Question

A chemist needs 80 milliliters of a 50% solution but she only has 20% and 60% solutions available. Calculate how many milliliters of each should be mixed to get the desired result. How many milliliters of the 20% solution should she use?

Answer 1

20 mL

Explanation:

Let x represent the amount of 20% solution added and y represent the amount of 60% solution added.

Our first equation would be

x + y = 80,

since the amount of 20% solution and the amount of 60% solution combine to make a total of 80 milliliters.

20% = 20/100 = 0.2; this means the 20% solution would be represented as 0.2x.

60% = 60/100 = 0.6; this means the 60% solution would be represented as 0.6y.

Together they make 80 milliliters of a 50% solution; this gives us the equation

0.2x + 0.6y = 0.8(50)

Simplifying,

0.2x + 0.6y = 40

This gives us the system

`\left \{ {{x+y=80} \atop {0.2x+0.6y=40}} \right.`

To solve this, we will use elimination. Multiply the top equation by 0.2 to make the coefficients of x the same:

`\left \{ {{0.2(x+y=80)} \atop {0.2x+0.6y=40}} \right. \\\\\left \{ {{0.2x+0.2y=16} \atop {0.2x+0.6y=40}} \right.`

Subtract the bottom equation from the first:

`\left \{ {{0.2x+0.2y=16} \atop {-(0.2x+0.6y=40)}} \right. \\\\-0.4y=-24`

Divide both sides by -0.4:

-0.4y/-0.4 = -24/-0.4

y = 60

There should be 60 milliliters of 60% solution.

Substitute this into the first equation:

x+60 = 80

Subtract 60 from each side:

x+60-60 = 80-60

x = 20

There should be 20 milliliters of the 20% solution.

Answer 2

We have 20% and 60% solutions and need 80 ml of 50%

A) x + y = 80

B) .20x + .60y = (80 * .5)

Multiplying A) by -.2

A) -.2x -.2y = -16 Then adding this to B) we get

.4y = 24

y = 60 ml of 60%

x = 20 ml fo 20%

Source: 1728.com/mixture.htm