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A military column of 1 km length marches with a constant speed of 6 kmph. A courier from the back of the column is sent to the head of the column by bike to deliver a message. Arrived at the front, he instantly returns. When he returns again at the back, the column has traveled a distance of 1 km since the start of his errand. The courier drove at a constant speed. How fast did the courier bike?
Explanation :
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As Let Point A and B is 1Km Distance.
Let Point A Soldier Column of 1 Km long standing. A Bike starts to enter from Back into this column, which means Biker is 1 Km Back from Point A. Now this Column reaches Point B( 1 Km far from A) and during this Biker reaches to again Back of Column which means at Point A.
Soldier Column will reach Point B in 1/6 hr.
1km ÷ 6 km/hr
and during this time Biker travelled = 1(1 km far from point A) + 1 ( Distance from A & B) + 1 ( Distance from B to A) = 3 km.
So Biker Speed = 3 ÷ 1/6= 18 Km/ hr.
The relative speed of the courier (compared to the column) is v-6 on the journey to the front, and v+6 on the way back. The distance travelled (relatively to the column) by the courier is in both cases 1 km, therefore the total time he is biking is 1 /(v - 6) + 1/(v + 6)
The column travels 1 km in 10 minutes (1/6 hour), therefore
1 /(v - 6) + 1/(v + 6) = 1/6
This can be written as a quadratic equation (v+6) + (v-6) = (v-6) × (v+6) / 6 with as only allowed solution v = 6 + 6 × sqrt(2) km/h (approximately 14.49 km/h).
