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Difficulty: Very hardRatios, Rates, and Percentages

A high-precision semiconductor fabrication plant operates two automated lithography processing lines, Line X and Line Y.

- Line X processes silicon wafers at a constant normal rate of rXr_X wafers per hour with a defect-free yield rate of 90%90\%.
- Line Y processes silicon wafers at a constant normal rate of rYr_Y wafers per hour with a defect-free yield rate of 80%80\%.
- When both lines operate simultaneously at their respective normal rates for 10 hours, they produce a combined total of 4,300 defect-free wafers.
- If Line X operates at 110%110\% of its normal processing rate while Line Y operates at 125%125\% of its normal processing rate, operating simultaneously for 10 hours yields a combined total of 4,970 defect-free wafers.

Which of the following paired values represents the normal processing rate for Line X (rXr_X) and the normal processing rate for Line Y (rYr_Y), in wafers per hour?

  1. A
    Line X: 250 wafers/hr; Line Y: 250 wafers/hr
  2. Line X: 300 wafers/hr; Line Y: 200 wafers/hrAnswer
  3. C
    Line X: 200 wafers/hr; Line Y: 300 wafers/hr
  4. D
    Line X: 350 wafers/hr; Line Y: 150 wafers/hr
  5. E
    Line X: 270 wafers/hr; Line Y: 160 wafers/hr

Answer

Line X: 300 wafers/hr; Line Y: 200 wafers/hr
The correct option gives Line X processing 300 wafers per hour and Line Y processing 200 wafers per hour. Under normal conditions, Line X outputs 300×0.90=270300 \times 0.90 = 270 defect-free wafers/hr and Line Y outputs 200×0.80=160200 \times 0.80 = 160 defect-free wafers/hr. In 10 hours, the total defect-free output is (270+160)×10=4,300(270 + 160) \times 10 = 4,300 wafers. Under the adjusted rates, Line X processes 300×1.10=330300 \times 1.10 = 330 wafers/hr (297297 defect-free) and Line Y processes 200×1.25=250200 \times 1.25 = 250 wafers/hr (200200 defect-free), yielding (297+200)×10=4,970(297 + 200) \times 10 = 4,970 defect-free wafers in 10 hours, satisfying both system constraints.

Step-by-Step Solution

1
Set up the linear equations for the defect-free hourly rates of Line X (rXr_X) and Line Y (rYr_Y).
The hourly defect-free output for Line X is 0.90rX0.90 r_X and for Line Y is 0.80rY0.80 r_Y. Over 10 hours, 10(0.90rX+0.80rY)=4,300  ⟹  0.90rX+0.80rY=43010(0.90 r_X + 0.80 r_Y) = 4,300 \implies 0.90 r_X + 0.80 r_Y = 430. Multiplying by 10 gives Equation (1): 9rX+8rY=4,3009 r_X + 8 r_Y = 4,300.
Converting the 10-hour total output into hourly rate expressions establishes the first system equation.
2
Set up the linear equation for the altered processing rates.
Line X operates at 1.10rX1.10 r_X with 90%90\% yield, so its hourly defect-free rate is 0.90(1.10rX)=0.99rX0.90(1.10 r_X) = 0.99 r_X. Line Y operates at 1.25rY1.25 r_Y with 80%80\% yield, so its hourly defect-free rate is 0.80(1.25rY)=1.00rY0.80(1.25 r_Y) = 1.00 r_Y. Over 10 hours, 10(0.99rX+1.00rY)=4,970  ⟹  0.99rX+1.00rY=49710(0.99 r_X + 1.00 r_Y) = 4,970 \implies 0.99 r_X + 1.00 r_Y = 497. Multiplying by 100 gives Equation (2): 99rX+100rY=49,70099 r_X + 100 r_Y = 49,700.
Applying the rate changes to each line yields the second system equation.
3
Solve the system of simultaneous linear equations.
From Equation (1), express rYr_Y in terms of rXr_X: 8rY=4,300−9rX  ⟹  100rY=12.5(4,300−9rX)=53,750−112.5rX8 r_Y = 4,300 - 9 r_X \implies 100 r_Y = 12.5(4,300 - 9 r_X) = 53,750 - 112.5 r_X. Substitute into Equation (2): 99rX+(53,750−112.5rX)=49,700  ⟹  −13.5rX=49,700−53,750=−4,05099 r_X + (53,750 - 112.5 r_X) = 49,700 \implies -13.5 r_X = 49,700 - 53,750 = -4,050. Thus, rX=−4,050−13.5=300r_X = \frac{-4,050}{-13.5} = 300. Substituting rX=300r_X = 300 into Equation (1): 9(300)+8rY=4,300  ⟹  2,700+8rY=4,300  ⟹  8rY=1,600  ⟹  rY=2009(300) + 8 r_Y = 4,300 \implies 2,700 + 8 r_Y = 4,300 \implies 8 r_Y = 1,600 \implies r_Y = 200.
Eliminating rYr_Y gives the exact unique values for the processing rates rX=300r_X = 300 wafers/hr and rY=200r_Y = 200 wafers/hr.

Key Concept

Simultaneous Rate and Percentage Equations in Two-Part Analysis
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Topics

Ratios, Rates, and Percentages

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