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Mathematics

A man covers a distance of 100 km, traveling with a uniform speed of x km/hr. Had the speed been 5 km/hr more, it would have taken 1 hour less. Find x, the original speed.

Quadratic Equations

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Answer

Original speed of man be x km/hr.

Given,

Distance to be traveled by man = 100 km

Time = DistanceSpeed\dfrac{\text{Distance}}{\text{Speed}}

Time taken by man at speed x km/hr = 100x\dfrac{100}{x}

Time taken by man at increased speed (x + 5) = 100x+5\dfrac{100}{x + 5}

Given, with increased speed the man takes 1 hour less.

⇒100x−100x+5=1⇒100(x+5)−100xx(x+5)=1⇒100x+500−100xx2+5x=1⇒500=(x2+5x)⇒0=x2+5x−500⇒x2+5x−500=0⇒x2+25x−20x−500=0⇒x(x+25)−20(x+25)=0⇒(x+25)(x−20)=0⇒(x+25)=0 or (x−20)=0….[Using zero-product rule]⇒x=−25 or x=20.\Rightarrow \dfrac{100}{x} - \dfrac{100}{x + 5} = 1 \\[1em] \Rightarrow \dfrac{100(x + 5) - 100x}{x(x + 5)} = 1 \\[1em] \Rightarrow \dfrac{100x + 500 - 100x}{x^2 + 5x} = 1 \\[1em] \Rightarrow 500 = (x^2 + 5x) \\[1em] \Rightarrow 0 = x^2 + 5x - 500 \\[1em] \Rightarrow x^2 + 5x - 500 = 0 \\[1em] \Rightarrow x^2 + 25x - 20x - 500 = 0 \\[1em] \Rightarrow x(x + 25) - 20(x + 25) = 0 \\[1em] \Rightarrow (x + 25)(x - 20) = 0 \\[1em] \Rightarrow (x + 25) = 0 \text{ or } (x - 20) = 0 \text{….[Using zero-product rule]} \\[1em] \Rightarrow x = -25 \text{ or } x = 20.

Since, speed cannot be negative.

Thus, x = 20.

Hence, the original speed of man = 20 km/hr.

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