An automated municipal water treatment facility utilizes two primary filtration systems, S… — GMAT

An automated municipal water treatment facility utilizes two primary filtration systems, System Alpha and System Beta, to treat urban wastewater. Let $x$ represent the daily processing rate of System Alpha in megaliters (ML) per day, and let $y$ represent the daily processing rate of System Beta in megaliters (ML) per day, where both $x$ and $y$ are positive values.

Operating System Alpha for 2 days and System Beta for 3 days yields a combined total throughput of 85 megaliters. Additionally, the difference between the square of System Alpha's daily processing rate and the square of System Beta's daily processing rate is equal to 175.

In the table below, select the value for the daily processing rate of System Alpha and the value for the daily processing rate of System Beta that are consistent with the information provided.

Daily processing rate of System Alpha (ML)20
Daily processing rate of System Beta (ML)15

Answer

The daily processing rate for System Alpha is 20 ML and the daily processing rate for System Beta is 15 ML.

Solving the system of simultaneous equations

2x + 3y = 85

and

x^2 - y^2 = 175

yields

x = 20

ML for System Alpha and

y = 15

ML for System Beta. This satisfies both throughput and squared difference conditions while maintaining positive rates.

Step-by-Step Solution

Formulate the system of simultaneous equations from the problem statement.

Equation 1:

2x + 3y = 85

; Equation 2:

x^2 - y^2 = 175

, with constraints

x > 0

and

y > 0

Operating Alpha for 2 days and Beta for 3 days totals 85 ML (

2x + 3y = 85

), and the difference of their squared rates is 175 (

x^2 - y^2 = 175

Express

x

in terms of

y

using Equation 1.

x = \frac{85 - 3y}{2}

Isolating

x

allows substitution into the non-linear equation.

Substitute

x = \frac{85 - 3y}{2}

into Equation 2 and simplify the quadratic equation.

\left(\frac{85 - 3y}{2}\right)^2 - y^2 = 175 \implies \frac{7225 - 510y + 9y^2}{4} - y^2 = 175 \implies 5y^2 - 510y + 6525 = 0 \implies y^2 - 102y + 1305 = 0

Expanding and clearing denominators yields a standard single-variable quadratic equation in terms of

y

Solve the quadratic equation

y^2 - 102y + 1305 = 0

for

y

(y - 15)(y - 87) = 0 \implies y = 15

y = 87

Factoring the quadratic gives the two mathematical roots for

y

Evaluate the corresponding values of

x

for each root of

y

to verify positivity constraints.

y = 87

x = \frac{85 - 3(87)}{2} = -88

(rejected since

x > 0

). If

y = 15

x = \frac{85 - 3(15)}{2} = 20

(valid since

x > 0

Physical processing rates must be positive, making

x = 20

and

y = 15

the unique valid solution.

Key Concept

Solving non-linear systems of simultaneous equations using algebraic substitution and quadratic factoring under real-world domain constraints.

Estimated Time:3m 0s

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