Physics, asked by Benny3824, 1 year ago

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 its original speed find the original speed of the aeroplane

Answers

Answered by Anonymous
117
 \large\sf{ \green{ \mathfrak{ \underline{ \underline{Question \ratio}}}}}

 \sf{ \red{An \: aeroplane \: left \: 30 \: mintues \: late \: due \: to }} \\ \sf{ \green{ \: bad \: weather \: and \: in \: order \: to \: reach \: the}} \: \\ \sf{ \red{ destination, \: 1500 km \: away in \: time, \: it \: had}} \\ \sf{ \green{\: to \: increase \: the \: speed \: by \: 250 \: km/her}} \\ \sf{ \red{\: from \: its \: usual \: speed.}} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ \blue{ Find \: the \: usual \: speed \: of \: the \: plane ?}}

 \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{7500 = {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.}}}}}

harshrajput26: wrong h step tum 7500 likhe ho par 750000 hoga sabhi jagah
TheBrainliestUser: Amazing color combination & superb answer brother :)
Anonymous: :-)↩
Anonymous: Thank You brother
Anonymous: :-)↩
Answered by Awesome98
65

-----+-+----

 \sf{ \implies{ \frac{1500}{x \times 250} - \frac{1500}{x} = \frac{1}{2}}} \\ \\ \sf{ \implies{ \frac{1500}{x} - \frac{1500}{x + 250} = \frac{1}{2}}} \\ \\ \sf{ \implies{2 \times 1500(x + 250) - 2x \times 250 = x(x + 250)}} \\ \\ \sf{ \implies{3000x + 75000 - 3000x = {x}^{2} + 250x}} \\ \\ \sf{ \implies{\green{\cancel{3000x }}+ 75000 \cancel{ - 3000x} = {x}^{2} + 250x}}} \\ \\ \sf{ \implies{7500 = {x}^{2} + 250}} \\ \\ \sf{ \implies{ {x}^{2} + 250 - 750000 = 0}} \\ \\ \sf{ \implies{ {x}^{2} + 1000x - 750x - 750000 = 0}}\\ \\ \sf{ \implies{(x - 750)(x + 1000) = 0}} \\ \\ \sf{ \implies{x = 750 \: \: or \: \:  -1000 \: km}} \\ \\ \mathsf{\implies{ \sf{ \: Therefore \: usual \: speed \: of \: the \: aeroplane \: is \: 750km/hr.}}}


harshrajput26: ans sahi h par step me problem h kahi 7500 to kahi 75000 to kahi 750000 likhe ho copy h answer
Similar questions