GMAT: Data sufficiency 40

We're on problem 151. If x and y are positive, is the ratio of x to y greater than 3? And they're telling us that x and y are positive. Fair enough. Statement one. x is 2 more than 3 times y. So x is equal to 3y plus 2. Let's rewrite the original question because I think it's going to look a lot like this. If we multiply y times both sides of the equation-- and we don't have to change the inequality because they told us that y is positive. They're both positive. So if you multiply both sides of the equation times y, you get x is greater than 3y. This is the same question as this. Well, if we look at statement number one, clearly x is greater than 3y. Because x is 2 more than 3y based on statement number one. So statement number one is sufficient to answer this question. x is definitely greater than 3y. x actually equals 3y plus 2. This plus 2 tells you that x is that much greater. 2 greater. Second statement. The ratio of 2x to 3y is greater than 2. Let's just do some algebraic manipulation. 3y, once again, we know that's positive. So when we multiply both sides of the equation by it, we don't have to switch the sign. So you get 2x is equal to 6y. Now we can divide both sides by 2. And you get-- oh, sorry, why did I put equals? It should be greater than. Divide both sides by 2 and you get x is greater than 3y. So this statement and this statement are the same thing, so 2 alone is also sufficient. So either statement alone is sufficient to answer this or this question. 152. OK, if arc PQ-- OK, let me see if I can draw this. So they've drawn a semicircle. So let me draw a circle. That's a circle, and I can make it a semicircle by drawing a line from there to there. That's about as good as I can do. And they have it connecting to some point. Let me see if I could do this. Looks like that. And then it looks like that. And then they actually have drawn an altitude coming down like that. And I think I've done all my line drawing. And then, let's see, they labeled this P. They labeled this Q. They labeled this R. And this is 2. And they say that this distance right here is a. This distance right here. And then they say that this distance right here is b. Fair enough. So they say if arc PQR-- so that's that arc right there-- is a semicircle, what is the length of diameter PR? So they essentially want to know what a plus b is, right? That's PR. So that's their question. So immediately-- well let's just look at their statements. What did they give us? Statement one, they tell us that a is equal to 4. So from knowing that a is equal to 4, if we can figure out what b is, then we would know what a plus b is and we would be done. Well, let's see if we can do that. Statement two is b is equal to 1. So if we know what b is, and then we can figure out a from it, then we know what a plus b is. So actually, I think if either of these is sufficient, the other one probably is. Because if statement one is sufficient-- if knowing a we can figure out b, then probably knowing b we could figure out a. But let's see if that holds. So one thing that immediately should pop out of your head, although I don't know if it will-- and this pops out of my head, really, from going to math competitions when I was a kid. But if you have a triangle inscribed in a semicircle, that triangle-- and one of its sides is a diameter, right? And that's the case with this one. Because we know it's a semicircle, so we know that this is the diameter-- then that triangle is a right triangle. This is a right triangle. Even the way they drew it, it kind of looks like that. But you that's something you should know. Even if this triangle was drawn like that, it's always going to be a right triangle. And I'll maybe do another YouTube video proving that. But that's just a good thing to know for the gmat. So let's see if we could use that information and the Pythagorean theorem to come up with a formula that relates a and b. So if we look at just this triangle right there, we know what? We know that a squared plus 4 is equal to-- let's call this side c squared-- is equal to c squared. Fair enough. If we look at this smaller triangle right here, we know that b squared plus 4 is equal to-- let's call this d squared, that's the hypotenuse of the smaller triangle-- is equal to d squared. Now, if we look at the larger triangle, which we also know is a right triangle, now c squared and d squared are the non-hypotenuse sides, and a plus b is a hypotenuse side. So we know that c squared plus d squared is equal to this entire side squared, which is a plus b squared. Well, we know what c squared and d squared are. We can substitute that. So c squared is a squared plus 4-- and I'll switch colors just to avoid the monotnony. So c squared is a squared plus 4. And d squared is b squared plus 4. And all of that's equal to a plus b squared. Let's just multiply that out because I think we can do some canceling. So we get a squared plus 2ab plus b squared. OK. So, we could subtract a squared from both sides. a squared, a squared. Subtract b squared from both sides. b squared and b squared. And this simplifies out quite nicely. Let's see. 4 plus 4 is 8, is equal to 2ab. Divide both sides by 2, and you get ab is equal to 4. So all of this crazy information, that this is inscribed in the semicircle and this is a, this is b, all of that reduces down to ab is equal to 4. So now we know. If we know a, if a is 4, then we know that b has to be equal to 1. And then we can figure out what a plus b is. It's 5. So statement one alone is sufficient. And similarly, if we know what b is, we know that a has to be 4. And so a plus b is equal to 5. So that was actually less hairy than I thought going into it. Next problem. And this really is the home stretch. 