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A train covered a certain distance at a uniform speed. If the train would have been 10km/h faster , it would have taken 2hrs less than the scheduled time. And ,if the train were slower by 10km/h ; it would have taken 3hrs more than the scheduled time . Find the distance covered by them


Let the speed, time, distance be s, t, d respectively related by st=d. We are given (s+10)(t-2)=d and (s-10)(t+3)=d. You have three equations and three variables showing that it is possible to solve them. You get s=50 kmph, t=12 h, d=600 km.
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Let the distance be x
& the speed be s
Time taken for traveling x distance =  \frac{x}{s} hr
When the speed is 10km faster time taken is  \frac{x}{s+10}
                                                              =  \frac{x}{s} -2
When the speed is 10km slower time taken is  \frac{x}{s-10} =s/x+3
equating both equations    2s²-10x+20s=3s²-10x+30s
                                     ⇒  s²-50s=0
substituting value in equation x=600
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