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Current time:0:00Total duration:12:27

We're on problem number 10. And it says, in a certain class,
one student is to be selected at random to read. What is the probability
that a boy will read? OK. So I guess we can assume that
everyone in the class can read, because they don't really
talk much about that. And I think we can assume that
everyone in the class is either a boy or a girl, that
we don't have any third genders in this reality. OK. So what are the statements? So we essentially just want to
know, what's the probability we pick a boy in the
class, right? Whether or not they read is
kind of -- so what's the probably that we pick a boy? So to figure out whether we're
going to pick a boy, we have to know, essentially,
what percentage of the class are boys. So what are the statements
they give us? Statement number one is,
2/3 of the students in the class are boys. Well, that's just what
I asked for. 2/3 boys. Well, that exactly tells
me the probability that I pick a boy. It's 2 out of 3, or 66.66
repeating percentage. So this alone is enough. So let's see if statement
two is at all useful. Statement two. 10 of the students in
the class are girls. 10 girls. Well, this, especially
by itself, is a fairly useless statement. Because maybe there are 10
girls, maybe there's 0 boys, in which case this probability
would be 0. Maybe there are 10 girls and
10 boys, in which case this would be 50%. So 10 girls, by itself, really
doesn't tell us much. So in this case, all we
need is statement one. And statement two alone
is not sufficient. So that is what? I always forget these. a, statement one alone is
sufficient, but statement two is not sufficient. That's a. Problem 11. Let me switch colors to green. Problem 11. If 5x plus 3y is equal to 17. They want to know what x is
equal to. x is equal to what? OK. The first statement
they give us is, x is a positive integer. Well, that doesn't
help us solve x. I mean, x could be any
positive integer. They gave us no constraints
on y. So we really could pick x
as any positive integer. So that doesn't help
us with that. Statement number two is,
y is equal to 4x. Well, this is a little
bit more interesting. y is equal to 4x. Well, here we could substitute
this in for y, and then solve for x. And remember, in these problems,
you really don't have to solve for x. You just have to know
that you could. So I, immediately just
recognizing it, I could say, well, this is all I need
to do to solve for x. And just to prove it to
you, let's do it. Although you shouldn't have
to do this on the GMAT. So 5x plus 3 times y,
where y is equal to 4x, is equal to 17. So you get 5x plus 12x
is equal to 17, or 17x is equal to 17. x is equal to 1. So all you needed was
statement two. And statement one was
fairly useless. So that is b, statement two
alone is sufficient, but statement one alone
is not sufficient. Problem 12. So they're asking whether the
product jkmn equal 1? Does that equal 1? So the first statement they give
us-- let me scroll down a little bit-- the first statement
they give us is, jk over mn is equal to 1. Let's see, we could do a little
algebraic manipulation. Multiply both sides by mn, you
get jk is equal to mn, right? I just multiplied both
sides by mn. But that really doesn't
help me. I mean, that just tells
me that this part is equal to this part. But it doesn't really tell me
that when you multiply them all together, you equal 1. Let's see if statement two
is it all helpful. Statement two tells us that j is
equal to 1 over k, and m is equal to 1 over n. So let's see if we could do
some interesting algebraic manipulations. Let's see. Multiply both sides
of this by k. You get jk is equal to 1. And do the same thing here. Multiply both sides times n. You get mn is equal to 1. Well, this is useful, because
if j times k is equal to 1, and m times n is equal to 1-- so
the whole product is going to be 1 times 1. Well, that's definitely going
to be equal to 1. So using just the information in
statement two, we were able to verify that j times
k, times m, times n is equal to 1. So this is all we needed. Statement one was
fairly useless. So that is what? That is b, statement two is
sufficient, and statement one alone is not sufficient. Let's do problem number 13. A certain expressway has exits
J, K, L, and M, in that order. All right. My guess is that it's going
to be some type of a number line problem. So let's see. The exits J, K, L, and M. That's the highway
in that order. And then they say, what is
the road distance from exit K to exit L. So they want to know this
distance right here. That's what we have to see, if
we have sufficient data to figure out. So statement one. They tell us the road distance
from exit J to L is 21. So we can just write that in. So line segment to JL
is equal to 21. So that tells us that this
