Question: A man is laying stones, from start to end, along the two sides of a 200-meter-walkway. The stones are to be laid 5 meters apart from each other. When he begins, all the stones are present at the start of the walkway. He places the first stone on each side at the walkway’s start. For all the other stones, the man lays the stones first along one of the walkway’s sides, then along the other side in an exactly similar fashion. However, he can carry only one stone at a time. To lay each stone, the man walks to the spot, lays the stone, and then walks back to pick another. After laying all the stones, the man walks back to the start, which marks the end of his work. What is the total distance that the man walks in executing this work? Assume that the width of the walkway is negligible.
Similarly, till last rock he had to travel 400 m.
Total sum would be 10 + 20 + 30 + ... + 400 = 40/2 × 410 = 8200 m.
Similarly, on the other side he will travel 8200 m.
Total 8200 + 8200 = 16400.
Hence, option (d).