GMAT: Math 34

We're on problem 173. The probability is 1/2 that a certain coin will turn up heads on any given toss. Fair enough. If the coin is to be tossed three times, what is the probability that on at least one of the tosses the coin will turn up tails? So probability, at least one tails. That might seem really hard for you. How do I know which tails it is and all of that? But let's think about the opposite circumstance. This is equivalent to saying what is the probability that I don't get all heads? These are equivalent things. At least one tails means that I don't get all heads. What is the probability that I don't get all heads? Well, that's 1 minus the probability that I get all heads. Figuring out the probability that you get all heads isn't that hard. Because you have 1/2 probability of heads on the first time, times 1/2 probability because you have to get a heads on the second time, times 1/2 probability because you have to get a heads on the third time. That's equal to 1/8. So 1 minus 1/8 is equal to, 8/8 minus 1/8, which is equal to 7/8. So that is choice D. A lot of these probability questions, you have to get at least 1 of something. If you think of it that way, it's very hard. You'll have to start thinking of combinatorics and binomial coefficients and all of that. But if you just think of the opposite problem, what would have to be true? That I don't get all heads. What's the probability that I do get all heads? And then that's a much easier thing to solve for. You just take that from 1, and you get the probability that it doesn't happen. Next question, 174. Of the final grades received by the students in a certain math course, 1/5 are A's, 1/4 are B's, 1/2 are C's, and the remaining 10 grades are D's. What's the number of the students in the course? So if we know what fraction 10 is, then we know the number of students in the course. So, essentially, what do these fractions add up to? So let's find a common denominator. Looks like a common denominator between 4, 5, and 2 is 20. So, 1/5 get A's. So that's four 20's plus 1/4. So, that's five 20's plus one 1/2. That's ten 20's. So the A's, B's, and C's are equal to-- what is this? 4 plus 5 is 9 plus 10 is equal to 19/20 of the class. They're saying that the remaining 10 grades are D's. So there are no F's in this class. So the remaining grades, that's 1/20 of the class. So that's equal to 1/20. Right? Because if the A's, B's and C's are 19/20, and all the other grades are D's, then all the other grades are going to be 1/20 of the class because that's what's left over. So if I say 1/20 of the class is equal to 10, multiply both sides by 20. You get the class is equal to 200 students. That's choice D. Next question, 175. As x increases from 165 to 166, which of the following must increase? Interesting. So 1, 2x minus 5. Well sure, as I go from 2 times 165 to 2 times 166, this is obviously going to increase, and this is going to stay the same. So this is going to increase. You can substitute the numbers in there if you don't believe me, but I think that should make some intuition. Statement 2, 1 minus 1 over x. So, 1/165 is greater than 1/166. This is a larger fraction than that is. 1 over 1 million is much smaller than 1/2. So 1/65 is greater than 1/66. So you might say, as I go from 165 to 166, this term right here gets smaller. Right? That term does get smaller. But we're subtracting that term. So if we're subtracting a term that's getting smaller, the whole value is actually going to get larger. We're subtracting a smaller number as we go to 166. 1 minus 1/166 is a larger number than 1 minus 1/165. So this, also, gets larger. Statement 3, 1 over x squared minus x. So the question is, what happens to x squared minus x as we go from 165 to 166? If you think about it, we are well beyond any type of x squared, any kind of vertex in this. If you know a little bit of calculus, you could figure out the vertex really fast. So, this is going to be some type of a parabola. This x squared minus x is going to look something like this. Right? Its vertex is actually-- 2x minus 1 is equal to 0-- its vertex is at x is equal to 1/2. It's a very small number. So we're in the point of this curve that's increasing upwards. Right? So when we go from 165 to 166, this right here is definitely increasing. This denominator is definitely going to increase from 165 to 166. You might say, we're going to be subtracting a larger number. But that's more than offset by squaring the larger number. So this denominator is going to increase. But if the denominator is increasing, then the whole fraction, which is the inverse of the denominator, that's going to decrease. So this one is not increasing. So, 1 and 2 only. So that's C. Next question, 176. I'll do it in a different color. A rectangular box is 10 inches wide, 10 inches long, and 5 inches high. So, it's 10 inches, 10 inches, and then 5 high. Fair enough. That's the box. What is the greatest possible distance in inches between any two points on the box? So, this is a bit of an old trick. So, the longest distance in this box is kind of a three-dimensional diagonal. So, a normal diagonal goes along one surface. But if you go from this point back here-- and I'll do that in a different color-- to this point over here, that's the longest dimension in the box. If you know how to figure out distance in three dimensions, then that's all you really need to know how to do. It's going to be the square root of 10 squared plus 10 squared plus 5 squared. So that's the fast way to do it. I'll show you the logic of it in a second. So, 10 squared plus 10 squared plus 5 squared. If we had a fourth dimension, which is hard to visualize, we would at that distance squared and take the square root of everything. So this is equal to the square root of what? 100? 225? 225, which is equal to 15. That's choice A. Let me give you the logic. So if I'm taking the distance from this point to this point, how do we figure that out? Well let's see, the easiest way to think about it is can we figure out this distance right here? So this is going along the top of the box. Let me do that in the magenta. Well sure, if we look at the box straight down, what I just drew-- if this is 10 and 10-- what I just drew goes like that. Right? It's going along the top of the box. So, this is going to be the square root of 100 squared, the square root of 10 squared plus the square root of 10 squared. So, this just the Pythagorean theorem. This distance right here is the square root of 200. I just used the Pythagorean theorem. I just said this is 10, this is 10, so this has to be the square root of 200. Now we have to do a little bit of visualization. To figure out that green line that's the longest diagonal in this box, think about it this way. We still have a right triangle although maybe it's a little harder to visualize. Well, maybe it isn't. So, the right triangle looks like this, where the side is the square root of 200, that's this up here. This side right here is 5. They tell us that. So the side right here is 5. Then the diagonal that we're trying to solve for is in green. That's the longest diagonal. Remember this line is the same thing, if you view from the top, as this line. It's just that you can't see what's below it. So here we just use the Pythagorean theorem. So, we could say c squared-- let's call this c-- is equal to 25 plus this squared plus 200. Then you get to the same answer that we just got to. So c is equal to the square root of 225, which is equal to 15. What problem was I on? That was choice A if I haven't already said it. Anyway. See you in the next video.