# GMAT: Math 40

## Video transcript

We're on problem 198. If 1/2 plus 1/3, plus 1/4 is equal to 13/x, which of the following must be an integer? Fascinating. So let's see what they're saying. I don't know. I'm tempted to just work through this. So if I get a common denominator here, the common denominator is going to be 12. 1/2 is 6/12 plus 4/12, plus 3/12. 6 plus 4, plus 3, is 13. So, that equals 13/12. So that tell us that x is equal to 13. I'm sorry. That tells us that x is equal to 12. It's the denominator. OK, so they give us a couple of choices. Choice one, x/8. Is that an integer? No, 12 divided by 8 is not an integer. It's 1 and 1/2. Choice two, x divided by 12. Is that an integer? Yeah, sure. I mean 12 divided by 12 is equal to 1. So, that works. Choice three, x divided by 24. No, x is 12. So, x divided by 24 is 1/2. So, two only is the answer, and that is choice B. Problem 199. And they have drawn a graph, which I will draw. So that's a y-axis. And there's an x-axis. And that's y. That is x. See, in the rectangular coordinate system above, the line, y equals x is the perpendicular bisector of segment AB, not shown. OK, so let me draw this y equals x graph. Start there. Nope. I thought I was using the line tool, but I'm not. Let me use the line tool. That's the y equals x graph. y is equal to x. And they draw point A. The place they drew it, it looks like it's roughly at the point 2, 3. I don't know if that's going to become relevant. 2, 1, 2, 3. So A looks like it's at the point 2,3. I don't know if we can assume that. The line, y equals x, is a perpendicular bisector of AB, and they don't show us B. And the x-axis is the perpendicular bisector of BC. All right. Not shown. If the coordinates of point A are 2 comma 3-- OK, we were right; A is 2 comma 3-- what are the coordinates of point C? OK, perpendicular bisector means there's some line that's exactly perpendicular to this line, where it splits it in half. So if you think about it, point A is at 2, 3. So where would B have to be in order for the line between A and B to be perpendicular to this line, and to be split directly in half? Well, it would be at the point-- I mean, you could just eyeball it-- it'd be at the point 3 comma 2. If we're at the point 3 comma 2, there's a lot of, I guess, intuitive reasons why you could think that this would be the perpendicular bisector. We know y equals x is a perpendicular bisector. The first is, you can see that the slope of this line-- you could calculate it if you want-- the slope of this line is negative 1. And I don't know if you remember in calculus class, but if you have the slope of one line-- so the slope of this line is 1-- the slope of a perpendicular line will be the negative inverse of this slope. That's something that, hopefully, you learned in calculus-- oh, in algebra a long time ago. So, the negative inverse of 1 is negative 1. So this line has a slope of negative 1. And you could make a bunch of arguments for why it's equidistant. Both of these points are on the same line, and they're equidistant from the point 2 comma 2. So that's why, you know, that it's the bisector, that this distance is equal to this distance. And, frankly, on the GMAT, you're doing it fast. You don't have to do a rigorous proof. You can eyeball it. You could say, oh, you know, those are going to be equal. I'm just going to take the mirror image and go on the other side. So this is point B. And they tell us that the x-axis is a perpendicular bisector of segment BC. OK, so x is going to be perpendicular to this new line. So this new line has to be parallel to the y-axis. So this new line is going to go like that. And whatever the distance is between B and the x-axis has to be the distance between x-axis and C. So B is two above the x-axis, so C is going to be two below the x-axis. So that's where C is. And x, at that point, is equal to 3. And so this y would be equal to negative 2. So what are the coordinates of point C? 3, negative 2. And that is choice D. Interesting problem. Problem 200. A store currently charges the same price for each towel that it sells. If the current price of each towel were to increase by $1, 10 fewer of the towels could be bought for$120, excluding sales tax. What is the current price of each towel? So let's say the current price of each towel. So if we take 120 and we divide it by the current price of each towel, we could buy, I don't know, n towels. Now, they're saying that if we had $120, and if we were to increase the price by$1-- so now the new price is p plus 1; now I can't buy n towels anymore-- they're saying I can only buy 10 fewer towels. So I can only buy n minus 10 towels. Now we have two equations and two unknowns. Well, they're not linear equations, but I think we should be able to solve them one way or the other. So here we get p times n is equal to 120. And here, let's see, we can multiply-- we have 120 is equal to p plus 1, times n minus 10. So that's equal to p times n, minus 10p, plus n, minus 10. And so both of these are equal to 120, so we could set them equal to each other. Both of these are equal to 120. So we could use both of these equations. Say pn is equal to pn minus 10p, plus n, minus 10. So let's see if we can make any headway here. I could subtract pn from both sides. So I get 0 is equal to minus 10p, plus n, minus 10. And let's see, I could add 10p to both sides, and I get 10p is equal to n minus 10. Well, what's n? Actually, I shouldn't have substituted like this. I'm kind of doing it a little messily. But we can do another substitution. n is equal to 120/p, so that should help us. The real way I should have done it is-- well, actually, no, this is I think the easiest way to do it. 10p. n is 120/p. A lot of times you just have to go forth with the algebra and see what happens. Minus 10. And let's multiply both sides of the equation by p. So you get 10 is equal-- sorry, 10p squared is equal to 120 minus 10p. Let's divide both sides by 10 just to get rid of them. So you get p squared is equal to 12, minus p. Let's take these on to the left-hand side. So you get p squared plus p, minus 12, is equal to 0. We have a quadratic. We can just factor it. That is p plus 4, times p minus 3 is equal to 0. What two numbers, when you add them, equal plus 1? The coefficient there. When you multiply them they equal negative 12. That's just practice you got in algebra doing that. So that's p plus 4, times p, minus 3. So p could be minus 4, or p could be equal to 3. And p can't be might minus 4 unless he was giving away the towels, he was paying people to take the towels, which isn't the case. We're assuming that price is positive. So the price had to be \$3, which, luckily enough, after doing all that algebra, is one of the choices. Choice C. Next question. 201. I think I have time to do this one. 201. If the sum of n consecutive integers is 0, --this is my ad hoc notation-- which of the following must be true? n consecutive integers is 0. n must be an even number. So let's think a lot about, like, our even and odd type of talk. So we know that an even plus an even is even. And we know that an odd plus an odd is also even. And you could prove it to yourself. You could try a bunch of numbers. Or if you just write an odd as 2k plus 1, and you add it to another 2m plus 1, 2k-- that becomes 2 times k, plus m, plus 2. So, that's an even number. It's divisible by 2. So an plus an odd is even. And so the only way you're going to get an odd number is if you have an even plus an odd. Now, the only way you're going to have an even plus an odd-- so we're summing n consecutive numbers. So, I don't know, let's call it number 1, plus number 2, plus number 3. Let's think about it a couple of different ways. If this number is odd, then this number is obviously even, and then this number is odd. Now, if we're taking the sum of an even number, if n is even, we're going to get the same number of evens and odds. So n could be 2. If n could be 2, we're going to have one even and one odd no matter what, right? This could be odd. This could be even. Or this could be even and this could be odd. And if n is 2, which is an even number, we can very easily-- you know an odd plus an even is an odd number. So, they say if the sum of n consecutive integers is 0-- oh, you know what? I completely missed that point. Let me continue this in the next video.