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### Course: Digital SAT Math > Unit 3

Lesson 2: Unit conversion: foundations# Units — Harder example

Watch Sal work through a harder Units problem.

## Want to join the conversation?

- This is an easier way to do it:

The word problem says that 1meter=3.28feet, but the answer needs to be per cubic meter, so you have to transform 1m=3.28ft in cubic:

(1m)^3 = (3.28ft)^3

1m^2 = 35.3ft^3

Then you just need to multiply 35.3 per 2.1 (max density):

35.3*2.1 = 74.13 ;)(108 votes) - Me the whole time . . so confused

o(40 votes)- 1meter-3.28feet

Cube both sides, (since we are asked to leave our answer in computers per cubicmeter)

.:1cubicmeter-35.29cubicfeet

and we were told that the room must not exceed 2.1 computers per cubicfoot, i.e,

2.1computers-1cubicfoot.

now, the question is... how many computers will contain 1cubicmeters and 1cubicmeters is equal to 35.29cubicfeet

so, translating the above we have, how may computers will contain 35.29cubicfeet,

and 2.1coputers-1cubicfoot

xcomputer-35.29cubicfeet (i.e 1cubicmeter)

so, when we cross-multiply, we have x to be equal to 35.29 X 2.1 = 74.10computers per 35.29cubicfeet

= 74.10computers per 1cubicmeter.(38 votes)

- Can you dumb this down for me(36 votes)
- Don't make a diagram!

Except just use this:

1m = 3.28ft

1m^3 = (3.28)^3 = 35.28

Now multiply it with given 2.1 as 2.1 x 35.28 = 74.10

Easy(26 votes) - shouldn't 3.28^3 be divided by 2.1 in order to see how many computers can fit in, instead of multiplying?(8 votes)
- Ok, so I was looking at Natalie's answer down below and it really helped me to understand. I was gonna copy and paste her answer here, but I didn't wanna take the credit so I think it'd be helpful if you looked at her answer below.(4 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).(6 votes)

- I get where he got 35.28ft^3 but why did he multiply it by 2.1ft^3 ?(5 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(6 votes)

- One way to look at this:

Given:

(1ft)^3 = 2.1 ----1

1m = 3.28ft ---- 2

Working:

1ft = (1/3.28)m ---- 2'

[(1/3.28)m]^3 = 2.1 ---- 2' into 1

[1/(3.28)^3]m^3 = 2.1

m^3 = 2.1(3.28)^3(7 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.(2 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.(4 votes)

## 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.