An empty pool being filled with water at a co | vedclass

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(D) The pool is filled at a constant rate.It takes $8$ hours to fill $\frac{3}{5}$ of the pool's capacity.Remaining capacity to be filled = $1 - \frac

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$5 	ext{ hours } 20 	ext{ minutes}$

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(D) The pool is filled at a constant rate.It takes $8$ hours to fill $\frac{3}{5}$ of the pool's capacity.Remaining capacity to be filled = $1 - \frac{3}{5} = \frac{2}{5}$ of the capacity.Let the total time to fill the pool be $T$ hours.Since the rate is constant,the time taken is proportional to the fraction filled:$\frac{3}{5} 	ext{ capacity} ightarrow 8 	ext{ hours}$$\frac{2}{5} 	ext{ capacity} ightarrow x 	ext{ hours}$Using the ratio method: $x = \frac{8 	imes (2/5)}{3/5} = \frac{8 	imes 2}{3} = \frac{16}{3} 	ext{ hours}$.$\frac{16}{3} 	ext{ hours} = 5 \frac{1}{3} 	ext{ hours}$.Since $1 	ext{ hour} = 60 	ext{ minutes}$,$\frac{1}{3} 	ext{ hour} = \frac{1}{3} 	imes 60 = 20 	ext{ minutes}$.Therefore,the remaining time is $5 	ext{ hours } 20 	ext{ minutes}$.

An empty pool being filled with water at a constant rate takes $8$ $hours$ to fill $\frac{3}{5}$ of its capacity. How much more time will it take to finish filling the pool?

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