Saturday, July 10, 2004


Mean machines

Aditya Rishi

THE argument reaches nowhere for a long time. Finally, they agree to travel as follows: Laurel rides the donkey for some time, then ties it down and continues walking in the same direction, while Hardy follows him on foot. Eventually, he finds the donkey and rides it until he overtakes Stan. Then he gives the donkey to Stan. At long last, they reach outside Brushwood Gulch. Stan and Olly are still not speaking to each other. Finally, Stan speaks: "Assuming that we walk with the same speed and ride faster than we walk, what is the speed of our travel?" Olly: "You think anyone can answer that?"

The answer is the harmonic mean (the reciprocal of the arithmetic average of the reciprocals) of the walking and biking speeds. This is because both walking and biking cover equal distances. If that doesn’t convince you, let us look at the various answers that we have.

Vimal Jit Kaur: "Hi Stan, the speed of your travel is equivalent to your walking speed per hour. Suppose you both started from point A at the crossroad and you have to reach point N at Brushwood Gulch. You rode the donkey first from point A and tied it down at point B, while Hardy followed you on foot upto point B. From point B, you started walking and Hardy overtook you at point C and gave the donkey to you to ride.

Since you both walk with the same speed, your travelling speed from point A to C is your walking speed from A to B, plus the walking speed from B to C. This way, you reached at point N by walking alternatively continuing from point C to D then D to E and so on. Since you walked N distance in H hours, your travelling speed is your walking speed, that is N/H miles per hour.

Arun Verma: "if Laurel ties the donkey at distance ‘d’ and by then Hardy has travelled ‘a’, the time taken ‘t’[let]=d/x=a/y; and they both eventually meet at ‘2d’, taking total time ‘t+t*’where t*=d/y so applying the formula:

velocity=total distance covered/time taken=2d/d/x+d/y=2xy/x+y.

The answer then is 2xy/x+y."

Brahm Kiran Singh: "Let the speed of riding = v and that of walking = u. Let S1 be the distance travelled in T1 time by the rider (Stan) = vT1. In the meantime, distance travelled by Olly = uT1. Therefore, distance between the positions of Stan and Olly after time T1 = (v-u) T1 = S2. Time taken by Olly to reach the position where donkey is tied = (v-u) T1/u=T2, in which time, both Stan and Olly cover a distance = S2. Also, while riding the donkey, Olly will overtake Stan in time T3 = T1 (found by simple calculations), and in that time, distance travelled by Stan = uT1 = S3. So, the average speed of the two = (S1 + S2 + S3) / (T1 + T2 + T3) = [v+ (v-u) +u]/T1 + T2 + T3 = 2uv/(u+v). Thus, the average speed of the two = harmonic mean of their speed of walking and riding."

Ankush Aggarwal (from Nahan): "The speed of their travelling is equal to their walking speed because all through their journey, one or the other was walking. They never rode together, yet reached the town at the same time, so it is clear that their travelling speed is equal to their walking speed."

Dr Vikas Handa (from Giessen0: "If the riding speed is R and is faster than the walking speed W, then the speed of travel is 2*W*R/(W+R). Riding for time t1 will cover distance equal to walking for t1 and then for t2 for both of them. This gives us the relation R*t1 = W*t2 + W*t1. Putting the value of t1 from this equation in the formula for average speed, (R*t1 + W*t2 + W*t1)/(2*t1 + t2) gives us the final result of twice the product of walking and riding speed divided by their sum." They are all mean machines, but Brahm is the meanest of them all. You know what I mean. (Write at Mind Games, The Tribune, or