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Mathematics

The time taken by a person to cover 150 km was 2.5 hours more than the time taken in return journey. If he returned at a speed of 10 km/hour more than the speed of going, what was the speed per hour in each direction ?

Quadratic Equations

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Answer

Let speed while going be x km/hour and speed while returning will be (x + 10) km/hour.

Time taken while going = DistanceTime=150x\dfrac{\text{Distance}}{\text{Time}} = \dfrac{150}{x} hours.

Time taken while returning = DistanceTime=150x+10\dfrac{\text{Distance}}{\text{Time}} = \dfrac{150}{x + 10} hours.

Given,

Time taken by a person to cover 150 km was 2.5 hours more than the time taken in return journey.

∴150x−150x+10=2.5⇒150(x+10)−150xx(x+10)=2510⇒150x+1500−150xx(x+10)=52⇒1500x2+10x=52⇒3000=5(x2+10x)⇒x2+10x=30005 ⇒x2+10x=600⇒x2+10x−600=0⇒x2+30x−20x−600=0⇒x(x+30)−20(x+30)=0⇒(x−20)(x+30)=0⇒x−20=0 or x+30=0⇒x=20 or x=−30.\therefore \dfrac{150}{x} - \dfrac{150}{x + 10} = 2.5 \\[1em] \Rightarrow \dfrac{150(x + 10) - 150x}{x(x + 10)} = \dfrac{25}{10} \\[1em] \Rightarrow \dfrac{150x + 1500 - 150x}{x(x + 10)} = \dfrac{5}{2} \\[1em] \Rightarrow \dfrac{1500}{x^2 + 10x} = \dfrac{5}{2} \\[1em] \Rightarrow 3000 = 5(x^2 + 10x) \\[1em] \Rightarrow x^2 + 10x = \dfrac{3000}{5} \\[1em]\ \Rightarrow x^2 + 10x = 600 \\[1em] \Rightarrow x^2 + 10x - 600 = 0 \\[1em] \Rightarrow x^2 + 30x - 20x - 600 = 0 \\[1em] \Rightarrow x(x + 30) - 20(x + 30) = 0 \\[1em] \Rightarrow (x - 20)(x + 30) = 0 \\[1em] \Rightarrow x - 20 = 0 \text{ or } x + 30 = 0 \\[1em] \Rightarrow x = 20 \text{ or } x = -30.

Since, speed cannot be negative.

∴ x = 20 km/hour and (x + 10) = 30 km/hour.

Hence, speed while going = 20 km/hour and speed while returning = 30 km/hour.

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