Question

An aeroplane left 30 minutes late due to heavy rain and in order to reach its destination 1500 km away in time it had to increase its speed by 250 km per hour from it's original speed find the original speed of the aeroplane ? ​

Answer 1

       $\sf{\green{\underline{\underline{ \large{Method \: of \: Solution}}}}}$$\sf{ \green{Let \: the \: usual \: speed \: of \: the \: aeroplane \: be \: x km/hr.}} \\ \\ \sf{ \red{Total \: Distance \: travelled \: by \: aeroplane=1500km.}} \\ \\\sf{ \blue{Time \: Taken \: by \: the \: aeroplane = \frac{1500}{x} }} \\ \\ \sf{ \green{According \: to \: the \: Question \: Statement:}} \\ \\ \sf{ \blue{Time \: taken \: by \: the \: aeroplane with \: New \: Speed \: = \frac{1500}{x + 250} km \: hr}}$$\sf{ \implies{ \red{ \frac{1500}{x \times 250} - \frac{1500}{x} = \frac{1}{2}}}} \\ \\ \sf{ \implies{ \red{ \frac{1500}{x} - \frac{1500}{x + 250} = \frac{1}{2}}}} \\ \\ \sf{ \implies{ \red{2 \times 1500(x + 250) - 2x \times 250 = x(x + 250)}}} \\ \\ \sf{ \implies{ \red{3000x + 75000 - 3000x = {x}^{2} + 250x}}} \\ \\ \sf{ \implies{ \red{\green{\cancel{ \green{3000x }}+ 75000 \cancel{ - 3000x} = {x}^{2} + 250x}}}} \\ \\ \sf{ \implies{ \red{\blue{75000 = {x}^{2} + 250}}}} \\ \\ \sf{ \implies{ \red{\green{ {x}^{2} + 250 - 750000 = 0}}}} \\ \\ \sf{ \implies{ \red{ {x}^{2} + 1000x - 750x - 750000 = 0}}} \\ \\ \sf{ \implies{ \red{\green{(x - 750)(x + 1000) = 0}}}} \\ \\ \sf{ \implies{ \blue{\blue{x = 750 \: \: or \: \: ( - 1000) \: km}}}} \\ \\ \mathsf{\implies{ \red{\boxed{ \sf{Since \: \: Speed \: can \: not \: be \: Negative, \: Therefore \: usual \: speed \: of \: the \: aeroplane \: is \: 750km/hr.}}}}}$