IIFT 2018 Question 105

Question 105

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


Speed of joseph = 50/60 = $$\frac{5}{6}$$ m/s

Let the radius of circular playground be r

Speed = distance/ time

Time to diametrically cross a semi-circular playground t1= $$\frac{2r}{\frac{5}{6}}$$ = $$\frac{12r}{5}$$

Time to cross the playground along the semi-circular path t2 = $$\frac{\pi\ r}{\frac{5}{6}}=\frac{\left(6\pi\ r\right)}{5}$$

t1-t2 = 48

$$\frac{12r}{5}-\frac{\left(6\pi\ r\right)}{5}\ =\ 48$$

=> r = 35m

hence diameter = 70m

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