A train traveled a certain route that consisted of 3 segments, all

Question

A train traveled a certain route that consisted of 3 segments, all having exactly the same length. If the average speeds of the train in these segments were 60km/h, 120km/h, 60km/h, what was the average speed of the train, in km/h, over the entire route?

A. 68
B. 72
C. 80
D. 82
E. Cannot be determined

A. 68
B. 72
C. 80
D. 82
E. Cannot be determined

Bunuel · Accepted Answer

Let's assume the length of each segment was 120km, so the entire route was 360km. Then covering these legs would take 120/60 = 2 hours, 120/120 = 1 hour, and 120/60 = 2 hours. Thus the average speed over the entire route would be (total distance)/(total time) = 360/5 = 72 km/h.

Answer: B.

Bunuel · Answer

If a, b, and c represent the speeds at which three   equal   distances are covered, then the average speed can be calculated using the formula:

\frac{3abc}{ab + bc + ca} .

However, I believe that depending solely on formulas may not be the most effective approach for GMAT preparation.

ruis · Answer

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)
Thanks in advance!

demawtf · Answer

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

nbharti · Answer

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?

egmat · 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:
 
 Total distance  = 3d  
 Total time  = \frac{d}{60} + \frac{d}{120} + \frac{d}{60} = \frac{2d + d + 2d}{120} = \frac{5d}{120}  
 Average speed  = \frac{3d}{\frac{5d}{120}} = 3d 	imes \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"

A train traveled a certain route that consisted of 3 segments, all

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