A man travels from point A in one state to point B in another state by a train moving at a speed of 90 kmph and he returns to the starting point A by another train moving at a speed of 72 kmph. What was the average speed of the trains for the whole journey?

Let the distance between point A and point B be $d$. Speed from A to B ($v_1$) = 90 kmph. Time taken from A to B ($t_1$) = Distance / Speed = $d/ 90$. Speed from B to A ($v_2$) = 72 kmph. Time taken from B to A ($t_2$) = Distance / Speed = $d/ 72$. The total distance for the round trip = $d + d = 2d$. The total time for the round trip = $t_1 + t_2 = d/90 + d/72$. Average speed = Total Distance / Total Time. Average speed = $(2d) / (d/90 + d/72)$. We can factor out $d$ from the denominator: Average speed = $(2d) / (d(1/90 + 1/72)) = 2 / (1/90 + 1/72)$. To calculate $1/90 + 1/72$, find the least common multiple (LCM) of 90 and 72. $90 = 2 \times 3^2 \times 5$ $72 = 2^3 \times 3^2$ LCM(90, 72) = $2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360$. $1/90 + 1/72 = 4/360 + 5/360 = 9/360 = 1/40$. Average speed = $2 / (1/40) = 2 \times 40 = 80$ kmph.