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Difficulty: MediumLinear Equations in One Variable

A water reservoir initially contains 450450 liters of water. Water drains out of the reservoir at a constant rate of rr liters per hour, while an inlet pipe supplies water at a constant rate of 1818 liters per hour. If the reservoir contains 390390 liters of water after 66 hours, what is the value of rr?

Answer: 28 liters per hour

Answer

The value of rr is 2828.
The reservoir starts with 450450 liters. Over 66 hours, water enters at 1818 liters/hour and leaves at rr liters/hour, giving a net volume equation of 450+6(18−r)=390450 + 6(18 - r) = 390. Simplifying yields 558−6r=390558 - 6r = 390, which subtracts to −6r=−168-6r = -168, giving r=28r = 28.

Step-by-Step Solution

1
Set up the linear equation representing the net change in water volume over time.
450+6(18−r)=390450 + 6(18 - r) = 390
The final volume equals the initial volume plus the net water added (inflow rate minus outflow rate multiplied by hours).
2
Expand and simplify the linear expression.
558−6r=390558 - 6r = 390
Distribute 66 across (18−r)(18 - r) to obtain 108−6r108 - 6r, then add to 450450.
3
Isolate the variable term −6r-6r.
−6r=−168-6r = -168
Subtract 558558 from both sides of the equation.
4
Solve for the rate rr.
r=28r = 28
Divide both sides by −6-6.

Key Concept

Formulating and solving a linear equation in one variable from a rate problem context.
Estimated Time:1m 30s
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Topics

Linear Equations in One Variable

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