29,682 views
11 votes
11 votes
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%?

User Adriano
by
2.2k points

1 Answer

13 votes
13 votes

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

The total amount of defective panels from plant A is:


(10)/(100)*2000=200

The total amount of defective panels from plant B is:


(3)/(100)* 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} (5)/(100)*(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=(100)/(0.02) \\ \Rightarrow B=5000 \end{gathered}

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


5000

User Kalpesh Dusane
by
2.8k points