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Difficulty: MediumLinear Equations in One Variable

An empty water tank is being filled by two hoses. Hose A fills the tank at a constant rate of 12 gallons per minute. Hose B fills the tank at a constant rate of 15 gallons per minute. Hose A is turned on first, and 5 minutes later, Hose B is also turned on. If both hoses run until the tank contains a total of 330 gallons of water, for how many minutes was Hose A running?

Answer: 15 minutes

Answer

15
Let mm be the number of minutes Hose A was running. Since Hose B was turned on 5 minutes after Hose A, Hose B was running for m−5m - 5 minutes. The volume of water contributed by Hose A is 12m12m gallons, and the volume of water contributed by Hose B is 15(m−5)15(m - 5) gallons. The sum of these volumes is 330 gallons, which gives the equation 12m+15(m−5)=33012m + 15(m - 5) = 330. Distributing 15 gives 12m+15m−75=33012m + 15m - 75 = 330. Combining like terms yields 27m−75=33027m - 75 = 330. Adding 75 to both sides gives 27m=40527m = 405. Dividing both sides by 27 yields m=15m = 15. Thus, Hose A was running for 15 minutes.

Step-by-Step Solution

1
Define variables for the running time of each hose.
Let Hose A run for mm minutes. Then Hose B runs for m−5m - 5 minutes.
Hose B started 5 minutes after Hose A, so it ran for 5 fewer minutes.
2
Set up a linear equation for the total gallons of water.
12m+15(m−5)=33012m + 15(m - 5) = 330
The sum of the water from both hoses must equal 330 gallons.
3
Simplify and solve the linear equation.
27m=40527m = 405, which gives m=15m = 15.
Distribute, combine like terms, and isolate the variable mm.

Key Concept

Setting up and solving a multi-step linear equation in one variable from a real-world context.
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Topics

Linear Equations in One Variable

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