How many milliliters of a 5% acid solution and a 13% acid solution should be used to make 200 ml of an 8% acid solution? Choose the system that would solve the application. Let x = number of liters of 5% acid solution and y = number of liters of 13% acid solution.
A) $\left\{ \begin{array} { l }
x + y = 200 \
.05 x + .13 y = 16
\end{array} \right.$
B) $\left\{ \begin{array} { l }
x + y = 200 \
.05 x + .13 y = 200
\end{array} \right.$
C) $\left\{ \begin{array} { l }
x + 200 = y \
.05 x + .13 y = 16
\end{array} \right.$
D) $\left\{ \begin{array} { l }
x + y = 200 \
.05 x + .13 y = 190
\end{array} \right.$

Question

How many milliliters of a 5% acid solution and a 13% acid solution should be used to make 200 ml of an 8% acid solution? Choose the system that would solve the application. Let x = number of liters of 5% acid solution and y = number of liters of 13% acid solution.
A) $\left\{ \begin{array} { l } 
x + y = 200 \
.05 x + .13 y = 16
\end{array} \right.$
B) $\left\{ \begin{array} { l } 
x + y = 200 \
.05 x + .13 y = 200
\end{array} \right.$
C) $\left\{ \begin{array} { l } 
x + 200 = y \
.05 x + .13 y = 16
\end{array} \right.$
D) $\left\{ \begin{array} { l } 
x + y = 200 \
.05 x + .13 y = 190
\end{array} \right.$

Quizplus · Accepted Answer

We use the system of equations where x represents the amount (in ml) of 5% acid solution used and y represents the amount (in ml) of 13% acid solution used. The total volume is 200 ml and the concentration of acid in the final solution is 8%.

So, the system of equations becomes:
x + y = 200 
0.05x + 0.13y = 0.08(200)

Simplifying the second equation, we get:
0.05x + 0.13y = 16

This is the same as the second equation in choice A. Therefore, the answer is A.

How Many Milliliters of a 5% Acid Solution and a 13