Question

Tap A and B can fill an empty tank in $3$ hours and $5$ hours, respectively. Tap C can empty the full tank in $7\dfrac{1}{2}$ hours. If all three taps are open at the same time, in how many hours will the tank be full ?

Answer 1

Hint: In this let us suppose the volume of tank will be equal to y then find out what volume it fill in one hour or empty the tank in one hour and then suppose that x time will be taken when all three pump are open at same time then equation will be seems like $\dfrac{y}{3} + \dfrac{y}{5} - \dfrac{{2y}}{{15}} = \dfrac{y}{x}$ get the value of x.

Complete step-by-step answer:
Firstly let us take the volume of the full tank as equal to y.
Hence it is given that the Tap A can fill the tank in $3$ hours
means that y volume of tank fill in $3$ hours
Then in $1$ hour the tank will fill $\dfrac{y}{3}$ volume.
Similarly for,
Tap B can fill the tank in $5$ hours
means that y volume of tank fill in $5$ hours
Then in $1$ hour the tank will fill $\dfrac{y}{5}$ volume.
Now for the Tank C
Tap C can empty the tank in $\dfrac{{15}}{2}$ hours
means that y volume of tank empty in $\dfrac{{15}}{2}$ hours
Then in $1$ hour the tank will empty $\dfrac{{2y}}{{15}}$ volume.

Hence if the Tap A Tap B and the Tap C will open at same time . Let us assume it will take in x hours to fill the tank.
In one hour it will fill $\dfrac{y}{x}$ volume.
then Tap A Tap B and the Tap C will open at same time then
$\dfrac{y}{3} + \dfrac{y}{5} - \dfrac{{2y}}{{15}} = \dfrac{y}{x}$
Divide by y in the whole equation
$\dfrac{1}{3} + \dfrac{1}{5} - \dfrac{2}{{15}} = \dfrac{1}{x}$
L.C.M of $3,5,15$ is $15$ change numerator according to that
$\dfrac{{5 + 3 - 2}}{{15}} = \dfrac{1}{x}$
$\dfrac{6}{{15}} = \dfrac{1}{x}$
Hence $x = \dfrac{{15}}{6}$ hours
Therefore $\dfrac{{15}}{6}$ hours or $2.5$ hours to fill the tank when all three tap are open at same time

Note: In this question in the place of x we will also take the volume of the tank is unity or $1$ because at last all are will cancel out then in solution less number of unknown are present.
In this type of question always try to find out what amount it will fill in one hour after that and solve it according to the question.