distance right here is 21. Well, that by itself-- because
we don't know where K is within that 21 miles-- so that
alone doesn't help us. Let's see if statement two
helps figure out things. The road distance from
K to M is 26. K to M is equal to 26. So this distance
is equal to 26. This doesn't really help us at
all, because if you think about it, the distance between
K and M is fixed, right? But K could be here, which
would put M here. And then, of course, that would
change the KL distance. Or K could be really close to
L, which would put M here. M is definitely 26 miles, or
whatever it is, 26 kilometers further from K. So we don't know where K is. This second constraint still
doesn't tell us where K is, relative to-- is it closer
to J, is it closer to L, we don't know. So both of them combined
are useless. So that is e. Problem number 14. Unless I missed something. Problem number 14. Is the integer k
a prime number? So k prime. So the first statement
they give us, is that 2k is equal to 6. Well, we could just do a
little algebra here. Divide both sides by 2, and
we get k is equal to 3. And 3 is prime. So this is good, that's
all we needed. We're done. Well, let's see what
two gives us. Maybe two also gives us that. Let's see, two says that
1 is less than k, which is less than 6. Well, this is kind of useless,
because this is just telling us that k-- we know it's an
integer, because they say, is the integer k a prime number?--
but k could be 2, in which case it wouldn't be--
well, 2 is prime-- but k could be 4, in which case
it's not prime. Or it could still be 3,
so we don't know. So this is kind of useless. So all we need is statement one,
and statement two is not sufficient. So that's a. Problem number 15. Let me switch colors. OK, so they've drawn some
type of x, y axes. That's the y-axis. That's is the x-axis. All right. x, y, and then they have
this, they drew 1, 1. And then they labeled the
quadrants, quadrant I, quadrant II, quadrant III. We're doing roman numerals. And they say quadrant IV. And they say if ab
does not equal 0. So a times b does not equal 0,
which essentially tells us that neither a or b-- neither
a nor b-- is 0. OK. ab does not equal 0. In what quadrant of the
coordinate system does point a, b lie? So they want to know
where a, b is. So quadrant of a, b. OK. Let's look at the data. All right. So we know a, b-- well, we know
it's not along the axes, because if one of these were
0, then we'd be along the axes, if either a or b were 0. It says, this is interesting,
it says b, a is in quadrant IV. b, a. So this is interesting. So quadrant IV tells us what? If something is in quadrant
IV-- let me do that in a different color. This is, I think, the first
interesting problem we've encountered. Quadrant IV tells us that x is
greater than 0, right, because we're in the positive x-axis. And tells is that y is less than
0, right, because we're in the negative y-axis, right? We're here. So if b, a is in quadrant IV, in
this case b is playing the role of x, right? This is the point b, a. So that tells us that b has
to be greater than 0. b is greater than 0. And this is telling us that a,
because it's playing the function of the y coordinate
here, a has to be less than 0. So if we talk about the point a
comma b, well, now we're in a situation where we know
that a is less than 0. So we know that x
is less than 0. And we know that b is
greater than 0. And now b is playing
the y coordinate. So we know that y is
greater than 0. So if x is less than 0, we're
in the [? plot ?] negative x, and y is greater
than 0 over here. So a comma b is going to have
to be there some place. So statement number one was
all we needed to know that it's going to be
in quadrant II. So statement number
one is good. Now let's take a look at
statement number two. Well, statement number two is
kind of the same thing. So we actually don't even
have to work it out. That statement two alone is
also sufficient, just eyeballing it, but let
me work it out. But we can apply the
exact same logic. Statement two says a comma
minus b in quadrant III. Well, if we're in quadrant III,
that means that x is less than 0, right? And a plays the x role. So this tells us that
a is less than 0. And quadrant III tells us that
y is also less than 0. And the y coordinate
here is minus b. So it tells us that minus
b is less than 0. This point is minus b, so minus
b has to be less than 0, which tells us-- you can
multiply both sides by negative 1, or divide-- and
that tells us that b is greater than 0. So you get the same information
from statement two as you get from statement one,
that a is less than 0, and b as greater than 0. And then you can use that to
figure out that a, b has to be in quadrant II. So each of them independently
are sufficient to answer the question. So the answer is d. See you in the next video.