A man travels 600km partly by train and partly by car.If he covers 120km by train and rest by car ; it takes him 8 hours.But if he travels 200km by train and and by car ; he takes 20 minutes longer.Find the speed of the train and that of the car.
speed of train ---- t km/h
speed of car ----- c km/h
120/t + 480/c = 8
15/t + 60/c = 1
times tc
tc = 15c + 60t
200/t + 400/c = 25/3
times tc
(25/3)tc = 200c + 400t
tc = (3/25)(200c + 400t) = 24c + 48t
24c + 48 = 15c + 60t
9c = 12t
c = 4t/3
sub into 15/t + 60/c = 1
15/t + 60/(4t/3) = 1
15/t + 45/t = 1
t = 60 , then
c = 4(60)/3 = 80
the car's speed is 80 km/h, the train's is 60 km/h
check my arithmetic
Thanks :)
To solve this problem, let's assume the speed of the train is "x" km/h and the speed of the car is "y" km/h.
In the first scenario, the man covers 120 km by train and the remaining distance (600 - 120 = 480 km) by car. The total time taken is 8 hours.
Using the formula: Time = Distance / Speed
Time taken by the train = 120 km / x km/h
Time taken by the car = 480 km / y km/h
Given that the total time taken is 8 hours, we can form the equation:
120/x + 480/y = 8 ... (Equation 1)
In the second scenario, the man covers 200 km by train and the remaining distance (600 - 200 = 400 km) by car. The total time taken is 8 hours and 20 minutes, which can be converted to 8.3333 hours.
Using the same formula, we have:
Time taken by the train = 200 km / x km/h
Time taken by the car = 400 km / y km/h
Given that the total time taken is 8 hours and 20 minutes, we can form the equation:
200/x + 400/y = 8.3333 ... (Equation 2)
Now, we have a system of equations (Equation 1 and Equation 2) that we need to solve simultaneously to find the values of x (train speed) and y (car speed).
We can use algebraic methods such as substitution or elimination to solve this system of equations. Once we find the values of x and y, those will be the speeds of the train and car, respectively.