Two trains starts simultaneously from two stations 300km apart and move towards each other. The speed of one train is more than the other by 20km/h. If the distance between the trains after two hours is 20km, find the speed of the trains.
Answers
SOLUTION:
Let us assume that the points A and B denote the two stations, I.e Station A and Station B respectively.
So,
From Station A the first train starts and from Station B the second train starts. They both start simultaneously towards each other (given).
Now,
Let the speed of the first train be 'a' km/h
So, according to the question,
The speed of the second train will be (a + 20)km/h
Finding the distances -
The distance the first train traveled in 1 h = a km
And, the distance of the first train traveled in 2 h = '2a' km
Also,
The distance of the second train in 1 h = (a + 20) km
So, distance traveled by second train in 2 h = 2(a + 20) = (2a + 40) km
⇒ The distance between two stations = 300km (given)
So,
The Distance of first train + Distance of second train + 20 = 300
⇒ 2a + (2a + 40) + 20 = 300
⇒ 4a + 60 = 300
⇒ 4a = 300 - 60
[ ∵ Transporting 60 to RHS]
⇒ 4a = 240
⇒ a = 240 ÷ 4
[ ∵ Transporting 4 to RHS]
∴ a = 60
Thus,
The speed of the first train = a km/h = 60 km/h
And The speed of the second train = a + 20 km/h = 60+20 = 80 km/h
- - - - -
Verification:
The distance traveled by first train in 2 hours = 2(60) = 120 km
And
The distance traveled by second train in 2 hours = 2(80) = 160 km
The distance between two trains after two hours = 300 - (120 + 160)
= 300 - 280
= 20 km
(As same as given in the question)
Hence, The speed of the trains are
1st train = 60 km/h
And 2nd train = 80 km/h
1st train be x km H
second train be X+20 km H
so
1st train 2h = 2x
2nd train 2h = 2x+40
now
2x + 2x+40 + 20 = 300
4x = 240
x = 60
1st train is 60kmh
and other 60+20=80kmh