Question

Jada walks up to a tank of water that can hold up to 10 gallons. When it is active, a drain empties water from the tank at a constant rate. When Jada first sees the tank, it contains 5 gallons of water. Six minutes later, the tank contains 3 gallons of water. Just type the number for your answer, and be prepared to explain or show your reasoning for all answers. 1. At what rate is the amount of water in the tank changing? Use a signed number for the rate. 2. How many more minutes will it take for the tank to drain completely? 3. How many minutes before Jada arrived was the water tank completely full?

Answer 1

1) Rate of change of water in the tank = -⅓ gallons per minute = -0.333 gallons per minute.

2) It would take the tank an extra 9 minutes to drain completely from the current level of 3 gallons of water.

3) THe tank must have been draining for 15 minutes before Jada first arrives at the water tank.

Step-by-step explanation

1) The rate of leakage or rate of change of the volume of water in the tank is given as

`\text{Rate of leakage = }\frac{Change\text{ in the volume of water in the tank}}{Time\text{ take for this change}}`

For this tank, Jada first sees the water in the tank at 5 gallons and then, she then sees the water in the tank at 3 gallons, 6 minutes later.

Change in the volume of water in the tank = (Final volume) - (Initial volume)

Change in the volume of water in the tank = 3 - 5 = -2 gallons

Time taken for this change = 6 minutes

Rate of change of water in the tank = (-2/6) = -⅓ gallons per minute = -0.333 gallons per minute.

2) How many minutes will it take for the tank to be completely empty?

`\begin{gathered} \text{Rate of change of water in the tank = }\frac{Change\text{ in the volume of water in the tank}}{Time\text{ take for this change}} \\ \text{Rate of change of water in the tank = }\frac{(final\text{ volume) - (Initial volume)}}{Time\text{ taken}} \end{gathered}`

If we want to find the time it would take for the tank to now be empty, from the current volume of 3 gallons that it is now,

Final volume = 0 gallon

Initial volume = 3 gallons

Time taken = t = ?

Rate of change of water in the tank = -0.333 gallons per minute.

`\begin{gathered} \text{Rate of change of water in the tank = }\frac{(final\text{ volume) - (Initial volume)}}{Time\text{ taken}} \\ -0.3333=(0-3)/(t) \\ -0.3333=(-3)/(t) \\ \text{Cross multiply} \\ t=(-3)/(-0.3333) \\ t=9\text{ minutes} \end{gathered}`

3) Before Jada's arrival, the tank must have been full at 10 gallons.

When she first arrives, she meets the tank at 5 gallons.

The time it takes the tank to drain from 10 gallons to 5 gallons is what we need to find now

Final volume = 5 gallon

Initial volume = 10 gallons

Time taken = t = ?

Rate of change of water in the tank = -0.333 gallons per minute.

`\begin{gathered} \text{Rate of change of water in the tank = }\frac{(final\text{ volume) - (Initial volume)}}{Time\text{ taken}} \\ -0.3333=(5-10)/(t) \\ -0.3333=(-5)/(t) \\ \text{Cross multiply} \\ t=(-5)/(-0.3333) \\ t=15\text{ minutes} \end{gathered}`

Hope this Helps!!!