Question

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Answer 1

$\textsf{Let the Original Duration of the Flight be : D hours}$

$\textsf{Given : Total Distance of the Flight (Journey) = 2800 km}$

$\bigstar\;\;\sf{We\;know\;that : \boxed{\sf{Average\;Speed = \dfrac{Total\;Distance\;traveled}{Total\;Time\;taken}}}}$

$\implies \sf{Original\;Speed\;of \;the\;Aircraft = \bigg(\dfrac{2800}{D}\bigg) km\;per\;hour}$

$\textsf{Given : Due to Bad Weather Conditions, The Average Speed of the Aircraft }\\\textsf{is reduced by 100 km per hour}$

$\implies \sf{New\;Speed\;of\;the\;Aircraft = \bigg(\dfrac{2800}{D} - 100\bigg)\;km\;per\;hour}$

$\textsf{Given : As the Average Speed is reduced by 100 km per hour, Time taken}\\ \textsf{to complete the journey increased by 30 Minutes = Half of an Hour}$

$\implies \sf{New\;Time\;taken\;to\;complete\;the\;Journey = \bigg(D + \dfrac{1}{2}\bigg) = \bigg(\dfrac{2D + 1}{2}\bigg)\;hours}$

$\textsf{Substituting the New values in the Average Speed Formula, We get :}$

$\implies \sf{\bigg(\dfrac{2800}{D} - 100\bigg) = \dfrac{2800}{\dfrac{2D + 1}{2}}}$

$\implies \sf{\bigg(\dfrac{2800 - 100D}{D}\bigg) = \bigg(\dfrac{5600}{2D + 1}\bigg)}$

$\implies \sf{\bigg(\dfrac{28 - D}{D}\bigg) = \bigg(\dfrac{56}{2D + 1}\bigg)}$

$\implies \sf{(28 - D)(2D + 1) = 56D}$

$\implies \sf{56D + 28 - 2D^2 - D = 56D}$

$\implies \sf{2D^2 + D - 28 = 0}$

$\implies \sf{2D^2 + 8D - 7D - 28 = 0}$

$\implies \sf{2D(D + 4) - 7(D + 4) = 0}$

$\sf{\implies (2D - 7)(D + 4) = 0}$

$\implies \sf{D = \dfrac{7}{2}\;\;(or)\;\;D = -4}$

$\sf{As : Time\;cannot\;be\;a\;Negative\;Value \implies D \neq -4}$

$\implies \sf{D = \dfrac{7}{2}\;hours}$

$\sf{\implies D = 3.5\;hours}$

$\implies \textsf{The Original Duration of the Flight = 3.5 hours}$