Total Walking Distance to Lay Stones

XAT 2020 · Quantitative Ability & Data Interpretation · Medium · Quantitative Aptitude

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.

  1. A.

    16400 metres

  2. B.

    4100 metres

  3. C.

    8050 metres

  4. D.

    16200 metres

  5. E.

    8200 metres

Answer

A

Explanation

Length = 200 m, stones placed 5 m apart. Number of spots per side = 2005+1=41\frac{200}{5} + 1 = 41. First stone at 0 m requires 0 walking. For remaining 40 stones on one side, distances to spots are 5,10,15,…,2005, 10, 15, \dots, 200. For each stone, round trip distance = 2×d2 \times d. Sum of distances for one side = 2×(5+10+⋯+200)=2×402×(5+200)=2×4100=82002 \times (5 + 10 + \dots + 200) = 2 \times \frac{40}{2} \times (5 + 200) = 2 \times 4100 = 8200 m. For both sides = 2×8200=164002 \times 8200 = 16400 m.

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