Weighted Average Problem: Quality Control

Percent & Mixture 9th-10th Grade
PROBLEM
Two factory plants are making TV panels. Yesterday, Plant A produced 12000 panels. 6% of the panels from Plant A and 2% of the panels from Plant B were defective. How many panels did Plant B produce, if the overall percentage of defective panels from the two plants was 5%?

What This Problem Teaches

  • Setting up weighted average equations where quantities and rates both vary
  • Understanding that percentages must be weighted by their corresponding quantities
  • Recognizing mixture problems disguised as manufacturing scenarios
  • Using algebraic manipulation to isolate unknown quantities in percentage contexts
  • Verification strategies for percentage mixture problems

Visualizing the Problem

Two factory plants are making TV panels. Yesterday, Plant A produced 12000 panels. 6% of the panels from Plant A and...

The visual shows how the overall 5% rate sits between the two plant rates. Since 5% is closer to Plant B's 2% rate than Plant A's 6% rate, we might expect Plant A to have produced more panels — which we'll verify algebraically.

Solution: Method 1 — The Weighted Average Approach

Step 1 — Define the unknown

Let x = number of panels produced by Plant B.

Step 2 — Understand what "overall percentage" means

The overall 5% defective rate means that out of all panels produced by both plants combined, exactly 5% are defective. This gives us our fundamental equation:

Total defective panels = 5% of total panels produced

Step 3 — Express each component

We need to count defective panels from each plant separately:

  • Plant A: 0.06 × 12000 = 720 defective panels
  • Plant B: 0.02 × x defective panels
  • Total panels: 12000 + x
  • Total defective: 0.05 × (12000 + x)

Step 4 — Set up the equation

(Defective from A) + (Defective from B) = (5% of total)
720 + 0.02x = 0.05(12000 + x)

Step 5 — Solve for x

Expand the right side:

720 + 0.02x = 600 + 0.05x

Subtract 0.02x from both sides:

720 = 600 + 0.03x

Subtract 600 from both sides:

120 = 0.03x

Divide by 0.03:

x = 120 ÷ 0.03 = 4000

Solution: Method 2 — The Alligation Method

Step 1 — Find the "distances" from target

In mixture problems, we can use the alligation shortcut. Find how far each rate is from the target rate:

  • Plant A: 6% - 5% = 1% above target
  • Plant B: 5% - 2% = 3% below target

Step 2 — Apply the inverse ratio rule

The quantities must be in the inverse ratio of their distances. Since Plant A is 1% away and Plant B is 3% away:

Plant A : Plant B = 3 : 1

This means for every 3 units Plant A produces, Plant B produces 1 unit.

Step 3 — Calculate Plant B's production

Since Plant A produced 12000 panels, and the ratio is 3:1:

12000 panels = 3 units
1 unit = 12000 ÷ 3 = 4000 panels

Therefore, Plant B produced 4000 panels.

Why alligation works: When mixing two components to achieve a target average, the quantities are inversely proportional to their distances from the target. The component farther from the target needs less quantity to balance out.
Plant B produced 4000 panels.

Verification

Let's check our answer by computing the total defective rate:

PlantPanels ProducedDefect RateDefective Panels
Plant A12,0006%720
Plant B4,0002%80
Total16,0005%800
Overall defect rate = 800 ÷ 16000 = 0.05 = 5% ✓

What Trips Students Up

✗ Averaging the percentages directly: Some students calculate (6% + 2%) ÷ 2 = 4% and think this should be the overall rate. This ignores that Plant A produced three times as many panels as Plant B — the weighted nature of the average.
✗ Setting up the wrong target: Writing something like 0.06 + 0.02x = 0.05, forgetting that we're dealing with weighted totals, not just rates. The percentages must be multiplied by their respective quantities.
✗ Misunderstanding "overall percentage": Thinking the 5% applies to each plant individually, rather than to the combined production from both plants.
✗ Unit confusion in alligation: In Method 2, getting the ratio backwards — thinking Plant A:Plant B = 1:3 instead of 3:1. The farther from target, the smaller the required quantity.

The Pattern Behind This

This is a classic weighted average problem. The general formula for any two-component mixture is:

w₁r₁ + w₂r₂ = (w₁ + w₂)r_target

Where w represents weights (quantities) and r represents rates (percentages). In our case:

  • w₁ = 12000 (Plant A panels), r₁ = 0.06
  • w₂ = x (Plant B panels), r₂ = 0.02
  • r_target = 0.05
Key insight: When one quantity is unknown, the problem becomes linear in that variable. When both quantities are unknown, you need additional information to solve.

How to Spot This Problem Type

Look for these telltale phrases:

  • "Overall percentage" or "combined rate"
  • Two or more sources with different rates/percentages
  • One quantity known, another unknown
  • Context involving mixing, manufacturing, or combining groups
  • Words like "defective," "success rate," "concentration," or "composition"

These problems appear in many disguises: drug concentrations in pharmacy, pollution levels in environmental science, quality control in manufacturing, and portfolio returns in finance.

