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A student reaches the bus station at speed of 8 km/hr on the cycle, boards the bus to the B point whose speed is 15 km/hr. From B to the rickshaw stand he gets a lift from the bike which travels at speed of 20 km/hr. From the rickshaw stand to the exam center he travels at speed of 12 km/hr. If the distance between home to the bus station, bus station to B, B to rickshaw stand, Rickshaw stand to exam center is equal then find student’s average speed. (Approximate values whenever felt necessary)


1. 12 km/hr
2. 15 km/hr
3. 16 km/hr
4. 18 km/hr

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Best answer
Correct Answer - Option 1 : 12 km/hr

Given:

Speed through cycle = 8 kmph

Speed through bus = 15 kmph

Speed through bike = 20 kmph

Speed  through rickshaw = 12 kmph

Formula used:

Speed = Distance/Time

Calculation:

LCM of 4 speeds = LCM (8, 12, 15, 20)

⇒ 4 × 2 × 5 × 3 = 120

⇒ Total distance covered = 120 × 4

⇒ Total distance covered = 480 km

⇒ Time taken to travel from home to bus station = 120/8

⇒ Time taken 1 = 15 hours

⇒ Time taken to travel from bus station to B = 120/15

⇒ Time taken 2 = 8 hours

⇒ Time taken to travel on bike = 120/20

⇒ Time taken to travel on bike = 6 hours

⇒ Time taken to travel on rickshaw = 120/12

⇒ Time taken to travel on rickshaw = 10 hours

⇒ Average speed = (Total distance/4)/(Total time/4)  (∵ average speed = average distance/average time)

⇒ Avg. Speed = 480/(15 + 8 + 6 + 10)

⇒ Avg. Speed = 480/39

⇒ 480/39 ≈ 12 km/hr

Required answer is 12 km/hr.

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