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# Matematyka GMAT: 52

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We're on problem 240. Seed mixture x is 40% rye grass and 60% bluegrass. Let's see, x is 40% rye grass and 60% bluegrass. Fair enough. Seed mixture y is 25% rye grass-- so it's 0.25 rye grass-- and 75% fescue. So I should look up what that is. 75% fescue. If a mixture of x and y contain 30% rye grass, what percent of the weight of the mixture is x? OK. Well, let's say that A is equal to the percent of weight of mixture that is x, what they're asking for. So how much rye grass-- what percentage of that mixture will be rye grass that comes from x? Well, we have A%, or you know, we could call that-- fraction A of the mixture is from x. And then x is 40% rye grass. So if we multiply the amount that we're getting from x times 40%, times 0.4, this'll tell us the amount of rye grass from x. So r from x, or the proportion of rye grass from x. And now, how much rye grass are we going to get from y? So first of all, what's our proportion of y? If this mixture is A of x, there's going to be 1 minus A of y. If this is 1/4, this is going to be 3/4, right? Makes sense. So this is how much y we have, and y is 25% rye grass. So our rye grass from y will be the amount of y we have times how much y is rye grass. So this is rye grass from y. And they tell us that this is equal to a mixture. The mixture should have 30% rye grass. So let's just solve for A. We get 0.4 times A plus-- let's see, multiply-- 0.25 minus 0.25A-- just distributed the 0.25-- is equal to 0.30. 0.4A minus 0.25A-- 40 minus 25-- that's 0.15A. And then subtract 0.25 from both sides, is equal to 0.5. So A is equal to 0.5 divided by 0.15, is the same thing as 5 over 15, which is equal to 1/3, which is equal to 33 and 1/3%, Which is choice B. Next problem. Problem 241. If the integer n has exactly three positive divisors, including 1 and n-- so it's divisors are, I don't know, 1, some number, and n-- how many positive divisors does n squared have? So one thing, I don't know if you've ever thought about this, but how can any number have an odd number of divisors? Right? Because normally, if you take a number, let's take the number 6-- and you shouldn't do this on the GMAT, it's a waste of time. But its divisors are 1 times 6, or 2 times 3. So every divisor tends to have a corresponding divisor. So how can you have an odd number of divisors? Well, only if one of the divisor's corresponding divisors is itself. If it's 1 times n, and x times x, which implies that if you have an odd number of divisors, you're dealing with a perfect square. Anyway, I just wanted to point that out. But that's actually not that relevant to that problem. It's just something, you know, it is useful sometimes. But anyway. They just want to know how many divisors n squared has. Let's just find a number that has this property. Well, we know any perfect square-- well, a small perfect square will have this property. So let's see. Number 4, it has its property. It's factors are 1, 2, and 4. So how many factors does 4 squared have? Well, 4 squared is 16. Its factors are 1 and 16, 2 and 8, and 4 and 4. So it has five factors, and that is choice B. And you could do it mathematically. You could say all the combinations of when I multiply all the factors of this times itself, and you could say, oh, they have five combinations. But in this case, it's just way easier to think of, OK, let me think of a small perfect square that's going to have three factors. A perfect square just tells you it has an odd number of factors, but a small one will have three. And let me just square it and just count its factors. Fastest way to do that problem, I think. Unless there is-- I don't know, maybe there is a faster one. 242. If n is a positive integer, then n times n plus 1, times n plus 2 is-- OK, they have a bunch of things. What happens when n-- let's see, even when n is even, the even only one is odd, odd [? 