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## SAT

### Course: SAT > Unit 10

Lesson 3: Problem solving and data analysis- Ratios, rates, and proportions — Basic example
- Ratios, rates, and proportions — Harder example
- Percents — Basic example
- Percents — Harder example
- Units — Basic example
- Units — Harder example
- Table data — Basic example
- Table data — Harder example
- Scatterplots — Basic example
- Scatterplots — Harder example
- Key features of graphs — Basic example
- Key features of graphs — Harder example
- Linear and exponential growth — Basic example
- Linear and exponential growth — Harder example
- Data inferences — Basic example
- Data inferences — Harder example
- Center, spread, and shape of distributions — Basic example
- Center, spread, and shape of distributions — Harder example
- Data collection and conclusions — Basic example
- Data collection and conclusions — Harder example

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# Units — Harder example

Watch Sal work through a harder Units problem.

## Want to join the conversation?

- Can you dumb this down for me(17 votes)
- The problem is essentially asking us "if 2.1 computers can fit in 1 cubic foot, then how many computers can fit in 1 cubic meter?" [Because it asks us what is equal to the maximum density of computers
**in cubic meters**.] So, it's just asking us to convert 2.1 computer/ft^3, one density, to the equivalence of that in cubic meters, another density. Or think of it this way: It's asking us to convert from 2.1 computers/ft^3 --> 2.1 computers/x amount of m^3 --> x amount of computers in 1 m^3.

If 2.1 computers can fit in 1 cubic foot, then that same amount of computers would be able to fit into the equivalent of 1 cubic foot in cubic meters. So, this means that we can just find out the equivalent of 1 cubic foot (ft^3) in cubic meters (m^3) and then just replace the ft^3 in*2.1 computers/ft^3*[because this is what we're trying to convert] with what ft^3 is in m^3.

From there, we can just simplify the fraction and this will tell us how many computers would fit in just 1 cubic meter. For the first step, the problem already tells us that 1 meter = 3.28 feet. So we can just manipulate the equation to turn ft to ft^3 so that we can substitute ft^3 in 2.1 computers/ft^3.

1 meter = 3.28 feet

1 meter/3.28 = feet

(1 meter/3.28)^3 = (1 feet)^3

meter^3/3.28^3 = feet^3

m^3/3.28^3 = ft^3

So, 1 cubic foot = (m^3)/(3.28^3).

Substitute ft^3 in the measurement we're trying to convert.

2.1 computers / ft^3

2.1 computers / (m^3/3.28^3) Substitute.

(2.1 computers) x (3.28^3/m^3) Multiply

(2.1 computers x 3.28^3)/(m^3) Condense

(74.1038592 computers)/(m^3) Simplify

We know this is correct because it has our conversion in the desired units of**computers per cubic meter**(or computers/m^3). So the density 74.10318592 is the maximum number of computers per cubic meter.(51 votes)

- Me the whole time . . so confused

o(14 votes) - I get where he got 35.28ft^3 but why did he multiply it by 2.1ft^3 ?(6 votes)
- He's multiplying 35.28ft^3 with 2.1
**computers per ft^3**, not just ft^3. There's a big difference. By multiplying these numbers, we can get how many computers can be put in 35.28ft^3(5 votes)

- im confused as to why he multiplied the 3.28^3 with 2.1, can anyone explain why ?(5 votes)
- The 3.28^3 is the volume of the server room in ft^3. He multiplied it by 2.1 because the question states that only 2.1 computers can fit in a cubic foot of the volume.

1 ft^3 = 2.1 computers

3.28^3 or approx. 35.29 ft^3 x 2.1 comp/ft^3 =

approx. 74.10 computers.

The answer is over cubic meters because (3.28^3) ft^3 = (1) m^3 (which is stated in the question itself).(2 votes)

- I don't understand why he multiplied by density?(5 votes)
- Good Question. The density in this problem gives a ratio for the number of computers to cubic feet.One of the reasons he multiplied by the density was to get the units
*(ft3)*to cancel, which leaves the cubic meters by themselves. Also, this method, known as the "factor label method" is a common means on unit conversion, which is what this problem involves (converting computers per cubic foot into computers per cubic meter.(1 vote)

- I do not understand diagrams like this too well...how do you manually convert everything (If that makes sense)(4 votes)
- I don't understand how he multiplied 2.1 x the volume. Could we have have divided the volume by how many computers we have to fit in??(3 votes)
- That would work if we knew the number of computers we needed to fit. But we don't, so instead of using

density = computers/volume

we use

computers = density*volume.

