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April 18, 2014

April 18, 2014

Posted by **Susannah** on Tuesday, February 21, 2012 at 3:54pm.

- algebra -
**Reiny**, Tuesday, February 21, 2012 at 4:58pmlet the amount of the 20% solution used be x ml

then the amount of the 50% solution used is 300-x m.

solve for x ....

.2x + .5(300-x) = .3(300)

times 10

2x + 5(300-x) = 3(300)

take it from there

- algebra -
**MathGuru**, Tuesday, February 21, 2012 at 5:00pmIt may help to set up a table for these kinds of problems:

............Amount.....%....HCl

Solution 1....x......0.2....0.2x

Solution 2...300-x...0.5....0.5(300-x)

Mixture......300.....0.3....0.3(300)

Equation:

0.2x + 0.5(300-x) = 0.3(300)

Solve for x. (Be sure to list the amount of both solutions!)

I hope this helps.

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