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Solving a percent mixture problem using a linear equationSamantha VEspañolTwo factory plants are making TV panels. Yesterday, Plant A produced twice as many panels as Plant B. Two percent of the panels from Plant A and 3% of thepanels from Plant B were defective. How many panels did Plant B produce, if the two plants together produced 490 defective panels?

Sagot :

We are given a question about factory plants making TV panels. The following information holds.

Plant A produces twice as many panels as plant B. This can be expressed mathematically as:

[tex]A=2B[/tex]

Also, 2% and 3% of the TV panels from Plant A and Plant B are defective, and both plants produced a total of 490 defective panels. This can be expressed mathematically as:

[tex]\frac{2A}{100}+\frac{3B}{100}=490[/tex]

We can clearly see that we have derived two equations. We will substitute the first equation in the second equation to get the number of panels Plant B produced.

[tex]\begin{gathered} \frac{2(2B)}{100}+\frac{3B}{100}=490 \\ \frac{4B}{100}+\frac{3B}{100}=490 \\ \frac{7B}{100}=490 \\ \text{Cross multiply} \\ 7b=490\times100 \\ B=\frac{490\times100}{7} \\ B=70\times100 \\ B=7000 \end{gathered}[/tex]

Therefore, the number of panels that plant B produced is:

ANSWER: 7000

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