GMAT: Math 49

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.