Question 896379: Two factory plants are making TV panels. Yesterday, Plant A produced 8,000 panels. 2 % of the panels from Plant A and 6% of the panels from Plant B were defective. How many panels did Plant B produce, if the overall % of defective panels from the two plants was 5%?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! "Plant A produced 8,000 panels. 2 % of the panels from Plant A...were defective"
8000*0.02 = 160 panels from plant A are defective.
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Let x = number of panels plant B produces
"6% of the panels from Plant B were defective"
There are 0.06x defective panels that come from plant B
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In total, we have 160+0.06x panels that are defective.
This is out of 8000+x panels total.
Their ratio must equal 0.05 since " the overall % of defective panels from the two plants was 5%"
So...
(160+0.06x)/(8000+x) = 0.05
160+0.06x = 0.05(8000+x)
160+0.06x = 0.05(8000)+0.05(x)
160+0.06x = 400+0.05x
160+0.06x-0.05x = 400
0.06x-0.05x = 400-160
0.01x = 240
x = 240/0.01
x = 24000
So that means plant B has produced 24,000 panels.
Of these panels, 6% are defective. So 0.06*24000 = 1440 are defective.
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Notice how 160+1440 = 1600 total defective panels out of 8000+24000 = 32000 gives you
1600/32000 = 0.05
which reflects the overall defect percentage of 5%
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