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Question
A passenger train takes 2 hours less for a journey of 300 km if its speed is increased by 5 km/hr from its usual speed. Find its usual speed.
Solution
Let the usual speed x km/hr.
According to the question:
`300/x-300/(x+5)=0`
⇒`( 300(x+5)-300x)/(x(x+5))=2`
⇒`(300x+1500-300x)/(x^2+5x)=2`
⇒`1500=2(x^2+5x)`
⇒`1500=2x^2+10x`
⇒`x^2+5x-750=0`
⇒`x^2+(30-25)x-750=0`
⇒`x^2+30x-25x-750=0`
⇒`x(x+30)-25(x+30)=0`
⇒`(x+30) (x-25)=0`
⇒`x=-30 or x=25`
The usual speed cannot be negative; therefore, the speed is 25 km/hr.
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