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3 August, 08:51

A paint that contains 26 % green dye is mixed with a paint that contains 16 % green dye. How many gallons of each must be used to make 90 gal of paint that is 22 % green dye?

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Answers (2)
  1. 3 August, 08:54
    0
    54 gallons of 26% green dye and 54 gallons of 16% green dye

    Step-by-step explanation:

    Let x be the quantity of gallons needed to mix both paints.

    The equation will be:

    0.26x + 0.16 (90gal - x) = 0.22 (90gal)

    0.26x + 14.4 - 0.16x = 19.8

    Making x subject of formular

    0.26x - 0.16x = 19.8 - 14.4

    0.1x = 5.4

    X = 5.4/0.1 = 54 gallons
  2. 3 August, 09:16
    0
    Number of gallons for paint 1 = 54

    Number of gallons for paint 2 = 36

    Step-by-step explanation:

    Let the amount of gallons for paint 1 = x

    amount of gallons for paint 2 = y

    Total gallons = 90

    So, x + y = 90 ... Equation 1

    26% green dye for paint 1 = 0.26*x = 0.26x

    16% green dye for paint 2 = 0.16*y = 0.16y

    22% green dye for the mixture = 0.22*90 = 19.8

    So, 0.26x + 0.16y = 19.8 ... Equation 2

    The system of equations is thus

    x + y = 90 ... Equation 1

    0.26x + 0.16y = 19.8 ... Equation 2

    Multiply equation 1 by 0.26

    0.26x + 0.26y = 90*0.26 = 23.4 ... Equation 3

    We then have,

    0.26x + 0.16y = 19.8 ... Equation 2

    0.26x + 0.26y = 23.4 ... Equation 3

    Subtract equation 2 from equation 3

    0.26x - 0.26x + 0.26y - 0.16y = 23.4 - 19.8

    0.10y = 3.6

    Divide both sides by 0.10

    0.10y/0.10 = 3.6/0.10

    y = 36 gallons

    Use y = 36 in equation 1

    x + 36 = 90

    x = 90 - 36

    x = 54 gallons
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