A fast train takes 3 hours less than a slow train for a journey of 900 km. If…

For two journeys covering the same distance, time = distance ÷ speed. When one speed is a fixed amount more than another and their time difference is given, express both times in terms of one variable, equate the difference to the given value, and solve the resulting quadratic — keeping only the positive (physically valid) root.

1. Let the fast train's speed be x km/hr; since the slow train is 15 km/hr slower, its speed is (x − 15) km/hr.

2. Time for the slow train over 900 km = 900/(x − 15) hours; time for the fast train = 900/x hours.

3. The fast train takes 3 hours less, so 900/(x − 15) − 900/x = 3.

4. Dividing throughout by 3: 300/(x − 15) − 300/x = 1.

5. Combining over the common denominator x(x − 15): 300x − 300(x − 15) = x(x − 15), i.e. 4500 = x2 − 15x.

6. Rearranging gives the quadratic x2 − 15x − 4500 = 0.

7. Factoring: (x − 75)(x + 60) = 0, so x = 75 or x = −60.

8. Speed cannot be negative, so x = 75 km/hr (the fast train); the slow train's speed is x − 15 = 60 km/hr.

Verification: at 60 km/hr the slow train covers 900 km in 900/60 = 15 hours; at 75 km/hr the fast train covers it in 900/75 = 12 hours. The difference, 15 − 12 = 3 hours, matches the condition given — confirming the slow train's speed is 60 km/hr.