Question : A person covers a distance of 300 km and then returns to the starting point. The time taken by him for the outward journey is 5 hours more than the time taken for the return journey. If he returned at a speed of 10 km/hr more than the speed of going, what was the speed (in km/hr) for the outward journey?
Option 1: 15
Option 2: 20
Option 3: 30
Option 4: 25
Correct Answer: 20
Solution :
Let the speed of the forward journey be $x$ km/hr.
Speed of return journey will = $(x + 10)$ km/hr
Time taken in return journey = $\frac{300}{(x + 10)}$ hours
Time taken in forward journey = $\frac{300}{x}$ hours
According to the question,
⇒ $\frac{300}{x} -\frac{300}{x+10} = 5$
⇒ $\frac{300(x + 10) - 300x}{x(x + 10)} = 5$
⇒ $300x + 3000 - 300x = 5x^2+50x$
⇒ $3000 = 5x^2 + 50x$
⇒ $5x^2 + 50x - 3000 = 0$
⇒ $x^2 + 10x - 600 = 0$
⇒ $x^2 + 30x - 20x - 600 = 0$
⇒ $x(x + 30) - 20(x + 30) = 0$
⇒ $(x - 20)(x + 30) = 0$
As speed cannot be negative, so $x=20$ km/hr
Hence, the correct answer is 20.
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