A train travellingat a uniform speed for 360 km,would have taken 48 minutes less to travel tge the same distance if its speed were 5 km/hour more.Find the original speed of the train.
Answers
Let the speed of train be x km/hr
Distance to be travelled = 360km
We know that,
begin mathsize 12px style s p e e d space equals space fraction numerator d i s tan c e space over denominator t i m e end fraction therefore space t i m e space equals space fraction numerator d i s tan c e over denominator t i m e end fraction end style
Time take by the train intially = begin mathsize 12px style 360 over x end style
If the speed was increased by 5km/hr
Time taken by train = begin mathsize 12px style fraction numerator 360 over denominator x plus 5 end fraction end style
Difference in the time taken is 48 minutes,
begin mathsize 12px style fraction numerator 360 over denominator x space end fraction minus space fraction numerator 360 over denominator x plus 5 end fraction space equals space 48 over 60 space space space space space left parenthesis c o n v e r t i n g space 48 m i n s space t o space h o u r s right parenthesis rightwards double arrow 360 open square brackets fraction numerator x plus 5 minus x over denominator x left parenthesis x plus 5 right parenthesis end fraction close square brackets equals 4 over 5 rightwards double arrow 90 open square brackets fraction numerator 5 over denominator x left parenthesis x plus 5 right parenthesis end fraction close square brackets equals 1 fifth rightwards double arrow 90 left square bracket 5 right square bracket 5 equals x squared plus 5 x end style
x2+5x -2250 = 0
(x+50)(x-45)
Speed = 45 km/hr as speed cannot be negative