Question

In the lab Jose has two solutions that contain alcohol and is mixing them with each other he uses twice as much solution a as solution b . Solution a is 16% alcohol and solution be is 19% alcohol how many milliliters of solution be he does he use if the resulting mixture has 480 mL of pure alcohol

Answer 1

Step1

Define the parameters for the given information

`\begin{gathered} \text{let} \\ A\text{lcohol content of solution A=x ml} \\ \text{ Alcohol content of solution B=y ml} \end{gathered}`

Step2

Write out the equation for the information given

He uses twice as much solution A as solution B is written as

`x=2y`

Then

Solution A is 16% alcohol and solution B is 19% we have

`\begin{gathered} \text{Solution A=}\frac{\text{16}}{100}x=0.16x \\ \text{Solution B=}\frac{\text{19}}{100}y=0.19y \end{gathered}`

For the mixture to have 480ml of alcohol, we have

`0.16x+0.19y=480`

Step3

Solve the equation simultaneously

`\begin{gathered} x=2y\ldots\ldots\text{.equation 1} \\ 0.16x+0.19y=480\ldots\ldots\text{equation 2} \end{gathered}`

substitute equation (1) into equation (2)

we have

`0.16(2y)+0.19y=480`

expand the parenthesis

`\begin{gathered} 0.32y+0.19y=480 \\ 0.51y=480 \\ divide\text{ both sides by 0}.51 \\ y=(480)/(0.51)=(16000)/(17) \end{gathered}`

Then the value of becomes

`\begin{gathered} x=2y \\ x=2*(16000)/(17)=(32000)/(17) \end{gathered}`

Therefore

`\begin{gathered} x=(32000)/(17) \\ \text{and } \\ y=(16000)/(17) \end{gathered}`

Hence

Jose will needs 1882.35ml of solution A and 941.18ml of solution B