Question

Three taps $A,B,$ and $C$ can fill an overhead tank in $6$ hours, $8$ hours, and $12$ hours respectively. How long would the three taps take to fill the empty tank, if all of them are opened together?

Answer 1

It is given that three taps $A,B,$ and $C$ can fill an overhead tank in $6$ hours, $8$ hours, and $12$ hours respectively.

If $1$ whole tank is filled by a tap in $x$ (suppose) hours.

Then, that tap can fill $\frac{1}{x}$ parts in $1$ hour.

Tap $A^{\prime }s$ one hour's work $=\frac{1}{6}$

Tap $B^{\prime }s$ one hour's work $=\frac{1}{8}$

Tap $C^{\prime }s$ one hour's work $=\frac{1}{12}$

$A,B$ and $C^{\prime }s$ together one hour's work

$=\frac{1}{6}+\frac{1}{8}+\frac{1}{12}$

$=\frac{4+3+2}{24}$ (LCM of $6,8,12=24)$

$=\frac{9}{24}=\frac{3}{8}$

$\therefore$ These three tap will fill the empty tank in $=\frac{8}{3}$ hour

$=2\frac{2}{3}$ hours

$=2$ hours $40$ minutes [$\because 1$ hour $=60$ minutes, $\frac{2}{3}$ hours $=\frac{2}{3}\times 60=40$ minutes]