Asked by Daughter of Jesus
Two pumps are being filled by two different pumps. The first vat has 141414 gallons is being filled at a rate of 202020 gallons per second. The second vat has 353535 gallons and is being filled at a rate of 161616 gallons per second. After how many seconds will both vats have the same amount of liquid?
Answers
Answered by
Daughter of Jesus
The numbers repeated. 14, 20, 35, 16
Answered by
GPT-4o mini
Let \( t \) be the time in seconds after which both vats will have the same amount of liquid.
For the first vat:
- Initial amount: 141414 gallons
- Filling rate: 202020 gallons per second
The amount in the first vat after \( t \) seconds can be represented as:
\[
A_1 = 141414 + 202020t
\]
For the second vat:
- Initial amount: 353535 gallons
- Filling rate: 161616 gallons per second
The amount in the second vat after \( t \) seconds can be represented as:
\[
A_2 = 353535 + 161616t
\]
We want to find the time \( t \) when both vats have the same amount of liquid:
\[
A_1 = A_2
\]
Setting the equations equal to each other:
\[
141414 + 202020t = 353535 + 161616t
\]
Now, we will isolate \( t \) by first rearranging the equation:
\[
202020t - 161616t = 353535 - 141414
\]
\[
40304t = 212121
\]
Now, we will solve for \( t \):
\[
t = \frac{212121}{40304}
\]
Calculating the division:
\[
t \approx 5.26
\]
To gain a more precise answer:
Performing the division using a calculator:
\[
t \approx 5.26
\]
Thus, the time after which both vats will have the same amount of liquid is approximately \( 5.26 \) seconds.
For the first vat:
- Initial amount: 141414 gallons
- Filling rate: 202020 gallons per second
The amount in the first vat after \( t \) seconds can be represented as:
\[
A_1 = 141414 + 202020t
\]
For the second vat:
- Initial amount: 353535 gallons
- Filling rate: 161616 gallons per second
The amount in the second vat after \( t \) seconds can be represented as:
\[
A_2 = 353535 + 161616t
\]
We want to find the time \( t \) when both vats have the same amount of liquid:
\[
A_1 = A_2
\]
Setting the equations equal to each other:
\[
141414 + 202020t = 353535 + 161616t
\]
Now, we will isolate \( t \) by first rearranging the equation:
\[
202020t - 161616t = 353535 - 141414
\]
\[
40304t = 212121
\]
Now, we will solve for \( t \):
\[
t = \frac{212121}{40304}
\]
Calculating the division:
\[
t \approx 5.26
\]
To gain a more precise answer:
Performing the division using a calculator:
\[
t \approx 5.26
\]
Thus, the time after which both vats will have the same amount of liquid is approximately \( 5.26 \) seconds.
Answered by
GPT-4o mini
It seems that you are emphasizing the repeating numbers in your scenario. Let's re-evaluate with the given specific values and their repetitions.
You have:
- First vat: 141414 gallons, filling at 202020 gallons per second.
- Second vat: 353535 gallons, filling at 161616 gallons per second.
Let’s find out when both vats will have the same amount of liquid using the provided repeated digits.
### Equations for the Amount of Liquid
1. For the first vat:
\[
A_1 = 141414 + 202020t
\]
2. For the second vat:
\[
A_2 = 353535 + 161616t
\]
### Setting the Equations Equal
We set the two amounts equal to each other to find \( t \):
\[
141414 + 202020t = 353535 + 161616t
\]
### Rearranging the Equation
Subtract \( 161616t \) from both sides:
\[
141414 + 202020t - 161616t = 353535
\]
This simplifies to:
\[
141414 + 40304t = 353535
\]
Now, isolate the term with \( t \):
\[
40304t = 353535 - 141414
\]
Calculate \( 353535 - 141414 \):
\[
40304t = 212121
\]
### Solve for \( t \)
Now divide both sides by 40304:
\[
t = \frac{212121}{40304}
\]
Calculating the division:
\[
t \approx 5.26 \, \text{seconds}
\]
So, the time after which both vats will have the same amount of liquid is approximately \( \text{5.26 seconds} \).
If I misunderstood the request regarding the emphasized digits, please clarify so I can assist further!
You have:
- First vat: 141414 gallons, filling at 202020 gallons per second.
- Second vat: 353535 gallons, filling at 161616 gallons per second.
Let’s find out when both vats will have the same amount of liquid using the provided repeated digits.
### Equations for the Amount of Liquid
1. For the first vat:
\[
A_1 = 141414 + 202020t
\]
2. For the second vat:
\[
A_2 = 353535 + 161616t
\]
### Setting the Equations Equal
We set the two amounts equal to each other to find \( t \):
\[
141414 + 202020t = 353535 + 161616t
\]
### Rearranging the Equation
Subtract \( 161616t \) from both sides:
\[
141414 + 202020t - 161616t = 353535
\]
This simplifies to:
\[
141414 + 40304t = 353535
\]
Now, isolate the term with \( t \):
\[
40304t = 353535 - 141414
\]
Calculate \( 353535 - 141414 \):
\[
40304t = 212121
\]
### Solve for \( t \)
Now divide both sides by 40304:
\[
t = \frac{212121}{40304}
\]
Calculating the division:
\[
t \approx 5.26 \, \text{seconds}
\]
So, the time after which both vats will have the same amount of liquid is approximately \( \text{5.26 seconds} \).
If I misunderstood the request regarding the emphasized digits, please clarify so I can assist further!
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