Question

in the lab, greg has two solutions that contain alcohol and is mixing them with each other. solution a is 60% alcohol and solution b is 10% alcohol. he uses 900 milliliters of solution a. how many milliliters of solution b does he use, if the resulting mixture is a 40% alcohol solution?
number of milliliters of solution b:

Explanation:

Step1: Define variable for Solution B

Let $x$ = milliliters of Solution B used.

Step2: Set up alcohol balance equation

Alcohol from A + Alcohol from B = Alcohol in mixture
$0.60 \times 900 + 0.10x = 0.40 \times (900 + x)$

Step3: Calculate alcohol from Solution A

$0.60 \times 900 = 540$
Equation becomes: $540 + 0.10x = 0.40(900 + x)$

Step4: Expand right-hand side

$0.40 \times 900 + 0.40x = 360 + 0.40x$
Equation: $540 + 0.10x = 360 + 0.40x$

Step5: Isolate variable terms

Subtract $0.10x$ and $360$ from both sides:
$540 - 360 = 0.40x - 0.10x$
$180 = 0.30x$

Step6: Solve for x

$x = \frac{180}{0.30} = 600$

Answer:

600