Will give brainliest answer to the most !
i worked this problem out, but i am not sure if i am correct. show your work to the solution. : )
determine the number of gallons of a 60% concentrated solution and the number of gallons of an 80% concentrated solution that are required to make 55 gallons of a 75% concentrated solution.

Step-by-step explanation:

I like to use the simple device shown in the attachment to work mixture problems. The source concentrations are listed on the left, and the desired mix concentration is shown in the middle. Differences are taken along the diagonals. Here, those differences are 15 and 5, representing the 3:1 ratio of solutions that is needed in the mix.

That is, 3/4 of the mix will be the 80% solution, and 1/4 of the mix will be the 60% solution. All that remains is to compute the corresponding number of gallons:

  60% concentration needed = (1/4) × (55 gal) = 13 3/4 gal

  80% concentration needed = (3/4) × (55 gal) = 41 1/4 gal

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If you write an equation to find the amount of solution needed, it is convenient to let the variable represent the amount of the higher-concentration solution. This has the effect of keeping the numbers positive as you solve the equation. Here, let x represent gallons of 80% solution, and (55-x) will be the gallons of 60% solution. Then the amount of material in the mix is ...

  0.80x + 0.60(55 -x) = 0.75·55

  0.20x + 33 = 41.25 . . . . . simplify

  0.20x = 8.25 . . . . . . . . . . subtract 33

  8.25/0.20 = x = 41.25 . . . divide by the coefficient of x

  55-x = 55 -41.25 = 13.75 . . . gallons of 60% concentration

41.25 gallons of 80% concentration are needed, and 13.75 gallons of 60% concentration are needed.

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step-by-step explanation:

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i dont know sorry

step-by-step explanation: