Question 1046826: Two factory plants are making TV panels. Yesterday, Plant A produced 8000 panels. 6% percent 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 3%?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! plant A produced 8000 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 3%.
if x is the number of panels produced by plant A and y is the number of panels produced by plant B, then:
.06 * x + .02 * y = .03 * (x + y)
you know that x equals 8000, so the formula becomes:
.06 * 8000 + .02 * y = .03 * (8000 + y)
simplify this equation to get:
480 + .02 * y = 240 + .03 * y
subtract 240 from both sides of this equation and subtract .02 * y from both sides of this equation to get:
240 = .01 * y
solve for y to get:
y = 240 / .01 = 24,000.
plant B produced 24,000.
together they produced 8,000 + 24,000 = 32,000 panels.
.06 * 8000 = 480 defective from plant A.
.02 * 24000 = 480 defective from plant B.
total defective = 960.
total percent defective from both plants was 960 / 32000 * 100 = 3 %.
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