# i need help with a word problem?

Darren needs to prepare 12 ounces of a 9% HCL solution. Find the amount of a 4% and the amount of 12% solution that he would need to mix to produce the needed solution.

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• Lucy
Lv 7

HINT: Write what you know in a table.

Type …………. Amount …………. Percent …………. Total

4% HCL …………. x …………. 0.04 …………. x * 0.04 = 0.04x

12% HCL …………. y …………. 0.12 …………. y * 0.12 = 0.12y

Total …………. x + y …………. 0.09 …………. (x + y) * 0.09 = 0.09(x + y)

The first 2 rows of the last column always add up to the third row in that column.

0.04x + 0.12y = 0.09(x + y)

Given: Mixture is 12 ounces

Means: x + y = 12

Replace (x + y) with 12 and simplify.

0.04x + 0.12y = 0.09(x + y)

0.04x + 0.12y = 0.09(12)

0.04x + 0.12y = 1.08

You now have 2 equations.

0.04x + 0.12y = 1.08

x + y = 12

Solve the 2nd one for y.

x + y = 12

y = 12 - x

Replace y with 12 - x in the 1st one.

0.04x + 0.12y = 1.08

0.04x + 0.12(12 - x) = 1.08

0.04x + 1.44 - 0.12x = 1.08

0.04x - 0.12x = 1.08 - 1.44

-0.08x = -0.36

x = -0.36 / -0.08

x = 4.5 = 4 1/2

Substitute x with 4.5 in y = 12 - x to find y.

y = 12 - x

y = 12 - 4.5

y = 7.5 = 7 1/2

ANSWER: Darren needs 4 1/2 ounces of the 4% HCL solution and 7 1/2 ounces of the 12% HCL solution to make 12 ounces of a 9% HCL solution.

Source(s): For more help with mixture word problems: http://www.purplemath.com/modules/mixture.htm

The easiest way to do this is to calculate how much HCL and how much water you are working with.

For example: in 12 ounces of a 9% solution, how much is HCL, and how much is water?

Then figure out how much of the other two solutions you need to make it match that.

There is likely a formula in your book, or a worked example, to show you how to combine two concentrations. It's a standard kind of problem that is often explained.

balancing the amount of pure acid:

pure HCL = (concentration as a decimal) * (volume)

since there are 12 ounces total, if there are x ounces of 4% solution, there must be (12 - x) ounces of 12% solution)

x ounces of 4% solution yields .04x ounces of HCL

(12 - x) ounces of 12% solution yields .12(12 - x) ounces of HCL

final solution has 12 ounces of 9% solution ==> .09(12) = 1.08 ounces of HCL

balancing the inputs and output:

.04x + .12(12 - x) = .09(12) = 1.08

.04x + 1.44 - .12x = 1.08

.36 = .08x

x = .36 / .08 = 4.5 ounces

therefore, 4.5 ounces of 4% solution (.18 ounces HCL)

and 12 - 4.5 = 7.5 ounces of 12% solution (.9 ounces HCL)

combine to give 12 ounces of 9% solution (1.08 ounces HCL)