108 passengers. During one of its flights, Anita noticed that not all the seats were occupied. There were twice as many occupied seats as empty ones. How many passengers were on the plane? And here's the poll for this one. That's not the right poll. Hang on. This poll. So between the answers I see in the chat and the answers that are coming up in the poll, you're doing very well on this. So let's share the results. I know not everyone has answered yet. But of those who answered, 86%. So that's a lot, and I see a lot correct in the chat as well. So we'll do this one pretty quickly. A plane can carry 108 passengers. Anita noticed not all the seats are occupied, twice as many occupied seats as empty ones. So that means for every two occupied, there's one emptied. So we'll call this occupied and this empty. So it's a two to one ratio. So that means that we have three parts, right? So if we think about that, it's three parts. Three parts total. So if I take my 108 and divide it by three, I get 36 people or seats per part. Two parts are occupied. So 36 times two does give me the 72 occupied seats and 36 empty seats. So just using some ratios, proportions, you should be able to do that problem. Okay. Now remember, today's webinar is being recorded. That is why we do not have you activating your videos or unmuting during the session. But your TA Jacob and I do like to see how you're doing. We do like to get your answers. So participate in those polls and send us some replies or questions in the chat. We do our best to help you as we're going through. We can't interrupt the flow of the webinar, but we do try our best. Okay. So today's webinar is on algebraic thinking. So there's our official title page. What is algebraic thinking? Algebraic thinking is a way of working through problems where we have some unknown things that we have to find out. So the first part of all of our problems will be to understand the problem, determine what it's asking, make our plan to solve the problem. Today, we're gonna be using that algebraic, what's missing type of problem solving. We'll work very carefully. And if we have time, we look back, we check our answers. Does our answer make sense? So frequently with algebra, once we get a solution to our algebraic equation, we can plug it back in and make sure that the problem works with that answer that we found. So key ideas in algebraic thinking, believe it or not, the biggest thing is going to be equality or perhaps inequality with greater than or less than, if I say the minimum number or it has to be greater than, but it's equality. Now, equal does not mean that it won't change. It just means that both sides of an equation have the same value. So if you have two children in a family, they always want to have the same of everything, right? They don't want you to give one side more than the other side. So you can kind of think of it that way. If one child has two candies and the other child has two candies, if I say I'm going to double it for both of you, they still have the same. If I say I'm going to give you each three more, I can add three to each of them and it's still the same. So we're always keeping it equal, always fair in a way. So you won't see too many negative numbers at this level, but definitely by sixth and seventh grade, you should be starting to think about what happens when you have negative numbers. We will be using variables today. I try to use variables either X and Y or variables that are letters that make sense from the problem. The only letters I try to avoid using are things like the letter O because it looks like a zero. There are a couple of others that sometimes get a little confusing. We don't like to use those letters. Jacob can probably warn you about that, right? Yeah. Okay, so let's see. Maybe we will be writing some equations from our problem. So that is taking the words and making them into symbols, numbers, operations, and equations. That's a really good concept here. What is the difference between the sum of the first 1,000 consecutive positive even numbers and the sum of the first 1,000 consecutive positive odd numbers? A lot of words there. Consecutive means in a row, not skipping any, right? Positive means has to be greater than zero and no negative numbers. I think everyone knows even and odd. And I do have a poll. I have a lot of polls today. So for those of you who like to do the polls, you'll be very happy. Okay. Anybody else wanna put the answer into the poll? Most of you participated. All right. So here we go. Kind of close between a few different answers. I see a few for 500, some people saying one, 1,000. All right, let's take a look. I think the hardest part in this one is reading and understanding the actual problem. So I'll admit that when I first read it, I thought they meant the first 1,000 numbers. So like one to 1,000, right? That makes sense. That's one reading of the problem, but they're saying the first 1,000 consecutive positive even numbers. So that is two, four, six. We need 1,000 of them. So we have to go all the way up to the number 2,000 in order to get 1,000 even numbers. And now we're gonna do the same thing with the odd numbers. So that's one, three, five, and we're gonna get all the way up to 1,999. Okay, so that gives me 1,000 even numbers and 1,000 odd numbers. If I take the difference, I kind of lined them up on purpose. If I take the difference here, that's one. If I take the difference here, that's one, right? And there's all those that I skipped in here. Altogether, we have 1,000 times one. So the correct answer is D, 1,000 of these ones. So not tricky to do 1,000 times ones, but tricky to figure out what is it asking? That's why math kangaroo problems require you to read the problem statement sometimes more than one time. Okay. At a fair, the tickets for four various types of rides cost two, three, four, and $5 respectively. A class took a field trip to the fair. They bought enough tickets for each student to go on each of the four rides once. The tickets cost $280 altogether. How many tickets did they buy? This is a little bit of a multi-step problem, at least the way I solve it. We'll see how you do it. And I do have a poll, but I'll give you a few seconds to have a good look at the problem before I launch that poll on top of your screen. Okay, so this is exactly what I would expect would happen with this problem. Students are either saying 20 or 80. And that makes perfect sense, because this problem has kind of a little trap in it. So most of you have done the problem correctly, except for the really tricky part Math Kangaroo does of reading, right? How many tickets did they