1n ?] is odd. OK, so they want us to see what happens when n is even or odd. So let's see what happens when n is even. Let's say that n-- if n is even, that means it's equal to 2 times some integer k. We don't know, some positive integer k. So let's substitute that in. So then this breaks down to 2k times 2k plus 1, times 2k plus 2. Interesting. And let's see, it looks like this is saying that e should be-- right. Because let's see, we could factor 2 out of this one. So this becomes 2k times 2k plus 1, times 2, times k plus 1. Right? I just factored the 2 out of there. And let's multiply these two. So you get 4k times 2k plus 1, times k plus 1. So this one is divisible by 4. And I just glanced at statement e, they talk about divisibility by 4, so I think we're done. This simplifies to this if we can assume that n is even. So if n is even, then we are definitely divisible by 4. 4 can definitely go into the number evenly. So statement E is right. Divisible by 4 when n is even. Statement E. Next problem. We're on the last page, 243. A straight pipe 1 yard in length was marked off in fourths and also in thirds. Maybe I'll give some dimensionality to our straight pipe. So let's say the pipe looks something like that. It's a straight pipe, and it was marked off in fourths and thirds. So if I mark it off in fourths, let's see, that's the middle, and that's a fourth, and that's another fourth, roughly. And if I mark it off in thirds, it looks something like this. Right? All right. If the pipe was then cut into separate pieces at each of these markings, which of the following gives all the different lengths of the pieces in fraction of a yard? Fascinating. OK, so we're going to cut it, actually, at here, and here, and here, and here, and at here. And actually, if we go halfway, then we can just use symmetry to say, OK, this piece is going to be the same as this piece, and this piece is going to be the same as this piece, and that piece is going to be the same as that piece. So we really just have to figure out up until the halfway point. So let's think about it. So this whole thing is 1 yard, so this distance right here, that distance right here is 1/4. That distance is 1/4. We know that this whole distance is 1/3. This distance is 1/4, so this distance right here is going to be 1/3, because that's 1/3, minus 1/4. Common denominator over 12. 1/3 is 4/12, minus 3/12 is equal to 1/12. So this distance is going to be 1/12. And then what's going to be this length? Let me get a different color. I want to make sure that you see that 1/12. Well, this distance is 1/2, and this distance is 1/3, so it's going to be 1/2 minus 1/3 is equal to common denominator of 6, is equal to 3/6 minus 2/6, is equal to 1/6. So this piece is 1/4. This piece is 1/12. This piece is 1/6. And then you could just say, oh, well, it's going to happen the same thing on the other side. This is 1/6, 1/12, and then this is 1/4. And so if I say-- let's see, 1/6, 1/12, and 1/4. Right, that is choice D. 1/6, 1/12, and 1/4. Choice D. Those are the different piece sizes we have. Problem 245. If 0.0015 times 10 to the m, over 0.03 times 10 to the k, is equal to 5 times 10 to the seventh, then what does m minus k equal to? And that's important to read, I was about to solve-- m minus k. So they want to know what m minus k. So if we immediately divide these, we should be able to come up with something interesting, right? Because this is the same thing as 0.0015/0.03 times 10 to the m divided by 10 to the k. Well, that's just times 10 to the m minus k, is equal to 5 times 10 to the seventh. And I don't know, let's see if we can do anything interesting to this decimal number. Let's see, if we were to-- let's just divide it. So if we just say 0.03, instead of multiplying by 10 and doing-- going into 0.0015. Let's add some 0's. OK. If we take this decimal two to the right, we're going to take this decimal two to the right. So then we get 3 goes into 1 zero times, 3 goes into 15 five times. So we get 0.05. So this whole thing simplifies to-- and I'll write it in magenta-- so that simplifies to 0.05 times 10 to the m minus k, is equal to 5 times 10 to the seventh. So let's divide both sides by this 0.05. What is 0.05 goes into 5? Let's add some decimal points here. So if I push this to the right two, I push this to the right two, and this is my new decimal point. So 5 goes into 5-- well, 5 goes into 5 one time, so it's 100. Right? So 5 divided by 0.05 is 100. So we get 10 to the m minus k, is equal to 100-- what's 100? That's the same thing as 10 squared. Times 10 to the 7th. So 10 to the m minus k is equal to 10 to the ninth. So m minus k must be equal to 9. And that is choice A. See you in the next video.