Question : A person travels 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 returns at a speed of 10 km/hr more than the speed of going, what is the average speed (in km/hr) for the entire journey?
Option 1: 20
Option 2: 15
Option 3: 24
Option 4: 30
Correct Answer: 24
Solution :
Let the speed of the outward journey as $v$ km/hr. Then, the speed of the return journey is $v + 10$ km/hr.
The time taken for the outward journey = $\frac{300}{v}$ hours
The time taken for the return journey = $\frac{300}{v+10}$ hours
Given that the time taken for the outward journey is 5 hours more than the time taken for the return journey.
$⇒\frac{300}{v} = \frac{300}{v + 10} + 5$
$⇒v^2+10v-600=0$
$⇒(v+30)(v-20)=0$
$⇒v = 20$ km/hr, as speed cannot be negative.
$\text{Average speed}=\frac{\text{Total distance}}{{\text{Total time}}}$
$=\frac{2 \times 300}{\frac{300}{20} + \frac{300}{30}} = \frac{600}{25} = 24 \, \text{km/hr}$
Hence, the correct answer is 24.
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