Then once we have the number of computers, we can use the first equation to find the density in cubic meters.(2 votes)

- How do you know the server is 1m×1m×1m big?(2 votes)
- The size of the server room doesn't affect the final answer in any way, because it is asking for computers per cubic meter, which will stay constant no matter how big the room is. Sal puts the room's size at 1x1x1 to make the math easier. In a lot of SAT problems you can add in things like that when talking about densities or percentages, and that makes the problem much more straightforward.(3 votes)

- Is there an easier way to do this?(3 votes)
- I'm really confused on this question ): How did he set up the proportions?(1 vote)

## Video transcript

- [Instructor] In order to
connect to the internet, dedicated computers are
kept in a server room. To prevent overheating, the density of computers in a server room must not exceed 2.1 computers per cubic foot. One meter is about 3.28 feet. So they're converting linear
meters to linear feet. They're not telling us how
many that one cubic meter. They're not saying how many
cubic feet is one cubic meter, so I suspect we're gonna
have to figure that out. Which of the following densities is equal to the maximum number of
computers per computers per cubic meter? So how would we think about this? Well, and if you're taking the SAT, you might not have time
to draw a nice diagram, but this is literally what
I am visualizing in my head when I first read that problem
is I imagine a cubic meter. I imagine a cubic meter. And along each of the three dimensions in the space we know of,
the three-dimensional space, a cubic meter, it's going to
be one meter in this dimension by one meter in this dimension by one meter in that dimension. That's what a cubic meter is. Now, one meter in this dimension, that's going to be 3.28 feet. So maybe, and this, I'm not
obviously drawing it perfectly, but that's maybe a foot. That's another foot. Actually, I drew 'em too small. That's a foot, that's a
foot, and that's a foot. So we have one, two, three,
and then a little bit, a little more than 1/4 of a foot. So that's 3.28 feet. And then let's see, one
foot, two foot, three foot. Then you have one foot, one foot, two foot, and three foot. And so a cubic foot will actually
look something like this. A cubic foot would look
something like this. So if you could put 2.1
computers per cubic foot, if you can put 2.1 computers in this small cubic foot right over here, surely you're going to
be able to put a lot more than 2.1 computers in the cubic meter. And just off of that, you can
rule out these two choices because these two are less than 2.1. But how would you actually calculate it? Well, you could just multiply
this volume, a cubic meter. To figure out how many cubic feet it is, you can just multiply by the
dimensions in terms of feet. So this right here is going to be 3.28 feet in that dimension. It's going to be 3.28 feet high. And it's going to be 3.28 feet, I guess we can call this
dimension maybe deep. And so what's the volume right over here? Well, it's going to be
3.28 times 3.28 times 3.28. So the volume here is 3.28 to the third power feet cubed. Cubic feet. And this is how many cubic
feet you have per cubic meter because this whole thing
is also a cubic meter. So this is the number of
cubic feet per cubic meter. And if you wanna know how many computers you can fit in that, you can then just multiply
by the density in cubic feet. So times 2.1, I'll write
comp for computers, computers per cubic feet. And you can see that these
units are going to work out. You have cubic feet divided by cubic feet. And if you multiply 'em,
you're going to get 3.28 to the third power times 2.1, that's this times this, computers per meters cubed or per cubic meter. So this is the maximum number of computers per cubic meter right over here. Now, we could try to solve this by hand or solve it some way,
but we can estimate it because these two remaining
choices are quite different. What's 3.28 to the third
power going to be roughly? Well, three to the third power is 27, so this thing is going
to be larger than 27. And if you multiply
something larger than 27 by something larger than two, you're going to get
something larger than 54. So this thing is definitely
going to be larger, way larger than 6.89. And it's completely reasonable. These numbers look right. 3.28 to the third power times 2.1 actually would be around 74.1. And so this is definitely
the choice that I would pick. Now, if you had a lot of time,
you could multiply this out, use your calculator, but
basically these choices, you could actually just deduce that this is going to be the right choice. This choice is interesting,
'cause if you look at it, this is kind of an answer to pick if you forgot to convert to cubic, if you forgot to convert how many cubic feet are per cubic meter and if you just kept things
in linear meters and feet, because then you might say, "Okay, if I have 2.1
computers per cubic foot," and then if you just say, "Hey, well, I'll have 3.28
as many feet per meter," not thinking in terms of
cubic feet and cubic meters. But if you take 2.1 times 3.28, that's probably this choice here. I haven't multiplied it out,
but that looks like 6.89. So that's where this
answer actually came from.