A train travels 360 kilometres at a uniform speed. If its speed had been 5 km/h more, it would have taken 1 hour less to cover the same distance. What is the actual speed of the train in km/h?

Introduction / Context:
This is a classic average speed and time relationship problem where changing the speed changes the time taken. The distance is fixed, and we compare the actual travel time to a hypothetical travel time at a slightly higher speed. Such questions test algebraic manipulation of the formula time = distance / speed.

Given Data / Assumptions:

• Actual distance travelled = 360 km.
• Actual speed of the train = v km/h (unknown).
• Hypothetical speed = v + 5 km/h.
• At speed v + 5 km/h, the time taken is 1 hour less than at speed v.
• Speeds are uniform and no stops are considered.

Concept / Approach:
We express the original time in terms of v using time = distance / speed. Similarly, we express the new time with speed v + 5. The condition that the new time is 1 hour less than the original provides an equation relating v. Solving this equation yields the actual speed. Finally, we verify by checking both travel times.

Step-by-Step Solution:
Step 1: Let original speed be v km/h. Step 2: Original time = 360 / v hours. Step 3: New speed = v + 5 km/h, new time = 360 / (v + 5) hours. Step 4: Given condition: 360 / v - 360 / (v + 5) = 1. Step 5: Multiply through by v(v + 5) to clear denominators: 360(v + 5) - 360v = v(v + 5). Step 6: Simplify left side: 360v + 1800 - 360v = 1800. Step 7: So equation becomes 1800 = v^2 + 5v. Step 8: Rearrange: v^2 + 5v - 1800 = 0. Step 9: Factorise: (v - 40)(v + 45) = 0, so v = 40 or v = -45. Step 10: Negative speed is not meaningful, so v = 40 km/h.

Verification / Alternative check:
At 40 km/h, time = 360 / 40 = 9 hours. At 45 km/h, time = 360 / 45 = 8 hours. The difference in time is 9 - 8 = 1 hour, which matches the problem statement exactly. Therefore, 40 km/h is confirmed as the correct actual speed.

Why Other Options Are Wrong:
Speeds such as 42, 43, or 45 km/h do not satisfy the exact condition of 1 hour difference when recalculated. For example, 42 km/h and 47 km/h or 45 km/h and 50 km/h will not give the required difference with the fixed distance of 360 km. The distractor 35 km/h also fails the algebraic equation derived above.

Common Pitfalls:
Learners may attempt to guess speeds instead of forming and solving the equation, which can be time consuming and inaccurate. Another mistake is to incorrectly handle algebraic manipulation or to forget that time difference is given as 1 hour, leading to misinterpretation. Writing out the equation step by step and solving carefully prevents such errors.

Final Answer:
The actual speed of the train is 40 kmph.