Real Applications

  • Manufacturing quality control: Exactly this scenario — combining output from multiple production lines with different defect rates to meet overall quality targets.
  • Financial portfolio management: Mixing investments with different return rates to achieve a target portfolio return.
  • Pharmaceutical compounding: Combining solutions of different concentrations to create a medication with the prescribed strength.

What If?

1
Different Target Rate
Plant A produced 8000 panels with 4% defective. Plant B has 3% defective. If the overall defect rate was 3.5%, how many panels did Plant B produce?
Step 1 — Set up the equation

Let x = panels from Plant B. Total defective = 0.04(8000) + 0.03x = 320 + 0.03x

Step 2 — Apply the overall rate

320 + 0.03x = 0.035(8000 + x)

Step 3 — Expand and simplify

320 + 0.03x = 280 + 0.035x

Step 4 — Solve for x

40 = 0.005x, so x = 8000

Step 5 — Verify

Check: (320 + 240) ÷ 16000 = 560 ÷ 16000 = 0.035 = 3.5%

Answer: Plant B produced 8000 panels.

2
Three Plants
Plant A made 9000 panels (6% defective), Plant B made 3000 panels (2% defective), and Plant C made an unknown number (1% defective). If the overall defect rate is 4%, how many panels did Plant C produce?
Step 1 — Count known defective panels

Plant A: 0.06 × 9000 = 540 defective
Plant B: 0.02 × 3000 = 60 defective

Step 2 — Set up equation with Plant C

Let x = Plant C panels. Total defective: 540 + 60 + 0.01x = 600 + 0.01x

Step 3 — Apply overall rate

600 + 0.01x = 0.04(9000 + 3000 + x) = 0.04(12000 + x)

Step 4 — Expand and solve

600 + 0.01x = 480 + 0.04x
120 = 0.03x
x = 4000

Step 5 — Verify

Total defective: 540 + 60 + 40 = 640
Total panels: 16000
Rate: 640 ÷ 16000 = 4%

Answer: Plant C produced 4000 panels.

3
Reverse the Known
Plant B produced 6000 panels (2% defective). The combined defect rate of both plants was 4.5%. If Plant A's defect rate was 7%, how many panels did Plant A produce?
Step 1 — Identify knowns

Plant B: 6000 panels, 2% defective = 0.02 × 6000 = 120 defective panels

Step 2 — Set up with Plant A unknown

Let x = Plant A panels. Plant A defective: 0.07x

Step 3 — Write the weighted average equation

120 + 0.07x = 0.045(6000 + x)

Step 4 — Expand and solve

120 + 0.07x = 270 + 0.045x
0.025x = 150
x = 6000

Step 5 — Verify

Plant A defective: 0.07 × 6000 = 420
Total: (420 + 120) ÷ 12000 = 540 ÷ 12000 = 4.5%

Answer: Plant A produced 6000 panels.

4
Cost-Weighted Analysis
Plant A made 9000 panels (6% defective), Plant B made 3000 panels (2% defective). Each defective panel from Plant A costs $8 to fix, from Plant B costs $12 to fix. What is the average cost per defective panel across both plants?
Step 1 — Count defective panels by plant

Plant A defective: 0.06 × 9000 = 540 panels
Plant B defective: 0.02 × 3000 = 60 panels

Step 2 — Calculate total repair costs

Plant A cost: 540 × $8 = $4320
Plant B cost: 60 × $12 = $720

Step 3 — Find totals

Total defective panels: 540 + 60 = 600
Total repair cost: $4320 + $720 = $5040

Step 4 — Calculate weighted average cost

Average cost per defective panel: $5040 ÷ 600 = $8.40

Step 5 — Verify reasonableness

Since Plant A has 90% of the defective panels at $8 each, the average should be close to $8. At $8.40, this makes sense.

Answer: The average cost per defective panel is $8.40.

Frequently Asked Questions

How do you solve weighted average problems with percentages? +
Set up an equation where the sum of (quantity × rate) for each group equals the total quantity times the overall rate. In this problem: (12000 × 0.06) + (x × 0.02) = (12000 + x) × 0.05, where x is the unknown quantity. The key is remembering that percentages must be weighted by their corresponding quantities.
What is the alligation method for mixture problems? +
Alligation uses the distances from each component rate to the target rate to find the ratio. In this example, Plant A is 1% above the 5% target, Plant B is 3% below, so they mix in a 3:1 ratio (inverse of the gaps). Since Plant A made 12000 panels, Plant B made 4000. This shortcut works because the component farther from the target needs less quantity to balance out.
How do you check your answer in percentage mixture problems? +
Calculate the total defective panels from each source and verify they equal the expected total. Here: Plant A contributes 720 defective panels, Plant B contributes 80, totaling 800 defective panels out of 16000 total, which is exactly 5%. Always verify both the individual contributions and the final percentage calculation.
NJ
Neven Jurkovic, PhD

Professor of Computer Science, Palo Alto College, Alamo Colleges District, San Antonio, TX

Developer of Algebrator

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This solution was prepared with AI assistance and reviewed by Dr. Jurkovic for mathematical accuracy and pedagogical clarity.

2026-07-13