Question

Two factory plants are making tv panels. Yesterday, plant A produced 2000 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 the two plants was 5%?

Answer 1

Let B be the amount of panels that plant B produced.

The total amount of defective panels from plant A is:

$\frac{10}{100}\times2000=200$

The total amount of defective panels from plant B is:

$\frac{3}{100}\times B=0.03B$

The total amount of panels produced by plants A and B is:

$2000+B$

Since 5% of the total of panels is defective, the total amount of defective panels is:

$\begin{gathered} \frac{5}{100}\times(2000+B) \\ =100+0.05B \end{gathered}$

On the other hand, the total amount of defective panels is the sum of the defective panels from plants A and B:

$200+0.03B$

Since these two expressions correspond to the total amount of defective panels, then:

$200+0.03B=100+0.05B$

Solve for B:

$\begin{gathered} \Rightarrow200-100=0.05B-0.03B \\ \Rightarrow100=0.02B \\ \Rightarrow B=\frac{100}{0.02} \\ \Rightarrow B=5000 \end{gathered}$

Therefore, the total amount of panels that plant B produced, is:

$5000$