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A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/h more. Find the original speed of the train.

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Let, the original speed of the train=x km/hr
Increased speed, of the train =(x+5) km/hr
distrance =360 km
`because"time"=("distance")/("speed")`
According to the question
`(360)/(x)-(360)/(x+5)=(4)/(5)implies(360(x+5)-360x)/(x(x+5))=(4)/(5)`
`implies(360x+1800-360x)/(x^(2)+5x)=(4)/(5)implies(1800)/(x^(2)+5x)=(4)/(5)`
`implies(x^(2)+5x)-5xx1800`
`impliesx^(2)+5x=2250`
`impliesx^(2)+5x-2250=0impliesx^(2)+50x-45x-2250=0`
`impliesx(x+50)-45(x+50)=0implies(x+50)(x-45)=0`
Now, `x+50=0impliesx=-50`
and `x-45=0impliesx=45`
since, speed cannot be negative. Hence, the original speed of the train =45 km/hr.
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