A woman carried water in a straight cylindrical bucket from the well to the kitchen. The side of the bucket had a hole at a distance of $\frac{1}{4}$ of the bucket's height from the bottom, and thus, half a bucket of water leaked out on the way from the well to the kitchen. The woman plugged the hole, but soon the bottom of the bucket also leaked and on this same path, $\frac{2}{3}$ of the bucket's contents leaked out. The woman plugged this hole as well, but later both plugs fell out and at this point, 1 meter before the kitchen, only $\frac{1}{40}$ of the water remained in the bucket. How far was the well from the kitchen? (For simplicity, we assume that the same amount of water leaks out at equal time intervals through each hole.)
I. solution. Let the sought distance be $x$ meters. Assuming that the woman always walked at a constant speed, the first time she lost $\frac{1}{2}: x=\frac{1}{2 x}$ barrels of water per meter, and the second time she lost $\frac{2}{3 x}$ barrels of water per meter. Therefore, initially, the third time $\frac{1}{2 x}+\frac{2}{3 x}=\frac{7}{6 x}$ barrels of water leaked out per meter, but this only lasted until $\frac{3}{4}$ of the barrel had leaked out. After that, only $\frac{1}{4}-\frac{1}{40}=\frac{9}{40}$ of the barrel leaked out through the lower hole. According to this, the water level in the barrel reached the side hole after $\left(\frac{3}{4}: \frac{7}{6 x}=\right) \frac{9 x}{14}$ meters, and then for the next $\left(\frac{9}{40}: \frac{2}{3 x}=\right) \frac{27 x}{80}$ meters, only the lower hole leaked.

According to the problem, the sum of these two distances is 1 meter less than the entire distance, i.e.,

$$
\frac{9 x}{14}+\frac{27 x}{80}+1=x
$$

from which

$$
360 x+189 x+560=560 x
$$

and thus

$$
x=\frac{560}{11}=50 \frac{10}{11} \approx 50.91 \text{ m}
$$

Marianna Katona (Bp. VIII., Széchenyi közg. g. techn. II. o. t.)

II. solution. According to the problem, $\frac{3}{4}$ of the barrel leaked out in $\left[\frac{3}{4}:\left(\frac{1}{2}+\frac{2}{3}\right)=\right] \frac{9}{14}$ of the entire distance. In the remaining $\frac{5}{14}$ of the distance, $\frac{5}{14} \cdot \frac{2}{3}=\frac{5}{21}$ of the barrel would have leaked out through the lower hole, so over the entire distance, $\frac{3}{4}+\frac{5}{21}=\frac{83}{84}$ of the barrel would have leaked out.

But at 1 meter before the kitchen, $\frac{1}{40}$ of the barrel was left. Therefore, over 1 meter, $\left(\frac{1}{40}-\frac{1}{84}=\right) \frac{11}{840}$ of the barrel leaks out through the lower hole, while over the entire $x$ meters, $\frac{2}{3}$ of the barrel leaks out. Thus,

$$
x=\frac{2}{3}: \frac{11}{840}=\frac{2 \cdot 840}{3 \cdot 11}=\frac{560}{11} \approx 50.91 \text{ m}
$$

Miklós Csapody (Bp. VIII., Piarista g. II. o. t.)

III. solution. The problem can also be solved graphically. On the horizontal axis, we measure the distance traveled by the woman, taking the sought distance $t$ as the unit; on the vertical axis, we measure the amount of water leaked out, taking 1 barrel of water as the unit. The labeling is shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_02_d29f030fd01c2cd4be55g-1.jpg?height=585&width=512&top_left_y=1732&top_left_x=796)

The water leakage on the 1st, 2nd, and 3rd trips is represented by the lines $O A_{1}, O A_{2}$, and $O A_{3}$, respectively (see figure). But on the 3rd trip, only $\frac{3}{4}$ of the barrel leaks out through both holes, and from point $B$ onwards, only the lower hole leaks water until point $C$. Thus, $B C \| O A_{2}$, $D C$ is parallel to the distance axis, $B D=\frac{9}{40} v$, and $D C=t-O B^{\prime}-1$ meters. Since

$$
\triangle B C D \sim \triangle O A_{2} A_{2}^{\prime \prime}
$$

we have

$$
D C: A_{2}^{\prime \prime} A_{2}=D B: A_{2}^{\prime \prime} O
$$

or

$$
D C: t=\frac{9}{40} v: \frac{2}{3} v
$$

from which

$$
D C=\frac{27}{80} t \text{ meters. }
$$

On the other hand,

$$
O B^{\prime}: t=\frac{3}{4} v: \frac{7}{6} v
$$

from which

$$
O B^{\prime}=\frac{9 t}{14}
$$

and thus

$$
D C=t-O B^{\prime}-1=\frac{5 t}{14}-1 \text{ meters. }
$$

Therefore,

$$
\frac{5 t}{14}-1=\frac{27}{80} t
$$

from which

$$
t=\frac{560}{11} \approx 50.91 \text{ meters. }
$$

Tibor Soós (Bp. I., Petőfi g. II. o. t.)