Question

when ( x ) liters of a 10% acid solution is mixed with ( y ) liters of a 40% acid solution, 12 liters of a 20% acid solution is produced. which system of equations represents this situation? (assume that the volume of the mixture is the sum of the volumes of the two solutions before they were mixed.)

a
( \begin{cases} x + y = 0.20(12) \\ 0.10x + 0.40y = 0.20(12) end{cases} )

b
( \begin{cases} x + y = 0.20(12) \\ 0.20(x + y) = (0.40 + 0.10)(x + y) end{cases} )

c
( \begin{cases} x + y = 12 \\ 0.10x + 0.40y = 0.20(12) end{cases} )

d
( \begin{cases} x + y = 12 \\ 0.40x + 0.10y = 0.20(12) end{cases} )

Explanation:

Step1: Analyze volume equation

The total volume of the mixture is 12 liters, and it's the sum of \(x\) and \(y\) liters. So \(x + y = 12\).

Step2: Analyze acid amount equation

The amount of acid in \(x\) liters of 10% solution is \(0.10x\), in \(y\) liters of 40% solution is \(0.40y\), and in 12 liters of 20% solution is \(0.20\times12\). So \(0.10x + 0.40y = 0.20(12)\).

Answer:

C. \( x + y = 12 \)
\( 0.10x + 0.40y = 0.20(12) \)