Question

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

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,

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

so that

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

and

$z=2\cdot\dfrac{79}6 = \boxed{\dfrac{79}3} \approx 26.33$