# GMAT: MathÂ 49

## Video transcript

We were on problem 227 when
I got a phone call. It was actually from Jonathan
who helps me with the Khan Academy, so I thought
it was worth taking. He works at LinkedIn and he
wanted let me know about all this news that was
occurring there. Anyway, I'm getting
distracted. Let's go back to the
GMAT problems. So problem 227. In the coordinate system above,
which of the following is an equation of line l? And they didn't put this 2 here
and this 3 here, but from the drawing you can assume that
this is the point x is equal to 3, and this is the
point x is equal to 2. And then frankly, you need that
information to figure out the equation of this line. So, first of all, what's
the slope of the line? Make sure my tool is right. The slope is equal to change
in y over change in x. See, when you go from this point
to this point, what's the change in y? It's 0 minus 2, which makes
sense, because we went down by 2, and the change in
x is 3 minus 0. Whenever you do slope, it's
important remember, if I'm taking this as the first point
on the y side, so 0 minus 2, I have to do the same on the
bottom, 3 minus 0. Otherwise, we'll get the
negative of the slope. Anyway, this is equal to minus
2/3, and what's the y-intercept? Well, it's pretty obvious. When x is equal to 0,
y is equal to 2. So we could say, y of
0 is equal to 2. So the equation of this line,
at least the mx plus b form that you learned in school, is y
is equal to minus 2/3 of the slope times x plus
the y-intercept. And that's not one
of the choices. They've written it in
a different form. They call it the ax plus by
equals c form, so let's see if we can get there. Let's get rid of this 3 in the
denominator, so let's multiply both sides of this
equation by 3. I'll switch colors. You get 3y minus 2x-- no, 3y is
equal to-- sorry-- minus 2x plus 6, right? I just multiplied
everything by 3. 3y is equal to minus
2x plus 6. Now let's add 2x to both sides,
and you get 2x plus 3y is equal to 6, which
is choice B. Next problem. Problem 200. No, what is this? Problem 228. If a two-digit positive
integer has its digits reversed, the resulting integer
differs from the original by 27. So let's say I had AB. That's a two-digit number. And I'm going to subtract
BA from it, and so the difference is 27. Since I'm assuming a positive
difference, AB has got to be a bigger number than BA. So at least the way I've
written it, what do we know about it? Well A has to be greater
than B, right? If AB is greater than BA, A has
got to be-- well, right. A has got to be greater
than B. I was going to say A could be
greater than or equal to B, but if A was equal to B,
then this thing right here would be 0. So if we assume that AB minus B
is a positive 27, then that tells us that A is greater than
B, or B is less than A. So if B is less than A, when
we're doing the subtraction problem, we're going to have
to-- when we do this part, when we do the ones place, if B
is less than A, we're going to have to borrow, right? So if we borrow, this A becomes
A minus 1 and the B becomes B plus 10, and then
we'd be ready to subtract. So what could we say? We could say that B plus 10
minus A is equal to 7, which tells us that B minus A
is equal to minus 3. And that's essentially what
they're asking for. They want to know by how much
to do the two digits differ. Well, B minus A is equal to
minus 3, that's saying that A minus B is equal to
positive 3, right? You just multiply both
sides by negative 1. Or you could say that the
absolute value of B minus A, or A minus B, or how much they
differ is equal to 3, right? The absolute value of
B minus A is plus 3. And just to make sure that this
all works out, if we do it with the tens place, A minus
1 minus B is equal to 2. Let's add 1 to both sides. You get A minus B is equal
to 3, so it works out. A and B differ from each
other by exactly 3, and that's choice A. Problem 229 The circle-- OK, let
me see if I can draw this. So I have one line, I have
another line, and then I have a circle. It starts right there, so it
looks something like that. There's a tangent. It has center c right there. And it says the circle with
center c shown above is tangent to both axes. If the distance from o to c
is equal to k-- I'll get a different color. So this is from the
origin to c. So they label the origin as o. But if that is equal to k,
what is the radius of the circle in terms of k? Let's think about this. So this is-- this is going to
deal with triangles, and how can we use the radius and relate
it to this right here? Well, the first thing that might
come to mind is if I just drop a line straight down
from here, that's equal to the radius, right? That's equal to the radius,
because it is a 90-degree angle right here, right? It touches that line. So if we could figure out this
distance right here, then we could use the Pythagorean
theorem to maybe figure out what the radius is. And luckily enough, this
distance is also the radius. And if that's not obvious
to you, think about it. The distance from here to here
is exactly the same thing as the distance from here
to there, right? And from here to here is the
radius of the circle. So this is also r. So now we can use the
Pythagorean theorem. r squared plus r squared
is equal to k squared. Or 2r squared is equal
to k squared. They wanted radius in
terms of k, right. So you get r squared is equal
to k squared over 2. Take the square root of both
sides. r is equal to k over the square root of 2. Do they want us to
rational-- nope, that's one of the choices. Sometimes they want you
to get the square root of 2 in the numerator. At least when I used to take
Algebra, they used to harp on that little bit. But anyway, this is choice B. Next question, problem 230. In an electric circuit, two
resistors with the resistances x and y are connected
in parallel. This is reminding me of
the Physics videos. So they're connected
in parallel. And this resistance is
x and resistance y. In this case, if r is the
combined resistance of these two resistors, then the
reciprocal of r is equal to the sum of the reciprocals
of x and y, and that's what they teach you. So if you want to know the
combined resistance of this, and let's say that's r, they
say 1 over the combined resistance is equal
to 1/x x plus 1/y. And if you took basic circuits,
this is one of the fundamental things about
parallel resistances. And then they want to know what
is r in terms of x and y? So let's see if we
can solve for r. So 1/r will be equal to--
let's get a common denominator, xy. 1/x x is y/xy. 1/y is x/xy, right? And then we can take the
reciprocal of both sides. So then r/1, or r, is equal
to xy over x plus y. I just switched those
two around. And that is -- xy over x plus
y-- that is choice D. Problem 231. Let me do it in this
brown color. 231. Xavier, Yvonne and Zelda, who
happen to go by X, Y and Z, at least in my world, each try independently to solve a problem. If their individual
probabilities for success are 1/4, 1/2, and 5/8, respectively,
what is the probability that Xavier and
Yvonne but not Zelda will solve the problem? So Xavier and Yvonne,
but not Zelda. So that's the case we're
looking for. So we're looking
for that case. So there's a 1/4 probability
that Xavier solves it, and then there's a 1/2 probability
that Yvonne does, but the probability that both of them
do is 1/4 times 1/2. That's the problem that
they both solve it. And then we want that case
and-- not or-- and Zelda can't solve it. So what's the probability that
Zelda doesn't solve it? 5/8 is the probability that she
does solve it, so what's 1 minus this? So there's a 3/8 probability
that Zelda does not solve the problem, right? And what does this
come out to? This is equal to 3 over--
4 times 2 is 8. 8 times 8 is 64, which
is choice E. And I'm out of time. I'll see you in the
next video. video.