Question

In a flight of 600 kms, an aircraft was slowed
down due to bad weather and the average
speed was reduced by 200 km/hour and the
time of flight increased by 30 minutes. What
is the total duration of flight?
​

Answer 1

Answer:-

$\red{\bigstar}$ Total duration of flight $\large\leadsto\boxed{\tt\purple{1 \: hr.}}$

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• Given:-

• Distance of the flight = 600 km

• Speed is reduced by = 200km/hr

• Time of flight is increased by = 30 min. = 1/2 hr.

• To Find:-

• Total duration of flight

• Solution:-

Let the duration of flight be 'x'.

Therefore,

✯ Reduced speed of flight = Orignal speed - New Speed

We know,

$\pink{\bigstar}$ $\large\underline{\boxed{\bf\green{Speed = \dfrac{Distance}{Time}}}}$

here,

✯ For Orignal Speed:-

• Distance = 600
• Time = x

✯ For New Speed:-

• Distance = 600
• Time = x + 1/2

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Now,

➪ $\sf 200 = \dfrac{600}{x} - \dfrac{600}{x + \frac{1}{2}}$

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➪ $\sf 200 = \dfrac{600}{x} - \dfrac{600}{\frac{2x+1}{2}}$

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➪ $\sf 200 = \dfrac{600}{x} - \dfrac{600 \times 2}{2x+1}$

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➪ $\sf 200 = \dfrac{600}{x} - \dfrac{1200}{2x+1}$

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➪ $\sf 200 = \dfrac{600(2x+1) - 1200x}{(2x+1)(x)}$

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➪ $\sf 200 = \dfrac{1200x + 600 - 1200x}{2x^2 + x}$

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➪ $\sf 200 = \dfrac{600}{2x^2 + x)}$

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➪ $\sf 2x^2 + x = \dfrac{600}{200}$

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➪ $\sf 2x^2 + x = 3$

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➪ $\sf 2x^2 + x - 3 = 0$

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➪ $\sf 2x^2 + 3x - 2x -3 = 0$

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➪ $\sf x(2x + 3) - 1(2x + 3) = 0$

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➪ $\sf (2x+3)(x-1) = 0$

Now,

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➪ $\sf 2x + 3 = 0$

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➪ $\sf 2x = -3$

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➪ $\sf x = \dfrac{-3}{2}$

Also,

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➪ $\sf x - 1 = 0$

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➪ $\large{\bf\pink{x = 1}}$

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Therefore, the duration of the flight is 1 hr.

Answer 2

Given :-

Distance of the flight = 600 km

Speed is reduced by = 200km/hr

Time of flight is increased by = 30 min. = 1/2 hr.

To Find :-

Total duration

SoluTion :-

${\huge {\fbox {let}}}$

Duration of flight be 'x'.

Reduced speed of flight = Orignal speed - New Speed

${\huge {\blue {\star {\huge {\boxed {\boxed { \bf Speed = \frac{Distance }{time} }}}}}}}$

Original Speed

Distance = 600

Time = x

New speed

Distance = 600

Time = x + ½

$\sf \mapsto \: 200 = \dfrac{600}{x} - \dfrac{600}{x + \frac{1}{2}}$

$\sf \mapsto \: 200 = \dfrac{600}{x} - \dfrac{ \dfrac{600}{2x + 1} }{2}$

$\sf \mapsto \: 200 = \dfrac{600}{x} - \dfrac{600 \times 2}{2x + 1}$

$\sf \mapsto \: 200 = \dfrac{600}{x} - \dfrac{1200}{2x + 1}$

$\sf \mapsto \: 200 = \dfrac{600(2x + 1) - 1200x}{(2x + 1)(x)}$

$\sf \mapsto200 = \dfrac{1200x + 600 - 1200x}{2 {x}^{2} + x }$

$\sf \mapsto 200 = \dfrac{600}{2x {}^{2} + x}$

$\sf \mapsto2 {x}^{2} + x = \dfrac{600}{200}$

$\sf \mapsto \: 2 {x}^{2} + x = 3$

$\sf \mapsto2 {x}^{2} + x - 3 = 0$

$\sf \mapsto \: 2x {}^{2} + 3x - 2x - 3 = 0$

$\sf \mapsto \: x(2x - 3) - 1(2x + 3) = 0$

$\sf \mapsto \: (2x + 3)(x - 1) = 0$

$\sf \mapsto \: 2x + 3 = 0$

$\sf \mapsto \: 2x = - 3$

$\sf \mapsto \: x = \dfrac{ - 3}{2}$

Also,

$\sf \mapsto \: x - 1 = 0$

$\sf \mapsto \: x \: = 1$

Time taken = 1 Hour