A man travels a total of 35 km partly at 4 km/h and partly at 5 km/h. If he instead travels the earlier portion at 5 km/h and the later portion at 4 km/h, he could cover 2 km more in the same total time. Find the original total time taken.

Introduction / Context:
This is a time–speed–distance puzzle using two speed segments. Swapping speeds between segments while holding the same total time allows forming equations based on time spent in each segment.

Given Data / Assumptions:

• Total distance originally = 35 km.
• Original speeds: first part at 4 km/h, second part at 5 km/h.
• When speeds are swapped, distance covered in the same total time increases by 2 km.

Concept / Approach:
Let t1 and t2 be times on the two parts. Then original distance = 4t1 + 5t2 = 35. If speeds are swapped for the same times, the distance becomes 5t1 + 4t2, which is 2 km more than before: 5t1 + 4t2 = 37. Solving these gives t1 and t2; the total time is T = t1 + t2.

Step-by-Step Solution:
Original: 4t1 + 5t2 = 35. Swapped: 5t1 + 4t2 = 37. Subtract: (5t1 + 4t2) − (4t1 + 5t2) = 37 − 35 ⇒ t1 − t2 = 2. Use 4t1 + 5t2 = 35 and t1 = t2 + 2: 4(t2 + 2) + 5t2 = 35 ⇒ 9t2 + 8 = 35 ⇒ 9t2 = 27 ⇒ t2 = 3 h. Then t1 = 5 h, so total time T = t1 + t2 = 8 h.

Verification / Alternative check:
Original distance = 4*5 + 5*3 = 20 + 15 = 35 km. Swapped distance = 5*5 + 4*3 = 25 + 12 = 37 km, which is exactly 2 km more in the same 8 hours.

Why Other Options Are Wrong:
9 h, 7 h, 4.5 h, 6 h do not satisfy both equations simultaneously; only 8 h fits both conditions.

Common Pitfalls:
Assuming the same distances rather than the same times after swapping speeds, or attempting to average speeds arithmetically instead of using segment times.

Final Answer:
8 hours