Question

A chemist has three different acid solutions. The first acid solution contains 25% acid, the second contains 45% and third contains 85%. They want to use all three solutions to obtain a mixture of 104 liters containing 50% acid, using 3 times as much of the 85% solution as the 45% solution. How many of each solution should be used?

Answer 1

Given:

There are given the chemist has three different acid solutions.

The first acid solution contains 25%.

The second contains 45%.

The third contains 85%.

Step-by-step explanation:

We need to set two-equation

So,

Let x is the amount of the 45% acid solution.

Then,

The amount of the 85% acid solution is 3x.

Since the total volume of the mixture is 104 liters.

So,

The amount of the 25% solution is:

`104-3x-x=104-4x`

Now,

The equation for the total amount of acid in the mixture is:

`0.50*104=0.45x+0.85*3x+0.25(104-4x)`

Then,

`\begin{gathered} 0.50*104=0.45x+0.85*3x+0.25(104-4x) \\ 52=0.45x+2.55x+26-x \\ 2x+26=52 \\ 2x=52-26 \\ 2x=26 \\ x=12 \end{gathered}`

Then,

For 45%, there are 13 liters, for 85%, there are 39 liters and for 25%, there is 52 liters

Final answer:

Hence, each solution should be used:

`\begin{gathered} 25\%\rightarrow52liters \\ 45\%\rightarrow13liters \\ 85\%\rightarrow39liters \end{gathered}`