## Subject III. (20 points)

A bathtub would be filled by the cold water tap in 5 minutes, and by the hot water tap in 7 minutes. The bathtub is equipped with a drain hole, through which, when the bathtub is full, the water can drain in 3 minutes. If we open both taps and do not close the drain hole, will the bathtub fill up? If so, in how much time?

Prof. Sorin Borodi, "Alexandru Papiu Ilarian" Theoretical High School, Dej
## Subject III.

Let $\mathrm{C}$ be the capacity of the bathtub and $t$ the filling time under the conditions given in the statement.

We use the simple rule of three for the cold water tap:

$5 \mathrm{~min}$. $\qquad$ C

$\mathrm{t}$ min........... $\mathrm{C}_{1} \quad$ (We denote $\mathrm{C}_{1}$ as the part of the capacity filled by this tap) $\mathrm{C}_{1}=\frac{\mathrm{tC}}{5}$

Similarly, we obtain $\mathrm{C}_{2}=\frac{\mathrm{tC}}{7}\left(\mathrm{C}_{2}\right.$ is the part filled by the hot water tap) and $\mathrm{C}_{3}=\frac{\mathrm{tC}}{3}$ (where $\mathrm{C}_{3}$ is the part of the capacity that drains through the hole).

Assuming that after $t$ minutes the bathtub is full, we have $\mathrm{C}_{1}+\mathrm{C}_{2}-\mathrm{C}_{3}=\mathrm{C} \Rightarrow$

$\frac{\mathrm{tC}}{5}+\frac{\mathrm{tC}}{7}-\frac{\mathrm{tC}}{3}=\mathrm{C} \Rightarrow \frac{\mathrm{t}}{5}+\frac{\mathrm{t}}{7}-\frac{\mathrm{t}}{3}=1 \Rightarrow \frac{21 t+15 t-35 t}{105}=1 \Rightarrow \frac{t}{105}=1 \Rightarrow t=105$.

Since $t$ is positive, it follows that the bathtub will be filled in 105 minutes.