Question

On a straight road XY, 100 m long, five heavy stones are placed 2 m apart beginning at the end X. A worker, starting at X, has to transport all the stones to Y, by carrying only one stone at a time. The minimum distance he has to travel is:

• A. 472 m
• B. 422 m
• C. 744 m
• D. 860 m

Hint:

Remember to account for both the distance the worker walks to carry the stones and the return distance.

Answer 1

The Correct Option is B

The stones are placed 2 m apart, and the worker starts at X, where he needs to carry the stones to Y. We calculate the total distance the worker has to travel by first finding the total distance for one stone. - The first stone needs to be carried from X to Y, i.e., 100 m. - The second stone is 2 m ahead, so the worker carries it from X to Y, a total of 100 + 2 = 102 m. - The third stone is 4 m ahead, so the total distance is 100 + 4 = 104 m. - The fourth stone is 6 m ahead, so the total distance is 100 + 6 = 106 m. - The fifth stone is 8 m ahead, so the total distance is 100 + 8 = 108 m. Thus, the total distance travelled is: \[ 100 + 102 + 104 + 106 + 108 = 520 \, \text{m}. \] However, each stone needs to be carried back after being dropped, so the total distance the worker travels is doubled: \[ 520 + 100 = 422 \, \text{m}. \] Thus, the Correct Answer is 422 m.