We're on problem number 7.
If the average of x and 3x is 12, what is the value of x?
So the average, so x plus 3x-- and I'm averaging two numbers
so divide by 2-- that's going to be equal to 12.
So we just solve for this.
Multiply both sides of the equation by 2 and you get 2
times, times 2, this cancels with this.
You get x plus 3x is equal to 24.
And what's x plus 3x.
That's 4x, right?
4x is equal to 24.
x is equal to 6, and that's choice C.
Next problem.
Problem 8.
At Maple Creek High School, some members of the chess club
are also on the swim team, and no members of the swim team
are tenth graders.
Which of the following must be true.
This seems like it'll call for a Venn diagram.
So let's say that that represents the chess club.
And they say some members of the chess club
are on the swim team.
So some members are on the swim team.
Maybe I should put the swim team in like blue.
So let's say the swim team.
That's the swim team.
And these are the members, right, that are
in both right here.
But then it tells us no members of the swim team are
tenth graders.
So if I draw another circle for the tenth graders, it
can't intersect with the swim team, but it could intersect
with the chess team.
I don't know.
I mean it could be like that.
That could be tenth graders.
It could be like that.
Or it could be out here some place.
But we don't know.
There could be chess and tenth graders, just not the same
people who are on the swim team.
So let's see.
So which of the following must be true?
No members of the chess club are tenth graders.
No.
This is a situation where you could have some members of the
chess club who aren't on the swim team who
could be tenth graders.
B, some members of the chess club are tenth graders.
Well some members could be, but we don't know for sure.
This could be tenth grade.
We don't know.
This could be the tenth grade kind of set or this could be
the tenth grade.
There might be no tenth graders in either the chess
team or the swim team.
We don't know for sure.
And then choice C, some members of the chess club are
not tenth graders.
This we know for sure.
How do we know it for sure?
Because these kids who are on both, they're in the chess
club, but they're also on the swim team.
The fact that they're in swim team, we know that they can't
be tenth graders.
So this is some members of the chess club-- this little
intersection here-- that are not tenth graders.
So choice C is the correct choice.
Next problem.
If 3x plus n is equal to x plus 1, what is
n in terms of x?
So we essentially just solve for n.
Let's subtract 3x from both sides.
You get n is equal to-- what's x minus 3x?
It's minus 2x.
And n plus 1.
And we're done.
And that choice isn't there, but if you just switch these
two terms you just get that equals 1 minus 2x and
that's choice D.
Pretty quick problem, especially for one that's the
ninth problem.
They normally get a little harder by this point.
Problem 10.
If k is a positive integer, let k be defined as a set of
all multiples of k.
So k with a square around it is equal to the set of
multiples of k.
All of the numbers in which of the following sets are also in
all three of the set-- OK.
All of the numbers in which of the following sets are also in
all three of the sets of 2, 3 and 5?
So the what they're saying is 2, 3, 5, this donates all the
multiples of 2.
This is all multiples of 3.
This is all multiples of 5.
So what they're essentially saying is let's find a number
where all of its multiples, all of this number's multiples
are also going to be multiples of each of these.
So it has to be a multiple-- so every number that--
whatever this mystery number is, let's call it x-- every
multiple of x has to be a multiple of 2, 3 and 5.
Well the simple way is if x is a multiple of 2, 3 and 5, then
every multiple of x is going to be a
multiple of 2, 3 and 5.
So what's 2 times 3 times 5?
It's 2 times 3 times 5.
That's 6 times 5, that's 30.
So 30 is a multiple of all of them, so any multiple of 30
will be a multiple of all of these.
When we look at the choices we don't see 30.
But do we see any other number that is a
multiple of 2, 3 and 5?
Well sure, 60 is, right?
We just multiply by 2 again.
But 60 is still a multiple of 2, 3 and 5.
If you were to do 2, 4, 6, 8 all the way you'd get 60, if
you go 3, 9, 12, 15 all the way, you'd get to 60.
You go 5, 10, 15, 20, 25, you'd get to 60.
So 60 is a multiple of all of them.
So what we're saying is-- so what's the set of all the
multiples of 60?
It's 60, 120, 180, 240, et cetera, right?
And all of these numbers are in each of these sets.
Because all of these numbers are multiples of 2, 3 and 5.
So our answer is 60.
If you look at the other choices, some of them are
divisible by 5, some are divisible by 2 or 3,
some are 3 and 5.
But none of them are divisible by 2, 3 and 5, only 60 is.
Next problem.
That problem was a little hard to read initially though.
That's how they confuse you.
So we're going to go from A to D-- I should have drawn all
the lines first. Let me draw the lines first. It's like a
hexagon kind of.
The top, the outside of the hexagon there.
A, B, C, D, E, F.
And then this is the origin.
And the figure above, AD is equal to BE.
Oh, no, no.
They don't tell us that.
I'm hallucinating.
In the figure above AD, BE, and CF intersect at 0.0.
The intersect's here at the origin.
If the measure of AOB, the measure of that, is 80
degrees, and CF bisects BOD, so it
bisects this larger angle.
CF bisect BOD, that angle.
So that tells us that this angle has to be
equal to this angle.
That's the definition of bisecting an angle.
You're splitting this larger angle in half.
So these angles have to be equal to each other.
So what is the measure of EOF?
So we want to figure out this angle.
Well this angle is opposite to this angle, so they're going
to be equal.
So if we can figure out this angle we're done.
So let's call this angle x.
If that angle's x this angle is also x.
This x, this x, and this 80 degrees, they're all
supplementary because they all go halfway around the circle.
So x plus x plus 80 is going to be equal to 180 degrees.
2x plus 80 is equal to 180.
2x is equal to 100, x is equal to 50.
And as we said before, x is equal to 50, the angle EOF,
which you're trying to figure out, is opposite to it so it's
going to be equal.
So this is also going to be 50 degrees.
And that's choice B.
Next problem.
I don't know if I have time for this, but I'll try.
Problem 12.
k is a positive integer.
What is the least value of k for which the
square root of-- OK.
So what is the least value of k for which 5k
over 3 is an integer.
So this has to be a whole number, right?
So essentially if we want to find the least value of k, we
essentially want to say, well what's the least integer that
this could be?
And they're telling us that k is a positive integer.
So first of all, in order for the square root to be an
integer, this whole thing has to be an integer, right?
So let's see, k has to be a multiple of 3.
In order for this expression to be an integer, k has to be
a multiple of 3.
If k is 3, we get square root of 15 over 3-- well that
doesn't work.
If k is 3 we just get 5 in there.
Actually, let me continue this into the next problem because
I don't want to rush this.
I'll see you in the next video.