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Two pumps are being filled by two different pumps. The first vat has 161616 gallons is being filled at a rate of 171717 gallons per second. The second vat has 232323 gallons and is being filled at a rate of 151515 gallons per second. After how many seconds will both vats have the same amount of liquid?

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Let \( t \) represent the time in seconds after both vats start being filled.

  1. For the first vat, we start with 161616 gallons and it is being filled at a rate of 171717 gallons per second. The amount of liquid in the first vat after \( t \) seconds can be expressed as: \[ \text{Amount in first vat} = 161616 + 171717t \]

  2. For the second vat, we start with 232323 gallons and it is being filled at a rate of 151515 gallons per second. The amount of liquid in the second vat after \( t \) seconds can be expressed as: \[ \text{Amount in second vat} = 232323 + 151515t \]

We want to find the time \( t \) when the amounts in both vats are equal. Therefore, we set the two equations equal to each other: \[ 161616 + 171717t = 232323 + 151515t \]

Now, we will simplify this equation. First, we can move all terms involving \( t \) to one side and constant terms to the other side: \[ 171717t - 151515t = 232323 - 161616 \]

Now simplify both sides: \[ (171717 - 151515)t = 232323 - 161616 \] \[ 20102t = 70707 \]

Next, we solve for \( t \): \[ t = \frac{70707}{20102} \]

Calculating \( t \): \[ t \approx 3.5 \]

Thus, \( t \) in seconds when both vats will have the same amount of liquid is approximately: \[ \boxed{3.5} \text{ seconds} \]