Archive | December, 2012

Army Of Ants

5 Dec

Question credited to original poster @FB

‎(QUANT.) There is a 50 cm long line of ants moving ahead. The last ant in the line had an urgent message to give to the queen ant, which is moving at first position. so, while the line is moving , the last ant runs ahead, reaches the first ant (the queen) and passes on the message. and without stopping it runs back to its original position.(assume that no time is lost in giving the message). during this time, the whole line has moved ahead by 50 cm. the question is how much distance did the last ant cover in that time? assume that it ran the whole distance at uniform speed.

Solution I could manage:

Travel-1 (time taken ‘t1’ time units)
1. Let ‘x’ be the distance travelled by the ant army for a period of ‘t1’ time units.
2. The single ant would’ve travelled (x+50)cm for the same ‘t1’ time units. And hence it managed to deliver the message.

Travel-2 (time taken ‘t2’ time units)
3. During the time ‘t2’ taken by the single ant during return journey, the ant army would’ve travelled (50-x)cm. This is because, during the whole time (t1+t2), the ant army managed to cross 50cm (from the problem).
4. Hence, the distance travelled by the single ant, on its return journey, ‘y’ should be equal to (50+x – 50), which is ‘x’cm.

Let the velocity of the army of ants is ‘v’ and the velocity of the single ant be ‘u’

Equation-1) vt1 = x
Equation-2) ut1 = x + 50

Equation-3) vt2 = 50 – x
Equation-4) ut2 = x
Equation-1/Equation-3 => t1/t2 = x/(50-x)
Equation-2/Equation-4 => t1/t2 = (x+50)/x

Hence, x/(50-x)=(x+50)/x
So,
x^2 = (x+50)(50-x)
x^2 = 50^2-x^2
2*x^2 = 50^2

Therefore, x = 50/√2cm

Hence, the single ant travelled 50+x+x distance in the overall run for (t1+t2) time units and that is, 50 + 50/√2 + 50/√2 = 120.7106781cm

equations-1 equations-2equations-3ant-diagram