Question
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
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.
Daughter of Jesus
The numbers repeated. 14, 20, 35, 16
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!