A manufacturing plant uses three assembly lines, L_1, L_2, and L_3, to produce three custo… — GMAT

A manufacturing plant uses three assembly lines, $L_1$ , $L_2$ , and $L_3$ , to produce three custom components, $X$ , $Y$ , and $Z$ .

- Producing one unit of component $X$ requires 2 hours on $L_1$ , 1 hour on $L_2$ , and 3 hours on $L_3$ .
- Producing one unit of component $Y$ requires 3 hours on $L_1$ , 4 hours on $L_2$ , and 2 hours on $L_3$ .
- Producing one unit of component $Z$ requires 1 hour on $L_1$ , 2 hours on $L_2$ , and 4 hours on $L_3$ .

During a given week, assembly lines $L_1$ , $L_2$ , and $L_3$ were operated for a total of 140 hours, 165 hours, and 235 hours, respectively, with zero idle time. Assuming full capacity utilization, how many units of component $Z$ were produced during that week?

Answer: 30 units

Answer

30 units of component Z were produced.

Translating the assembly line operational hours into a 3x3 system of linear equations yields

2X + 3Y + Z = 140

X + 4Y + 2Z = 165

, and

3X + 2Y + 4Z = 235

. Eliminating

X

results in two equations in

Y

and

Z

5Y + 3Z = 190

and

5Y + Z = 130

. Subtracting these equations gives

2Z = 60

, so

Z = 30

Step-by-Step Solution

Formulate a system of 3 linear equations representing total hours logged on each assembly line.

Line 1:

2X + 3Y + Z = 140

; Line 2:

X + 4Y + 2Z = 165

; Line 3:

3X + 2Y + 4Z = 235

Each component requires specific line processing time, and total time per line equals total available capacity.

Eliminate variable

X

by substituting

X = 165 - 4Y - 2Z

into the other two equations.

Equation A:

5Y + 3Z = 190

and Equation B:

5Y + Z = 130

Reducing a 3-variable system to a 2-variable system simplifies linear elimination.

Subtract Equation B from Equation A.

2Z = 60, so Z = 30.

Since the coefficients of

Y

in both reduced equations are identical (

5Y

), subtraction directly isolates

Z

Key Concept

Solving Systems of Three Linear Equations via Variable Substitution and Elimination

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