Question

In a flight of 2800km an aircraft was slowed down due to bad weather ,it's average speed is reduced by 100km per hour and time increased by 30min. find the original duration of flight?

Answer 1

$\huge\textbf\red{Answer :}$

Let the speed of craft = x km/hr.

Original Speed = 2800/x km/hr. [Given]

Average Speed = 100 km/hr.

Time = 30 min.

= $\frac{30}{60}\:hr.$

= $\frac{1}{2}\:hr.$

Time taken after the craft slow down is....

$\frac{2800}{x\:-\:100}$ hr.

A.T.Q.

$\frac{2800}{x\:-\:100}$ - $\frac{2800}{x}$ = $\frac{1}{2}$

$\frac{2800x\:-\:2800x\:-\:280000}{(x)(x\:-\:100)}$ = $\frac{1}{2}$

$\frac{-\:280000}{{x}^{2}\:-\:100}$ = $\frac{1}{2}$

Cross multiply them, then we get,

${x}^{2}\:-\:100\:-\:560000\:=\:0$

${x}^{2}\:-\:800x\:+\:700x\:-\:560000\:=\:0$

$x(x\:-\:800)\:+\:700(x\:-\:800)$

$(x\:-\:800)\:(x\:+700)$

x = 800 and x = - 700

Time can never be negative. So negative value will be neglected.

x = 800.

Original Duration = $\frac{2800}{800}$

= $\frac{7}{2}$

= $3.5\:hr.$

= 3 hrs. 30 min.
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Answer 2

Answer:

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