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A man travels by his office bus at the speed of 54 km/h and reaches 18 minutes early. If he goes at the speed of 36 km/h, he reaches 18 minutes late. What will be the speed of the bus to reach at normal time?

A man travels by his office bus at the speed of 54 km/h and reaches 18 minutes early. If he goes at the speed of 36 km/h, he reaches 18 minutes late. What will be the speed of the bus to reach at normal time? Correct Answer 216/5 km/h

Given:

Speed of bus before = 54 km/h

Speed of bus after = 36 km/h

Concept:

Speed = Distance/Time

Calculation:

Let be assume distance is D and time is T

Before

⇒ T - 18/60 = D/54     ---(1)

After

⇒ T + 18/60 = D/36     ---(2)

⇒ By equation (1) and (2)

⇒ 54 × (T - 18/60) = 36 × (T - 18/60)

⇒ 6T - 18/10 = 4T - 18/15

⇒ 2T = 18/10 - 18/15 = 18 × (1/10 - 1/15) = 18 × (3 + 2)/(30) = 90/30 = 3

⇒ T = 3/2

⇒ D = (3/2 - 18/60) × 54 = {(90 - 18)/60} × 54 = (72 × 54)/60 = 648/10

⇒ The original speed of bus = (648/10) ÷ (3/2) = (648/10) × (2/3) = 216/5 km/h

∴ The required result will be 216/5 km/h.

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