# GMAT: MathÂ 45

## Video transcript

We're on problem 214. Of the 50 researchers in a work group, 40% will be assigned to team A. So let's call that team A. Of the 50 researchers working, 40% percent will be assigned to team A. So 40% of 50, that's equal to 20 go to team A. And the remaining 60% go to team B. So B gets the remaining 60%, which is 30 people, right? 60% percent of 50 is 30. However, 70% of the researchers prefer team A, and 30% prefer team B. So 70%, which is what, 35, prefer A, and 30% prefer team B. So the remainder, which is 15, right, or 30% of 50, prefer B. What is the lowest possible number of researchers who will not be assigned to the team they prefer? So there's a lot of pseudo-double negatives here. What is the lowest possible number of researchers who will not be assigned to the team that they prefer? So this is -- I have to read that a couple of times. What is the lowest number of researchers who will not be assigned to the team they prefer? So essentially they're saying this is the case where most people get the team they prefer, right? Where as many people as possible to get the team they prefer, and then we want the lowest number of people who didn't get the team they prefer. So what's the situation? Well, the best situation is -- let's see, 35 prefer A, so let's say that 20 of them got A. And then 15 fall into this upset category. And likewise, out of these 15 that prefer B, well, I don't know, let's say that all of them got B. All of these got B, but then we have to put another 15 in B, and that's these 15 people. So the best case scenario in which all of these people get B, and 20 of these people get A, but then you've got to stick 15 people of these in B, so those 15 people are going to be upset. And so that is the lowest possible number of researchers who will not be assigned to the team they prefer, so that's these 15. And that's choice A. Problem 215. If m is the average of the first ten positive multiples of 5, m is equal to average of, I don't know, let's call it multiples of 5, and M is the median of the first ten -- and this is the first ten -- and M-- big M-- is equal to the median of the same thing, what is the value of big M minus small m? So let's just write them out. I think that's the simplest way to do it. The first ten multiples of 5. 5, 10 -- the first ten positive multiples of 5. Right, so we won't count 0. 5, 10, 15, 20, 25, 30, 35 -- actually, let me write it another way, I know this is a little confusing. So this is 5, 10, 15, 20, 25, 30, 35, 40, 45, and 50. And I just thought of writing it that way for a reason. So, first of all, they want a little m, which is the mean. Little m is equal to the average of this. Let me ask you a question. If you average-- so I paired these numbers from the lowest to the highest, second lowest to second highest. If you take the sum of this one and this one, you get 55. This one and this one, you get 55. This one and this one, you get 55. So I think you get the point. The sum of any two of them are 55. So if you take the average of any of these, the average of 5 and 50 is 27.5. And that's going to actually be the average for the whole set, because you can pair them up. The average of 25 and 30, 27.5. The average of 20 and 35 is 27.5. And if you don't believe me, I mean, there's a bunch of ways you can think about this. You could say, oh, this is the same thing as 55. If you wanted to figure out the sum of this, you can say if I sum these, it would be 55 times 5. So you'd have-- the sum of all of them would be 55 times 5 divided by the total of all of them, which is 10, which is 55 times 1/2, which is 27.5. Anyway, there's a bunch of ways to think about it, but I just wanted to show you a quick way of figuring out the average of the first ten multiples of 5. And then they want to know what is the median of the first ten positive multiples of 5? So the middle numbers-- so there's two middle numbers, these two. So when you have two middle numbers, you take the average of them to get the median. Well, guess what? The average of those two is 27.5. So big M is equal to 27.5. So big m minus small m is 0, and that is choice B. Next question, 216. If m is greater than 0, and x is m% of y-- so x is equal to m/100 times y, right? If m is 6%-- and this'll turns into 0.06, right? x is m% of y, then in terms of m, y is what percent of x? OK, so let's multiply both sides of this by 100/m. So you get 100/m x, and 100/m times m/100 is just 1 is equal to y, right? Let me switch the order to so it makes a little more-- y is equal to 100/m times x. So if I wanted to write this as a percentage, this right here is going to be some type of decimal number. And so if I have a decimal and I want to multiply, and if I want to express it as a percentage, I mean, if this right here is 1, if I wanted express it as a percentage, it would be 100%. If this right here ends up being 0.2, I would multiply it by 100 and get 20%. So to express this as a percentage, I'd multiply it by 100. So 100/m times 100 is equal to 10,000/m. And that is choice E. 10,000 over m%, which is the same thing as 100/m as a decimal. Next question, 217. I'll use green. A certain junior class has 1,000 students, so the junior class has 1,000 students, and a certain senior class has 800 students. So some people dropped out. Maybe they didn't. Maybe it started bigger or smaller. Among these students there 60 sibling pairs. 60 siblings-- I already sense a Venn diagram coming-- each consisting of 1 junior and 1 senior. If one student is to be selected at random from each class, what is the probability that the two students selected will be a sibling pair? If one student is to be selected at random from each class, what is the probability that the two students will be a sibling pair? OK, so how many sibling pair members are there in each class? Well, there have to be 30 in each class. So there's 30 sibling pair members in that class and there's 30 sibling pair members in that class. OK, so if one student is to be selected at random from each class, what is the probability that two students will be a sibling pair? OK, so let's say we pick a student from the junior class first, right? So first of all, what's the chances that they're even part of a sibling pair? Well, that's going to be-- there's 30 of them that are part of a pair, and there's 1,000 in the class, so the probability of that happening is 30/1,000, which is the same thing as 3/100, right? That's the probability that we got a part of a pair. Now we're going to go and pick someone from the senior class. If we want to find this guy's sibling, we can't just find another sibling pair, we've got to find this guy's actual sibling. So actually, only one person in the entire senior class is going to be-- whoever this person we pick from the junior class, if we did happen to pick one, the only way that we got an actual sibling pair is if we pick that one person in the senior class who is this person's sibling. So there's actually only a 1 in 800 chance that we picked them. So if we multiply that out, that is equal to 3 over-- let's see, what is it? 80,000? 3/80,000, which is not one of the choices. So what did I do wrong? Here. So I see 3/40,000, and for some reason, I think maybe I'm not counting two scenarios. There's probably two scenarios in which is this can occur, right? I could have picked-- if I go to the junior class first and there's a 30 in 1,000 chance that I pick the correct sibling, so that's 3 in 100, and then the odds that I take his or her sibling from the senior class is 1 in 800. But maybe then-- let me see if I'm missing some logic here. Maybe I could have picked-- well, instead of me stumbling because I'm already over time, let me continue this in the next video.