Question

Two factor plants are making tv panels. Yesterday, plant A produced 8000 panels. Ten percent of the panels from plant A and 3% of the panels from plant B were defective. How many panels did plant B produce, if the overall percentage of defective panels from two plants was 7%?

Answer 1

Let panels produced by plant A be represented by a

Let panels produced by plant B be represented by b

Let the total panels produced by both plants be represented by x

Generating the equations from the statements given;

Total panels

$\begin{gathered} a=8000 \\ a+b=x \\ 8000+b=x \end{gathered}$

Defective panels

$\begin{gathered} \frac{10}{100}a=\frac{10}{100}\times8000 \\ a=800 \\ \\ \frac{3}{100}b=\frac{3}{100}\times b \\ =0.03b \\ \text{Total defective panels = 800+0.03b} \end{gathered}$

Since the defective panel is 7% of the total panels, then;

$\begin{gathered} \frac{n\text{ umber of defective panels}}{\text{total panels}}\times100=7 \\ \\ \frac{800+0.03b}{8000+b}\times100=7 \\ \frac{800+0.03b}{8000+b}=\frac{7}{100} \\ C\text{ ross multiplying:} \\ 100(800+0.03b)=7(8000+b) \\ E\text{ xpanding the brackets,} \\ 80000+3b=56000+7b \\ 80000-56000=7b-3b \\ 24000=4b \\ \text{Dividing both sides by 4;} \\ b=\frac{24000}{4} \\ b=6000\text{panels} \end{gathered}$

Therefore, the number of panels produced by plant B is 6000panels