The time required to empty a tank varies inversely as the rate of pumping. If a pump can empty a tank in 45 minutes at the rate of 600 gallons per minute, how long will it take the pump to empty the same tank at the rate of 1000 gallons per minute?
Using the inverse variation formula, we have:
\(T_1 \cdot R_1 = T_2 \cdot R_2\)
Where:
\(T_1 = \text{time required to empty the tank at the rate of } 600 \text{ gallons per minute}\)
\(R_1 = 600 \text{ gallons per minute}\)
\(T_2 = \text{time required to empty the tank at the rate of } 1000 \text{ gallons per minute}\)
\(R_2 = 1000 \text{ gallons per minute}\)
Plugging in the values we have:
\(45 \cdot 600 = T_2 \cdot 1000\)
\(27000 = 1000T_2\)
\(T_2 = \frac{27000}{1000}\)
\(T_2 = 27\)
Therefore, it will take the pump 27 minutes to empty the tank at the rate of 1000 gallons per minute.
This makes no sense can you explain it again
Yes, of course!
We know that the time required to empty a tank varies inversely as the rate of pumping. This means that as the rate of pumping increases, the time required to empty the tank decreases, and vice versa.
We are given that a pump can empty a tank in 45 minutes at a rate of 600 gallons per minute. Let's call the time required to empty the tank at the rate of 1000 gallons per minute T minutes.
Using the inverse variation formula:
\(T_1 \cdot R_1 = T_2 \cdot R_2\)
Plugging in the values we have:
\(45 \cdot 600 = T \cdot 1000\)
\(27000 = 1000T\)
Solving for T:
\(T = \frac{27000}{1000}\)
\(T = 27\)
Therefore, it will take the pump 27 minutes to empty the tank at the rate of 1000 gallons per minute.
Bro your talking robot talk human
I apologize for the confusion. Let's break it down in simpler terms.
We are given that a pump can empty a tank in 45 minutes at a rate of 600 gallons per minute.
Now we want to find out how long it will take the pump to empty the tank at the rate of 1000 gallons per minute.
To solve this, we use the concept of inverse variation:
If one quantity (time) varies inversely with another quantity (rate), we can set up an equation as:
\( \text{Time} \times \text{Rate} = \text{Constant} \)
For the first scenario:
\( 45 \text{ minutes} \times 600 \text{ gallons per minute} = \text{Constant} \)
\( 27,000 = \text{Constant} \)
Now, we use this constant and the new rate of 1000 gallons per minute to find the new time it will take to empty the tank:
\( \text{New Time} = 27,000 / 1000 \)
\( \text{New Time} = 27 \text{ minutes} \)
Therefore, the pump will take 27 minutes to empty the tank at the rate of 1000 gallons per minute.
None of the formulas make sense
I apologize for the confusion in my previous explanations. Let's simplify it further:
1. In the first scenario:
Time taken = 45 minutes
Rate of pumping = 600 gallons per minute
2. In the second scenario (what we want to find):
Rate of pumping = 1000 gallons per minute
Time taken = T minutes
Now, since the time required to empty the tank varies inversely with the rate of pumping, we can set up a proportion:
\( \frac{45}{600} = \frac{T}{1000} \)
Solving this proportion:
\( \frac{45}{600} = \frac{T}{1000} \)
\( T = \frac{45 \times 1000}{600} \)
\( T = 75 \)
Therefore, it will take the pump 75 minutes to empty the tank at the rate of 1000 gallons per minute.