Q:

# Two factory plants are making TV panels. Yesterday, Plant A produced 16,000 panels. Five percent of the panels from Plant A and 2% of the panels from Plant Bwere defective. How many panels did Plant B produce, if the overall percentage of defective panels from the two plants was 4%?Number of panels produced by Plant B????

Accepted Solution

A:
Answer:Number of panels produced by Plant B is 8,000.Step-by-step explanation:The number of panels produced by Plant A  = 16,000The number of defective panels from A = 5%Now, 5% of 16,000 = $$\frac{5}{100} \times 16,000 = 800$$So, out of  16,000 panels produced by pant A , 800 are defective. .. (1)Let us assume number of panels produced by Plant B  = mThe number of defective panels from B = 2%Now, 2% of m = $$\frac{2}{100} \times m = 0.02 m$$So, out of total m panels produced by pant B , 0.02 m are defective. .. (2) Now, total panels produced over all = Panels by A  + Panels by B= 16,000 + m The percentage of defected panels over all = 4%Now, 4% of (16,000 + m) = $$\frac{4}{100} \times (16,000+m) = (0.04)(16,000 + m)$$ Also, the total number of defective panels = Defective from A  + Defective from B ⇒(0.04)(16,000 + m)  =  800 +   0.02 m   from (1) and (2)or, 640 + 0.04 m = 800 + 0.02or, 0.02 m = 160⇒ m = 160 /0.02 = 8,000or, m = 8,000Hence, Number of panels produced by Plant B is 8,000.