Question

plant a produced 12,000 panels two percent of the panels from plant a and six percent 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 1

SOLUTION

Plant A produce=12,000

Plant A= 2% defective,

Plant B produce=Y

Plant B=6% defective

Total defeative =3%

The total defeactive = Defeative from Plant A + Defeative of Plant B

Defeative from plant A=0.02 x 12,000=240

Defeative from Plant B= 0.06 x Y=0.06Y

Total defeactive=240 + 0.06Y...............eq2

Total Plant = Plant A + Plant B= 12,000 +Y

total Defeactive= 3%= 0.03(12,000+Y).................eq1

hence equating equation 1 and 2

$\begin{gathered} 240+0.06y=0.03(12,000\text{ +y)} \\ 240+0.06y=360+0.03y \end{gathered}$$\begin{gathered} 0.06y-0.03y=360-240 \\ 0.03y=120 \\ y=\frac{120}{0.03}=4000 \end{gathered}$

hence Y =4000

Therefore the number of Panels from Plant B is 4,000