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Mathematics

If the speed of an aeroplane is reduced by 40 km per hour, it takes 20 minutes more to cover 1200 km. Find the speed of the aeroplane.

Quadratic Equations

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Answer

Let speed be x km/hr.

Time taken to cover 1200 km = 1200x\dfrac{1200}{x} hr

Decreasing speed by 40 km/hr we get speed = (x - 40) km/hr

Time taken to cover 1200 km = 1200x−40\dfrac{1200}{x - 40} hr

According to question,

⇒1200x−40−1200x=2060⇒1200x−1200(x−40)x(x−40)=13⇒1200x−1200x+48000x2−40x=13⇒48000x2−40x=13⇒144000=x2−40x⇒x2−40x−144000=0⇒x2+400x−360x−144000=0⇒x(x+400)−360(x+400)=0⇒(x−360)(x+400)=0⇒x−360=0 or x+400=0⇒x=360 or x=−400.\Rightarrow \dfrac{1200}{x - 40} - \dfrac{1200}{x} = \dfrac{20}{60} \\[1em] \Rightarrow \dfrac{1200x - 1200(x - 40)}{x(x - 40)} = \dfrac{1}{3} \\[1em] \Rightarrow \dfrac{1200x - 1200x + 48000}{x^2 - 40x} = \dfrac{1}{3} \\[1em] \Rightarrow \dfrac{48000}{x^2 - 40x} = \dfrac{1}{3} \\[1em] \Rightarrow 144000 = x^2 - 40x \\[1em] \Rightarrow x^2 - 40x - 144000 = 0 \\[1em] \Rightarrow x^2 + 400x - 360x - 144000 = 0 \\[1em] \Rightarrow x(x + 400) - 360(x + 400) = 0 \\[1em] \Rightarrow (x - 360)(x + 400) = 0 \\[1em] \Rightarrow x - 360 = 0 \text{ or } x + 400 = 0 \\[1em] \Rightarrow x = 360 \text{ or } x = -400.

Since, speed cannot be negative

∴ x = 360 and x + 40 = 400

Hence, speed of aeroplane = 400 km/hr.

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