Thirty liters of a 40% acid solution is obtained by mixing a 28% solution with a 43% solution.How much of each solution is required to obtain the specified concentration of the final mixture? Use this system of linear equations there x and y represents the amounts of the 28% solution and 43% solution. $\left\{ \begin{array} { l }
x + y = 30 \
0.28 x + 0.43 y = 12
\end{array} \right.$
A) 28% solution: 6 L; 43% solution: 24 L
B)43% solution: 6 L; 28% solution: 24 L
C) 28% solution: 40 L; 43% solution: 24 L
D) 43% solution: 40 L; 28% solution: 24 L
E) 43% solution: 7 L; 28% solution: 24 L

Question

Thirty  liters of a 40% acid solution is obtained by mixing a 28% solution with a 43% solution.How much of each solution is required to obtain the specified concentration of the final mixture? Use this system of linear equations there x and y represents the amounts of the 28% solution and 43% solution.   $\left\{ \begin{array} { l } 
x + y = 30 \
0.28 x + 0.43 y = 12
\end{array} \right.$  
A) 28% solution: 6 L; 43% solution: 24 L
B)43% solution: 6 L; 28% solution: 24 L 
C) 28% solution: 40 L; 43% solution: 24 L
D) 43% solution: 40 L; 28% solution: 24 L
E) 43% solution: 7 L; 28% solution: 24 L

Quizplus · Accepted Answer

The system of equations correctly represents the problem. Solving it gives x = 6 L for the 28% solution and y = 24 L for the 43% solution.

Thirty Liters of a 40% Acid Solution Is Obtained {x+y=300.28x+0.43y=12\left\{ \begin{array} { l } x + y = 30 \\0.28 x + 0.43 y = 12\end{array} \right.{x+y=300.28x+0.43y=12​

Thirty Liters of a 40% Acid Solution Is Obtained $\left\{ \begin{array} { l } x + y = 30 \\0.28 x + 0.43 y = 12\end{array} \right.$