A train while going slower by 15 km reaches late by 45 hours, if it travels…

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

Then:

• Reduced speed = (x − 15) km/hr

• Increased speed = (x + 60) km/hr

Let the total distance be D km.

Step 1: Form equations using time difference

According to the question:
When speed is reduced by 15 km/hr, the train becomes late by 45 hours.
So,
D/(x − 15) − D/x = 45
Taking LCM:
D[x − (x − 15)] / x(x − 15) = 45
15D / x(x − 15) = 45
D = 3x(x − 15) —— (1)
When speed is increased by 60 km/hr, the train arrives 72 hours earlier.
So,
D/x − D/(x + 60) = 72
Taking LCM:
D[(x + 60) − x] / x(x + 60) = 72
60D / x(x + 60) = 72
5D = 6x(x + 60) —— (2)

Step 2: Solve equations

From (1):
D = 3x(x − 15)
Substitute into (2):
5[3x(x − 15)] = 6x(x + 60)
15x(x − 15) = 6x(x + 60)
Divide by 3x:
5(x − 15) = 2(x + 60)
5x − 75 = 2x + 120
3x = 195
x = 65

Step 3: Find distance

Using equation (1):
D = 3 × 65 × (65 − 15)
= 3 × 65 × 50
= 9750 km