You will have to work a little harder on this one

Let the time required to complete each of the two parts while covering the distance of BC be ‘x’. Therefore, BC= 80x + 100x=180x. Hence, AB=BC=180x. Time required to cover the distance of AB= 180x/60=3x. Hence, average speed= total distance/ total time So, the average speed = (180x+180x)/ (3x +x+ x) =360x/5x= 72 m/s.

Let the time taken for each of 2 parts of BC be $t$ seconds. Then first part of BC is of distance $80 m/s \times t seconds= 80t$ . Second part of BC = $100 m/s \times t seconds =100t$ . Entire BC = $180t$ . AB = BC = $180t$ , making entire AC = $360t$ . Since the particle travelled AB at $60 m/s$ , time taken for AB = $\frac{180t}{60} = 3t$ . Total time taken for AC = time taken for AB + time taken for BC = $3t+2t = 5t$ . Average velocity = $\frac{360t}{5t} =\boxed{72}m/s$

Now we have 3 velocities.

Now we take average velocity of 80m and 100m .

We have $\frac { 2xy }{ x+y }$ as formula to find average velocity .

Therefore we have

$\frac { 2\times 80\times 100 }{ 100+80 }$ =88.888888888888

Now we find average velocity of 88.888888888888 m\s and 60 m\s by above method and we get answer 71.684587 now rounding off

we get $\boxed{72}$ .