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A person travelled a distance of 60 km and then returned to the starting point. The time taken by him for the return journey was \(\frac{1}{2}\) hour more than the time taken for the outward journey, and the speed during the return journey was 10 km/h less than that during the outward journey. His speed during the outward journey (in km/h) was:
1. 24
2. 40
3. 36
4. 30

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Correct Answer - Option 2 : 40

Given :

The person traveled 60 km and returned to the same point

Time taken for return journey = 1/2 Hr + Time taken for outward journey

Speed during return journey = Speed during the outward journey - 10 km/h 

Formula used :

Speed = Distance/Time

Calculation :

Let the Speed of outward journey be S1 and speed for return journey be S2

Also, Let the time taken for outward journey be T1 and time taken for return journey be T2

 According to question,

S1 = 60/ T

⇒ T1 = 60/S1      ----(1)

Also, from given information in question

T1 + 1/2 = T2      ----(2)

S2 = S1 - 10      ----(3)

For return journey,

T2 = 60 / S2

From Eq (2) and (3)

⇒ T1 + 1/2 = 60 / (S1 - 10)

⇒ (60/S1) +1/2 = 60 / (S1 - 10)

⇒ \( {600 \over (S1-10)S1} = {1\over2}\)

⇒ S1 (S1 - 10) = 1200

⇒ S1 = 40

∴ Speed of the person during outward journey is 40 km/h

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