Places A and B are 396 km apart. Train X leaves from A for B and train Y leaves from B for A at the same time on the same day onparallel tracks. Both trains meet after $$5\frac{1}{2}$$ hours. The speed of Y is 10 km/h more than that of X. What is the speed (in km/h) of Y?
Let the speed of train X be s1.
Speed of train Y = s1 +Â 10
Relative speed in opposite direction = s1 +Â s1 +Â 10 = 2s1 + 10
Time =Â $$5\frac{1}{2} = \frac{11}{2}$$ hours
Distance = speed $$\times$$ time
396 = (2s1 + 10) $$\times \frac{11}{2}$$
2s1 + 10 = 72
s1 = 31 km/hr
Speed of train Y = s1 + 10 = 31 +Â 10 = 41 km/hr
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