Fourty liters of a 40% acid solution is obtained by mixing a 28% solution with a 50% solution.Write a system of equations in which one equation represents the amount of final mixture required and the other represents the percent of acid in the final mixture.Let x and y represent the amounts of the 28% and 50% solutions,respectively.
A) $$\left\{ \begin{array} { l }
x + y = 40 \
0.28 x + 0.5 y = 40
\end{array} \right.$$
B) $$\left\{ \begin{array} { l }
0.28 x + 0.5 y = 40 \
x + y = 16
\end{array} \right.$$
C) $$\left\{ \begin{array} { l }
x + y = 40 \
0.28 x + 0.5 y = 16
\end{array} \right.$$

Question

Fourty liters of a 40% acid solution is obtained by mixing a 28% solution with a 50% solution.Write a system of equations in which one equation represents the amount of final mixture required and the other represents the percent of acid in the final mixture.Let x and y represent the amounts of the 28% and 50% solutions,respectively. ​
A)​ $$\left\{ \begin{array} { l } 
x + y = 40 \
0.28 x + 0.5 y = 40
\end{array} \right.$$
B)​ $$\left\{ \begin{array} { l } 
0.28 x + 0.5 y = 40 \
x + y = 16
\end{array} \right.$$
C) $$\left\{ \begin{array} { l } 
x + y = 40 \
0.28 x + 0.5 y = 16
\end{array} \right.$$

Quizplus · Accepted Answer

Let x and y represent the amounts of the 28% and 50% solutions, respectively.

The amount of final mixture required is 40 L, so we can write:

x + y = 40

The percent of acid in the final mixture can be found by taking a weighted average of the percents of acid in the two solutions. The weight of the 28% solution is x/40, and the weight of the 50% solution is y/40, so we can write:

0.28x + 0.5y = 0.4(0.4)

Simplifying, we get:

x + y = 40
0.28x + 0.5y = 0.16

This system of equations can be solved using substitution or elimination. The solution is x = 4 and y = 36, so the amounts of the 28% and 50% solutions needed to make 40 L of a 40% solution are 4 L and 36 L, respectively.

Fourty Liters of a 40% Acid Solution Is Obtained by Mixing