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SAT Math 3.3 Numbers and Operations 179 Views
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Description:
SAT Math 3.3 Numbers and Operations
Transcript
- 00:02
Getting' Shmoopy with it...
- 00:04
William is diluting a 100% solution with water to create a solution that is 60%.
- 00:09
If he has 30 milliliters of the 100% solution,
- 00:12
how much water does he need to add to make the 60% solution?
- 00:17
And here are the potential answers...
Full Transcript
- 00:21
This problem wants us to find the amount of water that William needs to make a 60% solution.
- 00:27
To do so, we also need to know how many total milliliters of solution William will have
- 00:31
when he's done. Let's assign variables to the two unknowns,
- 00:35
we'll call the amount of water William needs to add w,
- 00:38
and the total amount of solution x.
- 00:41
Since there's a lot of information, let's make a chart of what we know.
- 00:44
The x milliliters of 60% solution is the result of adding the 100% solution and the water.
- 00:50
Our equation can be set up as 30 times 100% plus 0% times w equals 60% times x.
- 00:58
That's the same thing as 30 times 1, or 30...equals 0.6x
- 01:03
We solve for x by dividing both sides by 0.6, and we get x equals 50.
- 01:09
So 50 milliliters is the total amount of 60% solution
- 01:13
that William will have at the end of the day.
- 01:15
...the amount of 100% solution plus the amount of water put in should total 50 milliliters.
- 01:21
Our equation is 30 + w = 50.
- 01:24
The w, the amount of water William needs to add, is 20 milliliters...
- 01:29
...which makes D 100% of this problem's solution.
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