Express train

International express train drove from Kosice to Teplice. In the first 279 km, the track was repaired, and therefore it was moving at a speed of 10km/h less than it was scheduled to drive. The rest of the 465 km trip has increased the speed by 8 km/h than the timetable speed. It came to Teplice for a while. Find the average speed of the train to drive according to the timetable.

Correct result:

v =  64 km/h

Solution:

t1+t2=t  279v10+465v+8=279+465v   279(v+8)v+465 (v10) v=(279+465)(v+8) (v10)  279 (v+8) v+465 (v10) v=(279+465) (v+8) (v10)  930v=59520  v=64 km/ht_{1} + t_{2}=t \ \\ \ \\ \dfrac{ 279 }{ v-10 } + \dfrac{ 465 }{ v+8 }=\dfrac{ 279+465 }{ v } \ \\ \ \\ \ \\ 279(v+8)v + 465 \cdot \ (v-10) \cdot \ v=(279+465)(v+8) \cdot \ (v-10) \ \\ \ \\ 279 \cdot \ (v+8) \cdot \ v + 465 \cdot \ (v-10) \cdot \ v=(279+465) \cdot \ (v+8) \cdot \ (v-10) \ \\ \ \\ 930v=59520 \ \\ \ \\ v=64 \ \text{km/h}



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