Question

Two water tanks at a factory begin draining at the same time. The large tank contains 1198 liters of water and drains at a rate of 19 liters each minute. The small tank holds 120 liters of water but is only 5 6 full. It drains at a much slower rate of only 1 liter each minute. How many liters of water are in each tank at the moment that they have the same amount of water?

Answer 1

To determine the moment when both tanks contain the same amount of water, we can set up equations for the volume of water in each tank as a function of time, \( t \) (in minutes).

1. **Large Tank**:
- Initial volume of water = 1198 liters.
- Draining rate = 19 liters per minute.
- Volume of water after \( t \) minutes:
\[
V_{\text{large}}(t) = 1198 - 19t
\]

2. **Small Tank**:
- It is only \( \frac{5}{6} \) full, which means:
\[
\text{Initial volume of small tank} = \frac{5}{6} \times 120 = 100 \text{ liters}
\]
- Draining rate = 1 liter per minute.
- Volume of water after \( t \) minutes:
\[
V_{\text{small}}(t) = 100 - 1t
\]

Now, we want to find \( t \) when the two volumes are equal:
\[
1198 - 19t = 100 - 1t
\]

Next, we can solve this equation:

1. Rearranging the equation:
\[
1198 - 100 = 19t - 1t
\]
\[
1098 = 18t
\]

2. Dividing both sides by 18:
\[
t = \frac{1098}{18} = 61
\]

Now we will calculate the volume of water in each tank at \( t = 61 \) minutes.

**For the large tank**:
\[
V_{\text{large}}(61) = 1198 - 19 \times 61
\]
Calculating \( 19 \times 61 \):
\[
19 \times 61 = 1159
\]
Then,
\[
V_{\text{large}}(61) = 1198 - 1159 = 39 \text{ liters}
\]

**For the small tank**:
\[
V_{\text{small}}(61) = 100 - 1 \times 61
\]
Calculating:
\[
V_{\text{small}}(61) = 100 - 61 = 39 \text{ liters}
\]

Therefore, the amount of water in each tank when they contain the same volume is:
\[
\boxed{39 \text{ liters}}
\]