From station 130 km away started passenger train and after 1.9 hours after the express train, which travels 27 km an hour more. Express train finish journey 5 minutes early.

Calculate the average speed of these two trains.

Correct answer:

v1 =  192.2 km/h
v2 =  219.2 km/h

Step-by-step explanation:

v1t1=v2t2=130 v2=v1+27 t2=t15/60  v1t1=130 v1t10.083v1+27t1270.083=130  t1=0.003v1+0.083  v1(0.003v1+0.083)=130  0.0030864197530864x2+0.083x130=0  a=0.003086;b=0.083;c=130 D=b24ac=0.083240.003086(130)=1.611882716 D>0  x1,2=2ab±D=0.0061730.08±1.61 x1,2=13.5±205.675108 x1=192.175107876 x2=219.175107876   Factored form of the equation:  0.0030864197530864(x192.175107876)(x+219.175107876)=0  v1=192.2 km/h
v2=v1+27=219.2 km/h

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