Question

The distance between Akola and Bhusawal is 168 km. An express train

takes 1 hour less than a passenger train to cover the distance. Find the

average speed of each train if the average speed of the express train is

more by 14 km/hr than the speed of the passenger train.​

Answer 1

Here is the answer

Hope It helps you

Answer 2

Let the speed of Express train be "x", and that of Passenger train be "y".

Total distance = 168 KM.
(Distance between Akola and Bhusawal is 168 km)

As we know, Speed = Distance/Time
→ Time = Distance/Speed.

So, According to Question:
Express train takes 1 hour less than a passenger train to cover the distance
i.e., 168/y = 168/x - 1 _____(i)

and, The average speed of the express train is more by 14 km/hr than the speed of the passenger train.
i.e., x = y + 14 ________(ii)

Solving equations (i) and (ii), we get:
(Put x = y + 14 from equation (ii) in (i), we get)

168/y = 168/(y+14) - 1
→ 168/y - 168/(y+14) = -1
→ 168(y+14) - 168y]/y(y+14) = -1
→ 168y - 168.14 - 168y = -y(y+14)
→ -2352 + y(y+14) = 0
→ y² + 14y - 2352 = 0
→ y² + 56y - 42y - 2352 = 0
→ y(y + 56) - 42(y + 56) = 0
→ (y + 56) (y - 42) = 0
→ y = 42, -52

"y" is speed, so it can't be negative.
So, y = 42

Now, put y = 42 in equation (ii), we get:
x = 42 + 14 = 56

Hence, average speed of Express train is 56 KM/Hour, and that of Passenger train is 42 KM/Hour.

Thankyou!!!