Question

A sparkling water distributor wants to make up 300 gal of sparkling water to sell for $6.00 per gallon. She wishes to mix three grades of water selling for $10.00, $1.00, and $4.50 per gallon, respectively. She must use twice as much of the $4.50 water as the $1.00 water. How many gallons of each should she use?

Answer 1

Let's call x, y, and z the $10, $1, and $4.50 water respectively.

$x+y+z=300$

Because the total is 300 gallons.

"She must use twice as much of the $4.50 water as the $1.00 water" we express it like this

$z=2y$

We combine equations

$\begin{gathered} x+y+2y=300 \\ x+3y=300 \end{gathered}$

We know that we have to mix all three grades of water for $6.

$10x+y+4.5z=6(300)$

We combine the second equation with the last one

$\begin{gathered} 10x+y+4.5(2y)=1800 \\ 10x+y+9y=1800 \\ 10x+10y=1800 \\ 10(x+y)=1800 \\ x+y=\frac{1800}{10} \\ x+y=180 \end{gathered}$

Now, we combine the last question with x+3y = 300 to make a system

$\begin{cases}x+3y=300 \\ x+y=180\end{cases}$

We multiply the second equation by -1, then we combine the equations to solve for y

$\begin{gathered} \begin{cases}x+3y=300 \\ -x-y=-180\end{cases}\rightarrow x-x+3y-y=300-180 \\ 2y=120 \\ y=\frac{120}{2} \\ y=60 \end{gathered}$

This means she should use 60 gallons of the $1 water.

Then,

$\begin{gathered} x+y=180 \\ x+60=180 \\ x=180-60 \\ x=120 \end{gathered}$

She should use 120 gallons of the $10 water.

At last,

$\begin{gathered} x+y+z=300 \\ 120+60+z=300 \\ z=300-180 \\ z=120 \end{gathered}$

She should use 120 gallons of $4.50 water.