Two taps A and B can fill a tank in 10 hours | vedclass

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(B) Let the total capacity of the tank be the $LCM$ of $10$ and $12$,which is $60$ units.Efficiency of tap $A = 60 / 10 = 6$ units/hour.Efficiency of

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$3$

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(B) Let the total capacity of the tank be the $LCM$ of $10$ and $12$,which is $60$ units.Efficiency of tap $A = 60 / 10 = 6$ units/hour.Efficiency of tap $B = 60 / 12 = 5$ units/hour.The total time from $10 	ext{ am}$ to $4 	ext{ pm}$ is $6$ hours.Since tap $B$ remains open for the entire duration of $6$ hours,the work done by tap $B = 5 	imes 6 = 30$ units.The remaining work to be done by tap $A = 60 - 30 = 30$ units.The time for which tap $A$ must be open = $30 / 6 = 5$ hours.Since tap $A$ was opened at $10 	ext{ am}$,it should be closed after $5$ hours,which is at $3 	ext{ pm}$.

Two taps $A$ and $B$ can fill a tank in $10$ $hours$ and $12$ $hours$ respectively. If both taps are opened at $10 \text{ am}$,at what time (in $\text{pm}$) should tap $A$ be closed so that the tank is completely filled at exactly $4 \text{ pm}$?

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