Two taps can separately fill a cistern in 10 | vedclass

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(B) Let the capacity of the cistern be the Least Common Multiple $(LCM)$ of $10, 15,$ and $18,$ which is $90$ units.Efficiency of the first tap = $90

Vedclass · Accepted Answer

$9 	ext{ minutes}$

Vedclass · Answer

(B) Let the capacity of the cistern be the Least Common Multiple $(LCM)$ of $10, 15,$ and $18,$ which is $90$ units.Efficiency of the first tap = $90 / 10 = 9$ units/min.Efficiency of the second tap = $90 / 15 = 6$ units/min.Combined efficiency of both taps = $9 + 6 = 15$ units/min.When the waste pipe is open,the net efficiency = $90 / 18 = 5$ units/min.Efficiency of the waste pipe = (Combined efficiency of taps) - (Net efficiency) = $15 - 5 = 10$ units/min.Time taken by the waste pipe to empty the full cistern = $90 / 10 = 9$ minutes.

Two taps can separately fill a cistern in $10$ $minutes$ and $15$ $minutes$ respectively. When the waste pipe is open,they can together fill it in $18$ $minutes$. The waste pipe can empty the full cistern in:

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