Question

Hi guys, a question for maths solvers .

In a flight of 2800 km, an aircraft was slowed down due to bad weather. Its average speed is reduced by 200 km/ hr and time of flight increased by 30 minutes. Find the original duration of the flight.

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Answer 1

▶ Question :-

→ In a flight of 2800 km, an aircraft was slowed down due to bad weather. Its average speed is reduced by 200 km/ hr and time of flight increased by 30 minutes. Find the original duration of the flight.


▶ Answer :-

→ The original duration of the flight = 3 hours 30 minutes .


▶ Step-by-step explanation :-


→ Let the original speed of the aircraft be x km/hr .

→ Time taken to cover 2800 km = $\frac{2800}{x}$ hours .

→ Reduced speed = ( x - 100 ) km/hr .

→ Time taken to cover 2800 km at this speed = $\frac{2800}{x - 100 }$ hours .



$\huge \pink { \mid{ \underline{ \overline{ \tt Solution :- }} \mid}}$


▶ Now,


$\sf \therefore \frac{2800}{(x - 100)} - \frac{2800}{x} = \frac{30}{60} . \\ \\ \sf \implies \frac{1}{(x - 100)} - \frac{1}{x} = \frac{1}{2 \times 2800} . \\ \\ \sf \implies \frac{x - (x - 100)}{(x - 100)x} = \frac{1}{5600} . \\ \\ \sf \implies \frac{100}{( {x}^{2} - 100x) } = \frac{1}{5600} . \\ \\ \sf \implies {x}^{2} - 100x - 560000 = 0. \\ \\ \sf \implies {x}^{2} - 800x + 700x - 560000 = 0. \\ \\ \sf \implies x(x - 800) + 700(x - 800) = 0. \\ \\ \sf \implies (x - 800)(x + 700) = 0. \\ \\ \sf \implies x - 800 = 0. \: \: \green{or} \: \: x + 700 = 0. \\ \\ \sf \implies x = 800. \: \: \green{or} \: \: x = - 700. \\ \\ \huge{ \orange{ \boxed{ \boxed{ \sf \implies x = 800.}}}} \\ \\ \bigg[ \tt \because speed \: cannot \: be \: negative. \bigg]$



•°• Original speed of the aircraft = 800 km/hr .

And, original duration of the flight = $\frac{2800}{800}$ hours = 3 hours 30 minutes .



✔✔ Hence, it is solved ✅✅.



$\huge { \blue{ \boxed{ \boxed{ \mathscr{THANKS}}}}}$

Answer 2

There is mistake in the question , that it is 100 km/hr not 200 km/hr .


→ Let the original speed of the aircraft be x km/hr .

→ Time taken to cover 2800 km = $\frac{2800}{x}$ hours .

→ Reduced speed = ( x - 100 ) km/hr .

→ Time taken to cover 2800 km at this speed = $\frac{2800}{x - 100 }$ hours .




$\begin{lgathered}\tt \therefore \frac{2800}{(x - 100)} - \frac{2800}{x} = \frac{30}{60} . \\ \\ \tt \implies \frac{1}{(x - 100)} - \frac{1}{x} = \frac{1}{2 \times 2800} . \\ \\ \tt \implies \frac{x - (x - 100)}{(x - 100)x} = \frac{1}{5600} . \\ \\ \tt \implies \frac{100}{( {x}^{2} - 100x) } = \frac{1}{5600} . \\ \\ \tt \implies {x}^{2} - 100x - 560000 = 0. \\ \\ \sf \implies {x}^{2} - 800x + 700x - 560000 = 0. \\ \\ \tt \implies x(x - 800) + 700(x - 800) = 0. \\ \\ \sf \implies (x - 800)(x + 700) = 0. \\ \\ \tt \implies x - 800 = 0. \: \: \red{or} \: \: x + 700 = 0. \\ \\ \tt \implies x = 800. \: \: \red{or} \: \: x = - 700. \\ \\ \huge{ \orange{ \boxed{ \tt \implies x = 800.}}} \\ \\ \bigg[ \tt \because speed \: cannot \: be \: negative. \bigg]\end{lgathered}$



•°• Original speed of the aircraft = 800 km/hr .

And, original duration of the flight =$\frac{2800}{800}$ hours = 3 hours 30 minutes .