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CAHSEE Math 6.3 Algebra I
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CAHSEE Math: Algebra I Drill 6, Problem 3. If x ounces of a 10% salt solution are added with 20 ounces of a 50% salt solution to create a comb...

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CAHSEE Math 1.1 Algebra I
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CAHSEE Algebra I: Drill 1, Problem 1. What are some of the properties used to solve this equation? 

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CAHSEE Math 1.2 Algebra I
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CAHSEE Math Algebra I: Drill 1, Problem 2. Find the negative reciprocal.

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CAHSEE Math 6.3 Algebra I 182 Views


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CAHSEE Math: Algebra I Drill 6, Problem 3. If x ounces of a 10% salt solution are added with 20 ounces of a 50% salt solution to create a combined 35% salt solution, what would be the value of x, or the amount of the 10% salt solution?

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Transcript

00:02

And here's another shmoop du jour.

00:04

If x ounces of a 10% salt solution…

00:06

…are added with 20 ounces of a 50% salt solution to create a combined 35% salt solution,

00:13

what would be the value of x, or the amount of the 10% salt solution?

00:18

And here are the potential answers…

00:24

Basically, a highly diluted salt solution is being poured into a bucket with a more

00:27

concentrated solution…

00:29

…and the question wants to know how much we need of the highly diluted amount so that

00:33

the end result is a 35% salt solution.

00:37

So, said another way, x many ounces of the 10% diluted stuff is added to 20 ounces of

00:42

the 50% stuff; we want the total to be a 35% stuff.

00:47

The actual salt content of the 10% solution is 0.1x oz., while the salt content of the

00:55

50% solution is 0.5 times 20… or 10 oz.

00:59

These two amounts added together must equal the salt content in the combined solution.

01:04

So the combined equation looks like this:

01:06

0.1x plus 10 equals .35x plus .35 times 20.

01:12

If we simplify, we get .1x plus 10 equals .35x plus 7… which simplifies to 3 = .25x

01:22

Divide both sides by .25 and we get x equals 12.

01:25

So answer A is our guy.

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