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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

Language:
English Language

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...

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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