Question

Luis needs to produce 2000 milliliters of 64% alcohol solution. At his disposal he has 40% alcohol solution and 80% alcohol solution. How much of each does he need in order to produce his desired solution?

He needs _________ milliliters of 40% solution.

He needs _________ milliliters of 80% solution

Answer 1

Answer:

• 800 ml of 40%
• 1200 ml of 80%

Step-by-step explanation:

Alcohol content remains same at the input and output.

Let 40% solution is x ml and 80% solution is y ml.

We have:

• x + y = 2000
• 0.4x + 0.8y = 0.64*2000

Simplify the second equation and solve by elimination, subtract the equations:

• x + y = 2000
• x + 2y = 3200
• 2y - y = 3200 - 2000
• y = 1200

Find the value of x:

• x + 1200 = 2000
• x = 800

Answer 2

Answer:

He needs $\boxed{ \sf 800}$ milliliters of 40% solution.

He needs $\boxed{ \sf 1200}$ milliliters of 80% solution.

Step-by-step explanation:

Given information:

• Luis needs to produce 2000 ml of 64% alcohol solution.

Define the variables:

• Let x = number of ml of 40% alcohol solution
• Let y = number of ml of 80% alcohol solution

Create a system of equations with the given information.

$\begin{cases}x + y = 2000\\40x + 80y = 64(2000)\end{cases}$

Rewrite Equation 1 to make y the subject:

$\implies y=2000-x$

Substitute this into Equation 2 and solve for x:

$\implies 40x+80(2000-x)=64(2000)$

$\implies 40x+160000-80x=128000$

$\implies -40x=-32000$

$\implies x=800$

Therefore, 800 ml of 40% alcohol solution is needed.

Substitute the found value of x into Equation 1 and solve for y:

$\implies 800+y=2000$

$\implies y=1200$

Therefore, 1200 ml of 80% alcohol solution is needed.

Learn more about systems of equations here:

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