To determine after how many seconds both vats will have the same amount of liquid, we can denote the time in seconds as \( t \).
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First vat:
- Initial amount: 888 gallons
- Filling rate: 161616 gallons per second
- Amount after \( t \) seconds: \( 888 + 161616t \)
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Second vat:
- Initial amount: 383838 gallons
- Filling rate: 141414 gallons per second
- Amount after \( t \) seconds: \( 383838 + 141414t \)
To find when both vats have the same amount of liquid, we set the two expressions equal to each other:
\[ 888 + 161616t = 383838 + 141414t \]
Next, we rearrange the equation to isolate \( t \):
Subtract \( 141414t \) from both sides:
\[ 888 + 161616t - 141414t = 383838 \]
Combine like terms:
\[ 888 + (161616 - 141414)t = 383838 \] \[ 888 + 20102t = 383838 \]
Now, subtract 888 from both sides:
\[ 20102t = 383838 - 888 \] \[ 20102t = 382950 \]
Next, solve for \( t \) by dividing both sides by 20102:
\[ t = \frac{382950}{20102} \]
Perform the division:
\[ t \approx 19.04 \]
Since time is typically expressed in whole seconds, we can either round to 19 seconds or check both 19 seconds and 20 seconds for precision.
Calculate amounts at \( t = 19 \) and \( t = 20 \):
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For \( t = 19 \):
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First vat: \[ 888 + 161616 \times 19 = 888 + 3060704 = 3061592 \]
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Second vat: \[ 383838 + 141414 \times 19 = 383838 + 2686846 = 3060684 \]
They are not equal.
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For \( t = 20 \):
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First vat: \[ 888 + 161616 \times 20 = 888 + 3232320 = 3233208 \]
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Second vat: \[ 383838 + 141414 \times 20 = 383838 + 2828280 = 3212118 \]
Again they are not equal.
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Finding the exact second where they match is challenging without decimals, but the calculation indicates they will meet between 19 and 20 seconds.
Thus, to conclude: \[ \text{They will have the same amount of liquid at approximately } 19.04 \text{ seconds.} \]
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