Multiple choice

For a practice session, army personnel in a two-kilometre long group are walking one behind the other at a uniform speed. A commanding officer starts from the back, walks forward till he reaches the front. Then he returns with the same speed till he reaches the last army-man. When he gets to the last army-man, the last army-man has reached exactly the spot where the army-man at the front was when the officer started. What is the total distance travelled by the officer in this whole operation?

  1. 4.24 km

  2. 4.42 km

  3. 4.62 km

  4. 4.83 km

  5. 4.92 km
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let officer speed be v and group speed be u. Time to reach front: t1 = 2 / (v - u). Time to return to back: t2 = 2 / (v + u). Total time T = t1 + t2. In time T, the group moves distance uT = 2. So u = 2/T. Total distance = vT. Solving leads to v/u = sqrt(2) + 1. Total distance = 2 * (v/u) = 2 * (1 + sqrt(2)) = 4.828 km.

AI explanation

Let the group's speed be u and the officer's speed be v; the group moves 2 km in the total time T, meaning uT equals 2. The officer's time moving forward is 2/(v-u) and backward is 2/(v+u), so T equals 2/(v-u) plus 2/(v+u). Solving v squared equals u squared plus 2uv gives v/u equals 1 plus the square root of 2, so the officer's total distance of vT equals 2 multiplied by 2.414, which is 4.83 km.

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