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We're on problem 190. If the operation star with a circle around it is defined for all a and b by the equation, a star with a circle around it b is equal to a squared times b over 3, then what does 2 star with a circle around it 3 star with the circle around it minus 1 equal? So 3 star minus 1. That's equal to-- let's see, I'll do this in magenta-- 3 star minus 1. That's equal to 3 squared times minus 1 which is minus 3 squared, or minus 9. 3 squared, 9. Times minus 1. That's minus 9 divided by 3 which is equal to minus 3. So this reduces to 2 star minus 3 which is equal to 2 squared, which is 4, times minus 3. All of that over 3. Minus 3 divided by 3. That's just minus 1. So you have minus 1 times 4 is equal to minus 4. And that's choice E. Problem 192. No, 191! Don't want to skip problems. The inside dimensions of a rectangular wooden box are 6 by 8 by 10. A cylindrical canister is to be placed inside the box so it stands upright when the closed box rests on one of its 6 bases. Of all such canisters that could be used, what is the radius in inches of the one that has maximum volume? OK, so they're going to put one canister in this box. And how do you maximize its volume? So let's think about it this way. Let's think about what a volume of a cylinder is. If this is my cylinder, its volume is the area of the side times the height. If the area of the side is equal to-- if this is the radius-- it would be pi r squared times the height. Height times pi r squared. Now what's r going to be equal to? r has to go all the way around. So, r is going to be equal to half-- well, it obviously equals half the diameter of this circle. But if we look at the base that's 6 and 8, if we use that as the base, and the 10 comes out of the page, the radius can only be what? The radius can only be 3 because the diameter of the circle can only be 6. We don't get to use the 8. So, this is the case if we use the 6 and the 8 side. And so we'll kind of be-- I guess the best way to think about it is we'll be wasting this base over here. The other possibility, if we had the 8 and the 10 side, if this was our base then we could have a radius of 4. Let me think about it. Yeah, radius would be a 4. Because we would have the height 8, we'd have a radius of 4. Now when we look at this equation, if we had to maximize either the-- this is going to be the volume of the cylinder-- if we had to maximize either the height or the radius, because those are the two things that we can deal with, which one, at least to you, seems more important? Well, to me, the radius seems more important because I'm squaring it. So, these, in my mind, are the two contenders. The worst contender is where we use the 6 and the 10 side as a base. Because then we use a lot of space. But we can figure that one out, too. Because then we waste a lot of space and we have a height of 8 and the radius is still--. So, first of all, this one is clearly better than this one. And why is that? They have the same area of their base because the radius is 6. Sorry, the radius is 3, the diameter is 6. But in here the height gets to be 10, and here the height only gets to be 8. So this one is a non-option. So, really, we have to compare these two. And I've drawn their bases. So, if you think about this one, we have an area of the base of the radius is 4. So, 10 pi times the height of six. That's the volume of this cylinder. Here we have a radius-- we have a radius of 3. Half of the 6 diameter. So, the area is 9 pi. And we're going to multiply that times the height, which is 10. So, here we have 90 pi. And here we have 6 times 16, which is 96 pi. So, this is the best of our cases. And that was the correct intuition. We're maximizing the radius. And when you maximize the radius, we get a radius of 4. So what did they ask us in the question again? What is the radius in inches of the one that has the maximum volume? And so that radius is 4. Where one side is 8 and one side is 10. And that's choice B. Next question. 192. Square root. OK, they have all sorts of interesting things. They essentially want us to simplify. The square root of 2 plus 1 times the square root of 2 minus 1 times the square root of 3 plus 1 times the square root of 3 minus 1. So, this seems to be just an application of-- I'm sure you've learned this in algebra-- that a plus b times a minus b is equal to a squared minus b squared. You don't have to memorize this. You could actually multiply it all out and you'll get this. But it's a good thing to know if you want to do things quickly. So, square root of 2 plus 1, times square root of 2 minus one, is equal to square root of 2 squared, minus 1 squared. We're going to multiply that times-- using the same principle-- square root of 3 squared minus one squared. Square root of 2 squared. That's 2 minus 1 times 3 minus 1. That's equal to 1 times 2. Which is equal to 2. And that is choice A. Problem 193. In a certain Calculus class, the ratio of the number of math majors to the number of students who are not math majors is 2:5. So, math to not math is equal to 2:5. If 2 more math majors were to enter the class the ratio would be 1:2. So if I have 2 more math majors, then the ratio of them to the non-math majors is equal to 1:2. How many students are in the class? They want to know what m plus nm, non-math majors, is. Let's see. Let's see if we can solve for anything. We actually have two linear equations and two unknowns. You might not see it yet, but if we cross multiply we get 5m. I'll just call these n, non-math majors. I don't want to keep saying nm. 5m is equal to 2n. And here you get 2 times m plus 2 is 2m plus 4 is equal to n. We could substitute that. If n is equal to that, then that means that 5m is equal to 2 times n. n is equal to 2m plus 4. So, you get 5m is equal to 4m plus 8. Subtract 4m from both sides, you get m is equal to 8. If m is equal to 8, we could say 5m is equal to 2n. So, 5 times 8 is 40. And so n is equal to 20. Let's see, are they asking for the-- if 2 more students were to enter the class? OK. So they want to know how many are in the class right now. So they want to know m plus n. Well, that's equal to 8 plus 20 which is equal to 28 students. And that is choice D. Next question. I think I have time to fit this one in. I'll continue in the next if I don't. 194. What is the units digit of 13 to the fourth times 17 squared times 29 to the third? So, here you just have to realize that the units digit in any of these numbers, are just the units digit of each of these numbers times each other. And then you take the units digit of that. So, we really just have to worry about the units digits. The answer to this question would be the exact same thing as if I said the units digit of 3 to the fourth. We just have to worry about the units digits. 3 to the fourth times 7 squared times 9 to the third. It would be the exact same thing. So, 3 to the fourth, what's the units digit? 3 to the one is equal to 3. 3 squared is equal to 9. 3 to the third is 27, but the units digit is 7. And then 3 to the fourth is 81 and the units digit is 1. And just so you can see this principle that I just talked about-- and we could prove it if I had more time-- is that I could just multiply 3 times 7 and the units digit is just 1. Or I could just multiply 3 times 9 and the units digit is 7. We could just ignore everything else. So here the units digit is 1. 7 squared, what's the units digit? Well, 7 squared, the units digit is-- well, 7 squared is 49. The units digit is 9. And then 9 to the third, what's the units digit? 9 to the 1 is 9. 9 squared. The units digit is 1. It's 81. And then 9 to the third, we don't have to worry about anything else. When you multiply 81 times 9, the units digit is going to be 9 times that 1 in 81. So, that equals 9. So, it's 1 times 9 times 9. And then 9 times 9 is 81. We want the units digit, which is 1. So the answer is E. And I'll let you think about that a little bit. Maybe you want to play around and see why we only have to worry about the units digit and why I was able to make this statement right here to avoid doing a lot of harder math. Anyway, see you in the next video.