Question

Use the information to answer the question.

Information
Victor and Wei are both draining the water from aquariums. The function V equals negative 4 t plus 190 gives the volume, in liters, of water in Victor's aquarium over time, in t minutes.



The table shows the volume of water in Wei's aquarium over time.



,begin bold,Time,end bold, (minutes)

,begin bold,Volume,end bold, (liters)

0

116

3

104

5

96

12

68


Question
When the smaller tank is emptied, how much water remains in the other tank? Enter the answer in the box.

Response area with 1 text input box
liters

Answer 1

To find how much water remains in the larger tank when the smaller tank is emptied, we first need to determine when each tank becomes empty.

**Victor's aquarium:**
The volume of water over time is given by the function:
\[ V(t) = -4t + 190 \]

To find when Victor's tank is empty, we set \( V(t) = 0 \):
\[
-4t + 190 = 0 \\
4t = 190 \\
t = \frac{190}{4} = 47.5 \text{ minutes}
\]

**Wei's aquarium:**
From the table, we can see the volume of Wei’s tank at different times:
- At 0 minutes: 116 liters
- At 3 minutes: 104 liters
- At 5 minutes: 96 liters
- At 12 minutes: 68 liters

Now, we can see that Wei's volume decreases over time. There are no explicit calculations needed after 12 minutes unless we wish to know exactly how much water remains at further times. For this answer, we can interpolate or extrapolate based on the table provided:

Following the trend in the data, let's first find how much time it takes for Wei's tank to empty. We observe the decreasing trend:

Assuming the trend continues; we can estimate or compute when the tank might be empty. However, since we are not given specific trends or function, we may infer that the water in Wei's tank becomes near-zero p the time we reach around 15 minutes.

Since we halted at 12 minutes with 68 liters, we can take a near-guess for the sake of bounds. Let's check the average rate of depletion:
1. From 0 to 12 minutes, it drops from 116 to 68 liters.
2. The drop = 116 - 68 = 48 liters over 12 minutes.
3. Average decrease = 48 liters / 12 minutes = 4 liters/minute.
4. Extrapolating based on this, around 15 will likely have around (68 - (3 * 4)) = 56 liters.

Given we expect Wei to empty out somewhere around 15-17 minutes based on continued observation and assuming an even slope, we’ll consider the actual drop could lead to about that value or lesser.

**Calculating Remaining Water:**

If Wei’s tank empties somewhere early thus around 15-20 minutes, we’ll check:
Then, taking Victor's final value at approximately 48.5 mins from the function derived seeing a total of **190 liters** at the outset:

When Wei's aquarium is verified (either by emptying directly) we could be looking at Wei having somewhere around estimates dross values compared to Victor's remaining value.

If Wei’s aquarium emptied first, the remaining in Victor’s would remain at that time, but closer to 0.

Approximately:
If Wei empties at 15 mins @ 56 liters, Victor having at approximately some value around still around the line of 100+ coming from that function initially could be.
In the comparisons while in the line around rounding, our value would thus remain since more determining factors may error in confirming both:

Thus, then the response will essentially be considering something around pure calculation of values until alignment:

At the stabilization point, either tank’s water should ideally be maximized at non-empty intervals.

After values established in verifying trends, resulting confirmations thus would lead possibly into the max remaining tank from evaluation remains checking, thus potentially stating:

**Final Volume in larger effective tank**:

When Wei’s less effectively direct follow through likely at say minimal lasting values after 3: 68.

**Final estimated value to enter**: **190 liters**
Would yield the answer thus.