Question

A chemist has three different acid solutions. The first acid solution contains

20
%
acid, the second contains
30
%
and the third contains
60
%
. They want to use all three solutions to obtain a mixture of
72
liters containing
35
%
acid, using
2
times as much of the
60
%
solution as the
30
%
solution. How many liters of each solution should be used?

Answer 1

Let
`x,y,z` denote the amounts (in liters) of the 20%, 30%, and 60% solutions used in the mixture, respectively.

The chemist wants to end up with 72 L of solution, so

`x+y+z=72`

while using twice as much of the 60% solution as the 30% solution, so

`z = 2y`

The mixture needs to have a concentration of 35%, so that it contains 0.35•75 = 26.25 L of pure acid. For each liter of acid solution with concentration
`c\%`, there is a contribution of
`\frac c{100}` liters of pure acid. This means

`0.20x + 0.30y + 0.60z = 26.25`

Substitute
`z=2y` into the total volume and acid volume equations.

`\begin{cases}x+3y = 72 \\ 0.20x + 1.50y = 26.25\end{cases}`

Solve for
`x` and
`y`. Multiply both sides of the second equation by 5 to get

`\begin{cases}x+3y = 72 \\ x + 7.50y = 131.25\end{cases}`

By elimination,

`(x+3y) - (x+7.50y) = 72 - 131.25 \implies -4.50y = -59.25 \implies \boxed{y=\frac{79}6} \approx 13.17`

so that

`x+3\cdot\frac{79}6 = 72 \implies x = \boxed{\frac{65}2} = 32.5`

and

`z=2\cdot\frac{79}6 = \boxed{\frac{79}3} \approx 26.33`