1 Answer

Steve Lewis · Answer 1 · 2023-10-20T15:05:34+0000

A chemist has three different acid solutions.

Let x denotes the first acid, y denotes the second acid, and z denotes the third acid.

The first solution contains 20% acid the second contains 30% acid and the third contains 50% acid

She created 60 liters of a 40% acid mixture using all three solutions.

Mathematically,

$\begin{gathered} 0.20x+0.30y+0.50z=0.40\cdot60 \\ 0.20x+0.30y+0.50z=24\qquad eq.1 \end{gathered}$

The number of liters of 50% solution used is three times the number of liters of 30% solution used.

Mathematically,

$z=3\cdot y\qquad eq.2$

Also, we know that the sum of all solutions must be equal to 60 liters,

Mathematically,

$x+y+z=60\qquad eq.3$

Now let us substitute the eq.2 into the eq.1

$\begin{gathered} 0.20x+0.30y+0.50z=24 \\ 0.20x+0.30y+0.50(3\cdot y)=24 \\ 0.20x+0.30y+1.5y=24 \\ 0.20x+1.8y=24\qquad eq.4 \end{gathered}$

Now let us substitute the eq.2 into the eq.3

$\begin{gathered} x+y+z=60 \\ x+y+3y=60 \\ x+4y=60\qquad eq.5 \end{gathered}$

As you can see, we now have 2 equations with 2 unknowns

Let us solve them using the substitution method.

Re-arrange eq.5 for x and substitute it into eq.4

$\begin{gathered} x+4y=60 \\ x=60-4y \end{gathered}$

So eq.4 becomes

$\begin{gathered} 0.20x+1.8y=24 \\ 0.20(60-4y)+1.8y=24 \\ 12-0.80y+1.8y=24 \\ 12+y=24 \\ y=24-12 \\ y=12 \end{gathered}$

So y = 12 liters.

Now let us substitute y = 12 into eq.5

$\begin{gathered} x+4y=60 \\ x+4(12)=60 \\ x+48=60 \\ x=60-48 \\ x=12 \end{gathered}$

So x = 12 liters

Finally, substitute y = 12 into eq. 2

$\begin{gathered} z=3\cdot y \\ z=3\cdot12 \\ z=36 \end{gathered}$

So z = 36 liters

Therefore, the chemist used 12 liters of 20% solution, 12 liters of 30% solution, 36 liters of 50% solution.

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