Question

svlc algebra 1a - standard (15260) solving mixture problems 55:59 fifteen percent of the pigment in paint color a is black. sixty percent of the pigment in paint color b is black. an unknown amount of paint color b is mixed with 40 ml of paint color a, resulting in a paint that contains 25% black pigment. which equation can be used to solve for x, the total amount of paint in the mixture of the two colors? 0.15(40) + 0.6(x + 40) = 0.25(x) 0.15(40) + 0.6(x - 40) = 0.25(x) 0.15(40) + 0.6x = 0.25(40 + x) 0.15(40) + 0.6x = 0.25(40 - x) pretest mark and return 6 of 10 save & exit

Explanation:

Step1: Analyze paint A

Paint A is 40 mL, 15% black. So black pigment from A: $0.15\times40$.

Step2: Analyze paint B

Let amount of paint B be \( x \) mL (since total mixture is \( 40 + x \), but here we first find black from B: 60% of \( x \) is \( 0.6x \).

Step3: Analyze mixture

Total mixture volume: \( 40 + x \) mL, 25% black. So black pigment in mixture: \( 0.25\times(40 + x) \).

Step4: Set up equation

Black from A + Black from B = Black in mixture. So \( 0.15(40) + 0.6x = 0.25(40 + x) \).

Answer:

\( 0.15(40) + 0.6x = 0.25(40 + x) \) (the third option among the given equations)