I have two cans of paint. can a has 9 parts of blue paint to one part of yellow paint. can b is 20 percent blue paint and the rest is yellow paint. how much paint should i use from each can to obtain 8 liters of paint which is half blue and half yellow.

Respuesta :

Can A is ---------- > 9 parts of blue  1 part yellow= 90% blue, 10% yellow
Can B is 20% blue, 80% yellow. 
let
A + B = 8 Liters 

90% A + 20% B = 50% (Blue) 
10% A + 80% B = 50% (Yellow) 

resolving 

90% A + 20% B = 10% A + 80% B 
80% A = 60% B 

Go back to A + B = 8 and solve for one of the variables. 
B = 8 - A 

80% A = 60% (8 - A) 
80% A = 480% - 60% A 
140% A = 480% 
A = 3 43 liters

B = 8 - A = 4.57 liters

The answer is 3.43 liters of Can A, and 4.57 liters of Can B.

[tex]\dfrac{24}{7}\;\rm liters[/tex] of paint from Can A and [tex]\dfrac{32}{7}\;\rm liters[/tex] of paint from Can B should be used to obtain [tex]8\;\rm liters[/tex] of paint which is half blue and half yellow.

Let [tex]x\; liters[/tex] of paint from Can A and [tex]y\;liters[/tex] of paint from Can B should be used from each can to obtain [tex]8\;liters[/tex] of paint which is half blue and half yellow.

According to the question,

Can A has 9 parts of blue paint to one part of yellow paint and, Can B is [tex]20\%[/tex] blue paint and the rest is yellow paint so,

[tex]0.9x+0.2y=50\% \;\rm of \;total\;blue\;paint[/tex]

[tex]0.10x+0.80y=50\%\;\rm of\;total\;yellow\;paint[/tex]

Therefore,

[tex]0.9x+0.2y=0.10x+0.80y\\0.8x=0.6y\\4x=3y\\x=\dfrac{3y}{4}[/tex]

It is being given that the total volume of the final paint must be [tex]8\;liters[/tex] so,

[tex]x+y=8\\\dfrac{3y}{4}+y=8\\7y=32\\y=\dfrac{32}{7}\;\rm liters[/tex]

And,

[tex]x=\dfrac{3y}{4}\\x=\dfrac{3\times \frac{32}{7}}{4}\\x=\dfrac{24}{7}\;\rm liters[/tex]

Hence, [tex]\dfrac{24}{7}\;\rm liters[/tex] of paint from Can A and [tex]\dfrac{32}{7}\;\rm liters[/tex] of paint from Can B should be used to obtain [tex]8\;\rm liters[/tex] of paint which is half blue and half yellow.

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