153. Does the integer k have a factor p such that 1 is less than p is less than k. OK, so they're essentially saying, is k prime? This is another way of saying is k-- or you could say, is k not prime? Or k is composite. I think that's the word for it. A non-prime number. So that's all they're asking. Does k have some other factor other than 1 and itself. So let's see, statement number one. k is greater than 4 factorial. Well, 4 factorial is 4 times 3 times 2 times 1. So that's 12 times 2. That's 24. So this boils down to k is greater than 24. So I can give you both prime and non-prime numbers greater than 24. 29 is prime. 30 is not prime. So statement one, at least by itself, is pretty useless. k can be greater than 24 and be prime or not prime. Statement two. This looks interesting. 13 factorial-- which is a large number, so it's hard to multiply it out-- plus 2 is less than or equal to k, which is less than or equal to 13 factorial plus 13. So, essentially, what we need to prove, somehow, is that every number between this and this is not prime. And let's think about how we can do it. Well, let's take the initial example of 13 factorial plus 2. So 13 factorial, just so you know, is 13 times 12, times 11-- you keep going-- times 3 times 2 times 1. That's 13 factorial. So 13 factorial is definitely divisible by 2. So we could rewrite 13 factorial. We could write that as-- 13 factorial divided by 2 times 2 plus 2. I just rewrote 13 factorial plus 2. And I want you to see that this is the same thing, right? I just divided by 2 and multiplied by 2. And you'll see in a second why I did that. And this is a tricky problem. When I looked at this at first, which I actually did glance at before the video, because-- I don't know if this is something I would have gotten right if I was under time pressure on the gmat. But when I thought about it a little bit more it made sense. So if I divide by 2 and times 2, I could-- and this is going to be an integer. Because we know that 13 factorial is divisible by 2. 2 is actually one of its factors. So as long as you understand where I got-- you can divide anything by 2 and multiply by 2 and not change it So this is the same thing as this. And then if we factor the 2 out, this becomes 2 times 13 factorial over 2 plus 1. I just factored the 2 out. This isn't a fraction. That's just a line I did just to not make it look too messy. So what we show is that 13 factorial plus 2 is definitely divisible by 2. Whatever that number is, it's definitely divisible by 2. And how do I know that? Because this is just another way of writing 13 factorial divided by 2. I'm sorry, this is another way-- let me write this down-- this is the same thing as 13 factorial plus 2. It's just another way of writing it. I took 13 factorial divided by 2 and multiplying by 2, and then taking out the 2. And the reason why I know that 2 is divisible into this is because I know that this is an integer. We know that 13 factorial is divided by 2 because 2 is one of the things that goes into it. And so this is an integer. You add 1. So this is an integer. And so you're multiplying 2 times some integer. So this has to be divisible by 2. So 13 factorial plus 2 is divisible by 2. Now let's just extend that generally. 13 factorial plus x. Let's prove that as long as x is less than or equal to 13 that this is divisible by x. So we can do the same trick. You could say this is equal to 13 factorial divided by x, times x plus x. That's equal to-- you can factor the x out-- x times 13 factorial over x plus 1. And how do we know that this number, which is the same thing as this number, how do we know that this is divisible by x? Well, just a little thought experiment. We know as long as x is greater than or equal to 1 and less than or equal to 13, and x is an integer. As long as x is an integer that's greater than or equal to 1 and less than or equal to 13, it is one of the factors of 13 factorial. 13 factorial is 13 times 12 times 11. It's all of those numbers multiplied together. So we know that as long as x is less than or equal to 13, we know that this right here, this is an integer. If this is an integer, then the integer plus 1 is also an integer. So 13 factorial plus x can be rewritten as x times some integer. I'll call it I. But the bottom line is, the reason why I wanted to do that, it shows you that every number between 13 factorial plus 2 and 13 factorial plus 13 is actually divisible by whatever number you're adding to the factorial. So since it's divisible by those numbers, the numbers cannot be prime. So hopefully-- I've actually run way over time-- but hopefully I've shown you that that information alone, statement two alone, is sufficient to show that k is not prime or that k does have a factor p that's between 1 and k. Anyway, I hope didn't confuse you. I think that was one of the hardest problems we've done, so don't feel bad if you found that a little daunting. See you in the next video.