Question

A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km/hr more than its original speed. If it takes 3 hours to complete the total journey, what is the original average speed?

Answer 1

Let the original average speed of the train be x km/hr.

Time taken for the first part of the journey

Distance = 63 km

Speed = x km/hr

t₁ = `63/x` hrs

Time taken for the second part of the journey:

Second distance = 72 km

New speed = x + 6x km/hr

t₂ = `72/(x + 6)` hrs

According to the question,

Total time = 3 hours

t₁ + t₂ = 3

`63/x + 72/(x + 6)` = 3

Solve the equation:

Multiply both sides by x(x + 6):

63(x + 6) + 72x = 3x(x + 6)

63x + 378 + 72x = 3x² + 18x

135x + 378 = 3x² + 18x

3x² – 117x – 378 = 0

x² – 39x – 126 = 0

x² – 42x + 3x – 126 = 0

(x + 3)(x – 42) = 0

x = 42 or x = −3

As x can’t be negative, so x is 42 km/hr.

The original speed of the train is 42 km/hr.