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SAT Math 1.4 Numbers and Operations. Solve the equation.
SAT Math 1.1 Numbers and Operations. How many combinations of beverage and cereal can be made?
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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