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Question Find the number of TV panels produced by Plant B, if the two plants together produced 740 defective panels, where Plant A produced 3000 fewer panels than Plant B and the defective rate was 2%2\% for Plant A and 3%3\% for Plant B.

Studdy Solution

STEP 1

Assumptions
1. Let the number of panels produced by Plant B be xx.
2. Plant A produced 30003000 fewer panels than Plant B, so the number of panels produced by Plant A is x−3000x - 3000.
3. Two percent of the panels from Plant A were defective, which is 0.02(x−3000)0.02(x - 3000).
4. Three percent of the panels from Plant B were defective, which is 0.03x0.03x.
5. The total number of defective panels from both plants is 740740.

STEP 2

Set up the equation representing the total number of defective panels from both plants.
0.02(x−3000)+0.03x=7400.02(x - 3000) + 0.03x = 740

STEP 3

Distribute the 0.020.02 to both terms in the parentheses.
0.02x−0.02⋅3000+0.03x=7400.02x - 0.02 \cdot 3000 + 0.03x = 740

STEP 4

Combine like terms (0.02x0.02x and 0.03x0.03x).
0.05x−60=7400.05x - 60 = 740

STEP 5

Add 6060 to both sides of the equation to isolate the term with xx.
0.05x=740+600.05x = 740 + 60

STEP 6

Calculate the sum on the right-hand side of the equation.
0.05x=8000.05x = 800

STEP 7

Divide both sides of the equation by 0.050.05 to solve for xx.
x=8000.05x = \frac{800}{0.05}

STEP 8

Calculate the value of xx.
x=16000x = 16000
Plant B produced 16,000 panels.

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