SAT Prep: Test 8 Section 5 Part 2

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.