SAT Prep: Test 8 Section 8 Part 3

This might be the last video
if I can squeeze in three

problems. So problem 14.

The last video at least
for SAT prep.

Hopefully I will keep
doing videos.

OK, it looks something
like this.

And then it comes back
down like this.

And what are they telling us?

They're saying that this
is line l, line m.

This is x.

This is z degrees, this
is y degrees.

Looks like we're going to have
to play the angle game.

In the figure above, line
l is parallel to line m.

What does z equal in
terms of x and y?

OK.

So these two lines
are parallel.

If this is y degrees,
what other

angles are also y degrees?

Well, what's the corresponding
angle?

Right here.

Well, this is also going
to be y degrees, right?

These lines are parallel.

This is a transversal.

These are corresponding
angles.

This is going to be y.

And it makes sense too.

I mean, if you tilted this
angle, you would visually see

that these angles would
be the same.

If this is y, what
is this angle?

Well, they're opposite, so this
is also going to be y.

And then we have z plus x
plus y has to equal 180.

Because they're all in
the same triangle.

z plus x plus y is
equal to 180.

We want to solve for z.

So subtract x and y from both
sides, and you get z is equal

to 180 minus x minus y.

And that is choice E.

Next problem.

Problem 15.

If n over n minus 1 times
1/n times n over n plus

1 is equal to 5/k.

For positive integers n and
k, what is the value of k?

So these are positive
integers.

So before multiplying all of
this out, we can simplify a

little bit.

This n can cancel
out with this n.

And now let's see if we can.

What does the top-- what does
this left-hand side become?

And the numerator always left--
this is just a 1 now.

This is a 1.

So we're left with 1
times 1 times n.

So that's n over n minus 1,
times 1-- I can ignore that

1-- times n plus 1.

Is equal to 5 over k.

So what are they asking?

Well, what is the value of k?

Well, we can say that
n is equal to 5.

Or let's assume that
n is equal to 5.

Because we don't know
definitely that

n is equal to 5.

It could be some multiple--
I'll show you.

Let's assume that
n is equal to 5.

If n is equal to 5,
then what is k?

Well, then k would be
this denominator.

Right?

If n is 5, then k is this.

Then k would be-- so this could
be 5 over 5 minus 1

times 5 plus 1.

And that equals 5 over
4 times 6, which is

equal to 5 over 24.

So this could be 5 over 24.

And they're all positive
integers so k is 24.

k is 24.

And that's choice C.

So the trick here is really once
again-- simplify a little

bit, multiply it out and
then pattern matching.

Let me just set n
is equal to 5.

If n is equal to 5,
then what is k?

It's 5 minus 1 times 5 plus 1.

It's just pattern matching.

Next problem.

Problem 6.

I will do it in magenta because
this is the last

problem in the book.

To celebrate a colleague's
graduation the m coworkers in

an office agreed to contribute
equally to a catered lunch

that cost a total
of y dollars.

So there's m workers, and
the total price is y

dollars for the lunch.

If p of the workers fail to
contribute, which of the

following represents the
additional amount in dollars

that each of the remaining
coworkers must contribute to

pay for the lunch?

The additional amount
in dollars.

So if everyone paid, how much
would we have to pay?

Well, the total lunch
is y, right?

So if everyone was a good
coworker, we would each have

to pay y divided by the
number of coworkers.

This is the ideal situation.

But we know some of the
coworkers didn't pay.

p didn't pay.

So how many are we going to have
to divvy it up by now?

So then that means only
m minus p paid.

These are the deadbeats that
did not pay for the lunch.

So only the m minus p paid.

So now we have to actually
divide the y dollars between a

smaller group of people
who actually paid.

And the smaller group of
people who actually

paid is m minus p.

So if you wanted to figure out--
and this is going to be

a larger number.

Why?

Because its denominator
is smaller.

When the denominator is smaller
and you have the same

numerator, there's going
to be a larger number.

So if you want to know what is
the additional amount you have

to pay-- well, this is how
much we're having to pay,

which is a larger amount than
how much we would have paid if

everyone paid.

So how much are we
paying extra?

Well, we subtract
this from this.

This is how much we
end up paying.

And we subtract how much
we should have paid.

And we'll get the additional
amount.

Let me draw a line so we
don't get confused.

That doesn't look like one of
the choices, so let's actually

get a common denominator.

Common denominator would
be m times m minus p.

I just multiply the
denominators.

So y over m minus p.

That's the same thing as m
y over m times m minus p.

I just multiply the numerator
and the denominator by m.

And this is minus--
m minus p times y.

For this one I just multiplied
the numerator and the

denominator by m minus p.

I just found a common
denominator and added the

fractions, or subtracted
the fractions.

And so, the denominator stays.

m times m minus p.

Let me see if I can simplify
the numerator.

That becomes m y minus
m y plus py.

Minus times a minus
here is a plus.

Plus py.

These cancel out.

So you get py over m
times m minus p.

And that is choice E.

py over m times m minus p.

And we have now done something
like, what, 8 tests.

8 tests and 54 problems per
test. So that's 54 times 8.

54 times 8.

That's, what?

432 SAT problems. And I think
you're ready to go take the

SAT and get a perfect score.

And let me know if you do.

That would be very exciting.

All right.

I'll see you in, I guess,
other videos

that are not SAT related.

Or maybe when a new book
comes out I'll

have to do this again.