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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 9 liters of paint which is half blue and half yellow.



Answer :

The amount of paint that must be used from can A and can B is 3.86 liters and 5.14 liters respectively to obtain 9 liters of paint which is half blue and half yellow.

Can A has 9 parts of blue paint and 1 part of yellow paint, this can be expressed in percentage as;

Can A = 90% blue, 10% yellow

Similarly,

Can B = 20% blue, 80% yellow

Now consider an algebraic expression as follows;

A + B = 9 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

Solving the equation, A + B = 9, for one variable;

A + B = 9

B = 9 - A

80% A = 60% (9 - A)

80% A = 540% - 60% A

140% A = 540%

A = 3.86 Liters

Now solving for B;

B = 9 - A

B = 9 - 3.85

B = 5.14 Liters

Therefore 3.86 liters should be used from can A and 5.14 liters should be used from can B.

To learn more about algebraic expressions; click here:

https://brainly.com/question/2164351

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