Three taps are fitted in a cistern. The empty | vedclass

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(B) Let the capacity of the cistern be the $LCM$ of $3, 4,$ and $5$, which is $60 \, 	ext{units}$.Efficiency of the first tap $(A) = \frac{60}{3} = 2

Vedclass · Accepted Answer

$2 \frac{14}{23} \, h$

Vedclass · Answer

(B) Let the capacity of the cistern be the $LCM$ of $3, 4,$ and $5$, which is $60 \, 	ext{units}$.Efficiency of the first tap $(A) = \frac{60}{3} = 20 \, 	ext{units/h}$.Efficiency of the second tap $(B) = \frac{60}{4} = 15 \, 	ext{units/h}$.Efficiency of the third tap $(C) = -\frac{60}{5} = -12 \, 	ext{units/h}$ (since it empties the cistern).When all three taps are opened simultaneously, the net efficiency $= 20 + 15 - 12 = 23 \, 	ext{units/h}$.Time taken to fill the cistern $= \frac{	ext{Total Capacity}}{	ext{Net Efficiency}} = \frac{60}{23} \, h$.Converting this to a mixed fraction, we get $2 \frac{14}{23} \, h$.

Three taps are fitted in a cistern. The empty cistern is filled by the first and the second taps in $3 \, h$ and $4 \, h$, respectively. The full cistern is emptied by the third tap in $5 \, h$. If all three taps are opened simultaneously, the empty cistern will be filled up in?

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