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Difficulty: HardLinear Equations in One Variable

Two water pumps, Pump A and Pump B, were used to drain a reservoir containing 12,000 gallons of water. Pump A operates at a constant rate that is 50 gallons per hour greater than the rate of Pump B. Pump A worked alone for 4 hours, after which both pumps worked together for another 8 hours to completely empty the reservoir. What is the pumping rate of Pump B, in gallons per hour?

Answer: 570 gallons per hour

Answer

570
Let rr represent the rate of Pump B in gallons per hour. Pump A's rate is (r+50)(r + 50) gallons per hour. In the first 4 hours, Pump A drains 4(r+50)=4r+2004(r + 50) = 4r + 200 gallons. In the next 8 hours, both pumps operate together at a combined rate of (r+50)+r=2r+50(r + 50) + r = 2r + 50 gallons per hour, draining 8(2r+50)=16r+4008(2r + 50) = 16r + 400 gallons. Adding both quantities gives total volume drained: (4r+200)+(16r+400)=12,000(4r + 200) + (16r + 400) = 12,000. Simplifying gives 20r+600=12,00020r + 600 = 12,000, so 20r=11,40020r = 11,400, which yields r=570r = 570 gallons per hour.

Step-by-Step Solution

1
Define variables for the rate of each pump
Rate of Pump B = rr gal/hr; Rate of Pump A = r+50r + 50 gal/hr
Establishing a single unknown variable allows setting up a one-variable linear equation.
2
Write expressions for water drained during each time period
Period 1 (Pump A alone for 4 hrs): 4(r+50)=4r+2004(r + 50) = 4r + 200; Period 2 (Both pumps for 8 hrs): 8(2r+50)=16r+4008(2r + 50) = 16r + 400
Work done equals rate multiplied by time for each phase of operation.
3
Sum the work done in both periods to equal total volume and solve for rr
20r+600=12,000⇒20r=11,400⇒r=57020r + 600 = 12,000 \Rightarrow 20r = 11,400 \Rightarrow r = 570
Solving the linear equation yields the exact rate of Pump B.

Key Concept

Linear Equations in One Variable
Estimated Time:2m 0s
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Linear Equations in One Variable

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