Average Speed Over Multiple Stages

Average speed equals total distance divided by total time, and it is crucial to recognise that this is generally not the arithmetic mean of the two speeds when equal distances are covered at different rates. Because the van spends more time on the slower half, the slower speed is weighted more heavily, so the correct measure is the harmonic mean. Let the total route length be $2d$, so each half is $d$. The time for the first half is $d/40$ and for the second half is $d/60$. The total time is $d/40+d/60=d(3+2)/120=5d/120=d/24$. The average speed is therefore $\frac{2d}{d/24}=2\times 24=48\text{ km/h}$. The option 50 km/h is wrong because it naively averages 40 and 60. The option 52 km/h and 45 km/h are arbitrary values not supported by any correct time-weighting. This applies the NCERT harmonic-mean result for equal-distance segments, $v_{avg}=\frac{2v_1{v}_2}{v_1+v_2}$. As a sanity check, 48 km/h lies between the two speeds but below their arithmetic mean of 50, exactly as expected when the slower leg consumes more time.