Joseph diametrically crosses a semi-circular playground and takes 48 seconds less than if he crosses the playground along the semi-circular path. If he walks 50 metres in one minute, the diameter of playground is
Explanation:
Since Joseph takes less time while crossing the path diametrically rather than in a semi-circular manner, the distance travelled along the diameter is less.
Let the radius of the path be r m.
∴ Difference in distance = πr – 2r
= (π – 2)r m
Also, difference in distance = difference in time × constant speed
∴ (π – 2)r = 48 × (50/60)
∴ 1.14r = 40 i.e. r = 40/(8/7) = 35 m
∴ Diameter = 2r = 70 m
Hence, option (b).
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