478,162 views
13 votes
13 votes
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%?

User Draca
by
2.5k points

1 Answer

13 votes
13 votes

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} (10)/(100)a=(10)/(100)*8000 \\ a=800 \\ \\ (3)/(100)b=(3)/(100)* 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}}*100=7 \\ \\ (800+0.03b)/(8000+b)*100=7 \\ (800+0.03b)/(8000+b)=(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=(24000)/(4) \\ b=6000\text{panels} \end{gathered}

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

User Ammar Sabir Cheema
by
2.8k points