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Question
A bus covers a distance of 90 km at a uniform speed. Had the speed been `(15"km")/"hour"` more it would have taken 30 minutes less for the journey. Find the original speed of the bus
Solution
Let the original speed of the bus be “x” `"km"/"hr"`
Time taken to cover 90 km = `90/x`
After increasing the speed by `(15"km")/"hr"`
Time taken to cover 90 km = `90/(x + 15)`
By the given condition
`90/x - 90/(x + 15) = 1/2`
`(90(x + 15) - 90x)/(x(x + 15)) = 1/2`
90x + 1350 – 90x = `(x^2 + 15x)/2`
1350 = `(x^2 + 15x)/2`
2700 = x2 + 5x
x2 + 15x – 2700 = 0
(x + 60) (x – 45) = 0
x + 60 = 0 or x – 45 = 0
x = – 60 or x = 45
The speed will not be negative
∴ Original speed of the bus = `(45"km")/"hr"`
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