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In problems where we talk about 2 different speeds, we can calculate the average speed with the following formula: 2*(v1*v2) / (v1+v2). Why does it not work when we talk about three different speeds?
3* (v1*v2*v3) / (v1+v2+v3)
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AVG SPEED = Total Distance / Total Time

T=D/S
Given that the rout consists of 3 segments, each of length d
t1=d/60
t2=d/120
t3=d/60

TOT time = d/24 (d/60+d/120+d/60)
TOT Distance =3d because each segment counts d and we have 3 of them (d+d+d)

AVG SPEED= 3d *24/d --> 72 km/h­
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Why do we consider here that there is no time interval where the train is stationary between each successive segment? In a DS question would it not have led to E as the answer?
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Why do we consider here that there is no time interval where the train is stationary between each successive segment? In a DS question would it not have led to E as the answer?
nbharti

Why stationary time isn't considered:

The problem states the route "consisted of \(3\) segments" and gives the train's speed "in these segments." On the GMAT, we work exclusively with the information provided. Unless stationary time is explicitly mentioned, we don't introduce it.

Here, we have everything needed:
  • Each segment has the same length (call it \(d\) km)
  • Speeds in segments: \(60\) km/h, \(120\) km/h, \(60\) km/h

Calculation:
  1. Total distance \(= 3d\)
  2. Total time \(= \frac{d}{60} + \frac{d}{120} + \frac{d}{60} = \frac{2d + d + 2d}{120} = \frac{5d}{120}\)
  3. Average speed \(= \frac{3d}{\frac{5d}{120}} = 3d \times \frac{120}{5d} = 72\) km/h

Regarding your DS question:

This would NOT lead to answer \(E\) even in DS format. The information provided is sufficient to determine a unique answer. For \(E\) to be correct, you'd need genuinely missing information, like: "The train stopped for an unknown duration between segments."

Hope this helps! If it helps further, you can practice time and work problems here under "Word Problems"
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