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?

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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?

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Given:ToFindFormulaApplied:[tex]\underline{\boxed{\sf Reduced_{(Speed)}=Original_{(Speed)}-New_{(Speed)}}}[/tex]

[tex]\underline{\boxed{\sf Speed = \dfrac{Distance}{Time \:taken}}}[/tex]

Solution:Let the total duration covered by the flight be x.

formulatobeused,[tex]\boxed{\sf Reduced_{(Speed)} =Original_{(Speed)}-New_{(Speed)}}[/tex]

Substitutingvaluesinformula,[tex]{\Rightarrow \sf 200 = \dfrac{600}{x}-\dfrac{600}{x+\dfrac{1}{2}}}\\[/tex]

[tex]{\Rightarrow \sf 200=\dfrac{600}{x}-\dfrac{600}{\dfrac{2x+1}{2}}}\\[/tex]

[tex]{\Rightarrow \sf 200 = \dfrac{600}{x}-\dfrac{600\times 2}{2x+1}}\\[/tex]

[tex]{\Rightarrow \sf 200=\dfrac{600}{x}-\dfrac{1200}{2x+1} }\\[/tex]

[tex]{\Rightarrow \sf 200 = \dfrac{(2x+1)(600)-(1200)(x)}{(x)(2x+1)} }\\[/tex]

[tex]{\Rightarrow \sf 200=\dfrac{(1200x+600)-(1200x)}{2x^2+x}}\\[/tex]

[tex]{\Rightarrow \sf 200 = \dfrac{{\cancel{1200x}}+600-{\cancel{1200x}}}{2x^2+x}}\\[/tex]

[tex]{\Rightarrow \sf 200 = \dfrac{600}{2x^2+x}}\\[/tex]

[tex]{\Rightarrow \sf 2x^2+x= \dfrac{6{\cancel{00}}}{2{\cancel{00}}}}\\[/tex]

[tex]{\Rightarrow \sf 2x^2+x=3}\\[/tex]

[tex]{\Rightarrow \sf 2x^2+x-3=0}[/tex]

Bysplittingmiddleterms,[tex]{\longrightarrow \sf 2x^2+x-3=0}\\\\{\longrightarrow \sf 2x^2-2x+3x-3=0}\\\\{\longrightarrow 2x(x-1)+3(x-1)=0}\\\\{\longrightarrow (2x+3)(x-1)=0}\\\\\\{\begin{array}{c|c} \sf 2x+3=0&\sf x-1=0\\\\\sf 2x=-3&\sf x=0+1\\\\\sf x=\dfrac{-3}{2}&\sf x=1 \end{array}}\\\\\boxed{\bf x = 1\:hour}[/tex]

Note:–RequiredAnswer:The total duration of the flight is [tex]\underline{\sf 1 \: hour}[/tex]

Answer:[tex]\red{\bigstar}[/tex] Total duration of flight [tex]\large\leadsto\boxed{\tt\purple{1 \: hr.}}[/tex]

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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,

[tex]\pink{\bigstar}[/tex] [tex]\large\underline{\boxed{\bf\green{Speed = \dfrac{Distance}{Time}}}}[/tex]

here,

✯ For Orignal Speed:-

Distance = 600

Time = x

✯ For New Speed:-

Distance = 600

Time = x + 1/2

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Now,

➪ [tex]\sf 200 = \dfrac{600}{x} – \dfrac{600}{x + \frac{1}{2}}[/tex]

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➪ [tex]\sf 200 = \dfrac{600}{x} – \dfrac{600}{\frac{2x+1}{2}}[/tex]

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➪ [tex]\sf 200 = \dfrac{600}{x} – \dfrac{600 \times 2}{2x+1}[/tex]

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➪ [tex]\sf 200 = \dfrac{600}{x} – \dfrac{1200}{2x+1}[/tex]

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➪ [tex]\sf 200 = \dfrac{600(2x+1) – 1200x}{(2x+1)(x)}[/tex]

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➪ [tex]\sf 200 = \dfrac{1200x + 600 – 1200x}{2x^2 + x}[/tex]

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➪ [tex]\sf 200 = \dfrac{600}{2x^2 + x)}[/tex]

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➪ [tex]\sf 2x^2 + x = \dfrac{600}{200}[/tex]

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➪ [tex]\sf 2x^2 + x = 3[/tex]

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➪ [tex]\sf 2x^2 + x – 3 = 0[/tex]

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➪ [tex]\sf 2x^2 + 3x – 2x -3 = 0[/tex]

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➪ [tex]\sf x(2x + 3) – 1(2x + 3) = 0[/tex]

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➪ [tex]\sf (2x+3)(x-1) = 0[/tex]

Now,

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➪ [tex]\sf 2x + 3 = 0[/tex]

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➪ [tex]\sf 2x = -3[/tex]

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➪ [tex]\sf x = \dfrac{-3}{2}[/tex]

Also,

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➪ [tex]\sf x – 1 = 0[/tex]

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➪ [tex]\large{\bf\pink{x = 1}}[/tex]

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Therefore, the duration of the flight is 1 hr.