A manufacturing facility uses Machine A and Machine B to process standard orders of raw ma… — GRE General Test

A manufacturing facility uses Machine A and Machine B to process standard orders of raw materials. Operating independently at its constant rate, Machine A requires $x$ hours to process one standard order, where $x > 0$ . Machine B operates at a constant rate and requires $x + 2$ hours to process one standard order. When both machines operate simultaneously at their respective constant rates for $6$ hours, the total number of standard orders processed is $2$ fewer than the number of standard orders Machine A would process operating alone for $18$ hours. What is the value of $x$ ?

A
2
B
3
4Answer
D
6
E
8

Answer

The rate of Machine A is

\frac{1}{x}

orders per hour, and the rate of Machine B is

\frac{1}{x+2}

orders per hour. In 18 hours, Machine A processes

\frac{18}{x}

orders. Working together for 6 hours, both machines process

6\left(\frac{1}{x} + \frac{1}{x+2}\right)

orders. Setting up the difference:

\frac{18}{x} - 6\left(\frac{1}{x} + \frac{1}{x+2}\right) = 2

, which simplifies to

\frac{12}{x} - \frac{6}{x+2} = 2

. Multiplying both sides by

x(x+2)

yields

12(x+2) - 6x = 2x^2 + 4x

, leading to

2x^2 - 2x - 24 = 0

x^2 - x - 12 = 0

. Factoring gives

(x-4)(x+3) = 0

. Because time must be positive,

x = 4

Step-by-Step Solution

Express the individual work rates of Machine A and Machine B.

Machine A completes

\frac{1}{x}

orders per hour; Machine B completes

\frac{1}{x+2}

orders per hour.

Work rate is the reciprocal of the total time required to complete one unit of work.

Formulate the equation based on the total orders processed in the given time frames.

18 \cdot \frac{1}{x} - 6\left(\frac{1}{x} + \frac{1}{x+2}\right) = 2

Machine A alone in 18 hours processes

\frac{18}{x}

orders. Together in 6 hours, they process

6\left(\frac{1}{x} + \frac{1}{x+2}\right)

orders, which is 2 orders less.

Simplify the algebraic equation.

\frac{12}{x} - \frac{6}{x+2} = 2

Subtracting

6 \cdot \frac{1}{x}

from

18 \cdot \frac{1}{x}

yields

\frac{12}{x}

Clear the denominators by multiplying through by

x(x+2)

and solve the resulting quadratic equation.

12(x+2) - 6x = 2x(x+2) \implies 6x + 24 = 2x^2 + 4x \implies 2x^2 - 2x - 24 = 0 \implies x^2 - x - 12 = 0 \implies (x-4)(x+3) = 0

Clearing denominators transforms the rational equation into a standard quadratic equation.

Select the physically meaningful solution for time

x

x = 4

hours (rejecting

x = -3

since

x > 0

Time must be positive.

Key Concept

Algebraic Work-Rate Modeling and Quadratic Solution

Estimated Time:2m 0s

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