 300 km by train and 200 km by taxi takes 5hr 30min

(covert the unit of minutes into hours)

(5hr 30min = 5hr + 0.5hr = 55/10)

$\frac{300}{x} + \frac{200}{y} = \frac{55}{10}$

$\frac{300}{x} + \frac{200}{y} = \frac{11}{2}$

.....eq (1)

If they cover 260 km by train and 240 km by taxi, it will take 6minutes more to complete the journey.

time taken before + 6 min

(coverting the units into hours)

= (55/10 + 6/10)hr

$\frac{260}{x} + \frac{240}{y} = \frac{55}{10} + \frac{6}{10}$

$\frac{260}{x} + \frac{240}{y} = \frac{56}{10}$

......eq (2)

taking eq (1)

$\frac{300}{x} + \frac{200}{y} = \frac{11}{2}$

$\frac{1}{x} = ( \frac{11}{2} - \frac{200}{y}) \frac{1}{300}$

......eq (3)

substituting equation 3 in equation 2

we get :

$\frac{260}{300} ( \frac{11}{2} - \frac{200}{y} ) + \frac{240}{y} = \frac{56}{10}$

$\frac{26}{30}( \frac{11}{2} - \frac{200}{y} ) + \frac{240}{y} = \frac{28}{5}$

$\frac{286}{60} - \frac{5200}{30y} + \frac{240}{y} = \frac{28}{5}$

$\frac{143}{30} - \frac{520}{3y} + \frac{240}{y} = \frac{28}{5}$

$\frac{ - 520}{3y} + \frac{720}{3y} = \frac{28}{5} - \frac{143}{30}$

$\frac{200}{3y} = \frac{5}{6}$

200 X 6 = 5 X 3y

1200 = 15y

1200/15 =y

y = 80 km/hr

Therefore, speed of taxi is 80 km per hour.

Substituting the value of x in equation 3 we get

1/80 = (11/2 - 200/y)1/3

x = 100km/hr