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We're on problem number five.
Let's see, we have a pie graph here of survey results.
And let's see, that looks like almost a straight line.
Dividing it in half there.
And then this gets divided in half again.
It looks like that, roughly.
It looks like that.
And then this is slightly bigger like that.
All right, and this tells us x is less than 20
for 30% of the time.
I don't even know what these represent.
x is greater than or equal to 60 25% of the time.
x is between 40 and less than 60, this is 25% of the time.
So it can equal 40, but it's less than 60.
And then it's greater than or equal to 20 and less than 40
20% of the time.
That's the survey results.
The chart above shows the results when 1,000 people were
asked how old are you?
The age they gave is represented by x.
How many people said their age was less than 40?
So x is less than 40.
Well, this category right here is x is between 20 and 40, so
it's this one plus all the people who said that they were
younger than 20 because if you're less than 20, you're
definitely less than 40 as well, so it's
both of these combined.
And this is 30%, this is 20%.
So if you combine it, it's 30% plus 20% equals 50% of the
entire population asked.
And 1,000 people were asked, so 50% times 1,000, well,
that's straightforward.
That equals 500 people.
And that's choice D.
Next problem, problem 6.
Which of the following could be the remainders when four
consecutive positive integers are each divided by 3?
Interesting.
So let's say we have x plus 1, x plus 2, x plus 3, right?
These are four consecutive positive integers.
Well, let's assume that this first one is
divisible by 3, right?
Let's say that x divided by 3, there's no remainder, so the
remainder is 0.
So when this number's divisible by this number, x is
divisible by 3, so now when you divide x plus 1 by 3,
you're going to have 1 left over.
Similarly, you're going to have 2 left over.
Here, you're not going to 3 left over, right?
Because if x is divisible by 3, then x plus 3 is also
divisible by 3, right?
You're going to have a cycle.
It's going to go back to a remainder of zero.
So this is a possible situation.
And similarly, if x had a remainder of 1, then x plus 1
would have a remainder of 2.
Then x plus 2 would have a remainder of 0, and then this
would have a remainder of 1 again.
So I don't know, let's look at the choices.
Are any of these out there?
Well, sure.
This is choice D, actually.
The first thing we did, choice D was 0, 1, 2, 0, which is D.
And the key here is realizing that you can't have a
remainder of 3 or 4 when you're dividing by 3, right?
Your remainder can only be 0, 1 or 2.
So with that realization alone, you can cancel out all
the choices that have a 3 or a 4 in it.
You can't have a remainder of 3 or 4 if
you're dividing by 3.
And you can try that out, right?
Because if you have a remainder by 3, that means you
could divide one more 3 into the number and have a
remainder of 0.
So you could cancel out A, B, C and E.
You can actually cancel out everything.
So you would have just had to have that one realization, and
you would have said the choice is D because you can't have a
remainder of 3 or 4 if you're dividing by 3.
Next problem, problem 7.
If y is inversely proportional to x, so that means that y is
proportional to the inverse of x, so it's equal to some
constant times 1/x.
Because that's what inversely proportional means.
If we said proportional, it'd be y equals k times x, but
it's inversely proportional.
And they tell us that y is 15 when x is 5.
So y is 15 is equal to k times 1/5, right?
y is 15 when x is equal to 5.
Let's see, let's multiply both sides of this by 5.
So you get 5 times 15 is 75 is equal to k, right?
So y is equal to 75/x.
I just rewrote this.
What is the value of y when x is 25?
So y is equal to 75.
x is now 25.
What's 75 divided by 25?
Well, it's 3.
And that's choice C.
Problem 8.
If 2x plus z is equal to 2y, and 2x-- and then they also
tell us that 2x plus 2y plus z is equal to 20.
What is the value of y?
y is equal to what?
So there's something interesting here.
we can rewrite this second equation as subtract 2y from
both sides of this equation right here, and you get 2x
plus z is equal to 20 minus y, right?
So essentially, you have a 2x plus z here, and you have a 2x
plus z here.
Let's make a new variable.
You don't have to do this step, but I think this'll
simplify things.
Let's call the variable Q.
Let's say Q is equal to 2x plus z.
That's where they're trying to confuse you.
They're giving you two equations with three unknowns.
And you're like, how can I solve for one of them?
Well, what's interesting is they have-- you're solving for
one of them, and then the relationship has a 2x plus z
in both equations.
So if you say Q is equal to 2x plus z, everything starts to
make sense because then this top equation will become Q is
equal to 2y.
And what will this bottom equation be?
That would be Q is equal to 20 minus y, right?
And now you could set these equal to each other.
2y is equal to 20 minus y.
Add y to both sides, you get 3y is equal to 20.
Am I doing that right?
3y is equal to 20?
You add y to both sides. y is equal to 20/3.
And they don't have that choice, so I
must have made a mistake.
Let me redo the problem.
I must've made a mistake.
Problem 8.
So they're telling us 2x plus z is equal to 2y, and then
they tell us 2x plus 2y plus z is equal to 20.
What is the value of y?
OK, the second equation can be written as 2x plus z is equal
to 20 minus 2y.
And this top equation is still 2x plus z is equal to 2y.
Oh, I see.
I had dropped a y someplace.
So this must equal this, because they both
equal 2x plus z.
I don't even have to do all that substitution Q.
I think that you can see that 2x plus z equals this, 2x plus
z equals this, so this must equal this.
So 2y is equal to 20 minus 2y.
Add 2y to both sides, you get 4y is equal to 20,
y is equal to 5.
And that is choice A.
I'll see you in the next video.
Oh, whoops!
I didn't delete.
Sorry.