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.

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Solution

let 'x' be the original speed of the bus
$\frac{90}{x} - \frac{90}{x = 15} = \frac{1}{2} \frac{90 (x + 15) - 90 x}{x (x + 15)} = \frac{1}{2} \frac{90 x + 1350 - 90 x}{x^{2} + 15 x} = \frac{1}{2} x^{2} + 15 x - 2700 = 0 B y s o l v i n g Q u a d r a t i c e q u a t i o n m e t h o d, x = \frac{- 15 \pm \sqrt{225 + 10800}}{2} = \frac{- 15 \pm \sqrt{11025}}{2} = \frac{- 15 \pm 105}{2} \Rightarrow \frac{- 15 + 105}{2} = \frac{- 15 - 105}{2} \Rightarrow 90 / 2, - 120 / 2 \Rightarrow 45, - 60 T h e r o o t s a r e r e a l a n d e q u a l$

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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.

let 'x' be the original speed of the bus 90x−90x=15=1290(x+15)−90xx(x+15)=1290x+1350−90xx2+15x=12x2+15x−2700=0By solving Quadratic equation method,x=−15±√225+108002=−15±√110252=−15±1052⇒−15+1052=−15−1052⇒90/2,−120/2⇒45,−60The roots are real and equal