buy? So a lot of you, very correctly, have either found the average price of a ticket or found out a whole set of tickets because they bought enough tickets for each student to go on each of the four rides once. So for each student, they used 2 plus 3 plus 4 plus 5, right? So that's how much it would cost for each student to ride one time. So it's $9, $10, $11, $12, $13, $14 for each student. So a lot of you then took the $280, divided it by $14 per student, and you get $20. And so a lot of you clicked off, oh, I think $20 is the answer. But that is how many students were on the field trip or how many sets, packages of tickets did they buy. But if we want to know how many tickets they bought, well, each one of these was actually four tickets. They were of the four different prices. So the correct answer that Math Kangaroo was looking for was that they bought 80 tickets. But the 40, the 20, is a logical answer as a partway response. Yeah? Does that make sense? So hopefully, next time you see a problem like this, you're going to read extra carefully and make sure you say, oh, it's tickets, not sets of tickets or how many students were there. Okay. I do have a poll for this one, too. There are five containers in a treasure chest. In each container, there are three boxes. And in each box, there are 10 gold coins. The treasure chest, the container, and the boxes are all locked. How many locks do you need to open to get 50 coins? Be careful on this one. The answer choices are not in numerical order. They kind of scrambled them a little bit, so make sure you choose carefully. OK, anybody else want to answer? We have most of the answers in the poll, and I've seen good answers in the chat as well. You're doing a really good job. So most of the students have said the answer is 8. That's 3 4ths of you are saying that. So that is correct. That's a really good job. The thing that catches most students on this is they forget to open the treasure chest to start with. So you can do this from inside the smallest boxes, or you can do it from the outside. You're going to get the same answer. So there are five containers. Sorry, if I don't have the correct thing clicked, it does that. There are five containers in a treasure chest. So I have to open one chest. I have to open the chest to get to anything. And then I'm going to open one of those containers because I can't get to any coins if I don't open a container. In each container, there are three boxes. And in each box, there are 10 gold coins. So from one container, I can open three boxes. And I can get 10 gold coins in each. So that gives me 30 coins. But I want 50 coins. So I need two more boxes. To get those two more boxes, I have to open one more container. Because I already used up the three. So this is 1, 2, 5, 6, 7, 8. Eight locks that I have to open. You can do this, like I said, from the top, counting the chests, the two containers, and the boxes. Or you can say, I need five boxes to get five boxes. That would take two containers. And I have to open the chest. And I guarantee the one that most students forget is to open the chest to start with. The sum of four natural numbers is 39. The product of two of these numbers is equal to 80. And the product of the other two numbers is also equal to 80. What is the largest of these four numbers? So natural numbers are the same as counting numbers. They are 1, 2, 3, 4, 5. So there are always positive integers, no 0, no negative numbers, no fractions. I've taken the freedom of trying to write this in a little bit of algebraic statements. The sum of four numbers, a, b, c, and d, is 39. The product means multiply two of them and you get 80. And if I multiply the other two, I also get 80. So here's one way to look at it if you were using some algebraic variables. And I'm going to launch the poll now. Anybody else want to put an answer in the chat or into the poll? The poll is not finished yet. I know I say this every time but sometimes I have students who have joined the webinar series just this week. Math kangaroo contests are multiple choice. We do not subtract for wrong answers. So do not leave a blank on a math kangaroo contest since there's no penalty for a wrong guess but a right guess might just get you some extra points. All right, I'm gonna end the poll. Most students have answered, thank you. And most of the students who answered have it correct. The correct answer is 16, with 70% of you answering that. So thank you. Hopefully, for those of you who got it, you understand what you needed to do. If you didn't get it, let's see if this helps you. I know that A times B equals 80 and C times D equals 80. So I need factor pairs, two numbers that when multiplied together equal 80, but then I need the sum of all of those numbers to be 39. So factor pairs, I can do 80 times one. Ah, I'm all wobbly. 80 times one, but of course, 80 is greater than 39, so that's not a good choice. I can do 40 times two, and I have the same problem. How about times three? 80 is not divisible by three, but 80 is divisible by four. I can do 20 times four. So these two options are gonna cross out. I can do 20 times four. I could do 16 times five. It's not divisible by six. It's not divisible by seven. It is divisible by eight, eight times 10. And I think that gets me back to the turnaround fact. So I think these are all the factor pairs that I can use to get 39. I've already eliminated two of them because 80 and 40 are too large. So I just have to see, some of these numbers must line up with these. If I make A be eight and B 10, I do get 80. And then I have eight plus 10 is 18. If I make C 16 and D five, of course I have 80. And if I put plus 16 and plus five here, let's see what that sum is. 10, 18, 24, 34, 39. That works exactly right. If you had tried it with the 20 and the four, you would have correct products, but your sum would be too large. So these are the four numbers that I've used. And it's asking me what is the largest of these four. So the largest of those four is in fact the 16 right here. And it is okay to go ahead and to try 10 and four. It's a logical thing to try. So don't be afraid to make that little error to try the wrong thing and say, okay, too large, try again, no problem. The scale shown in the picture is in balance. There is a 20 gram weight and certain solids, cubes and cylinders. All the solids, the cubes and the cylinders together weigh 500 grams. How much does one cube weigh? So just to go over this, I see three cubes and one cylinder on the left. I see the 20 gram weight, two cubes and two cylinders on the right. We want to know how much does one cube weigh? There is no pole because I wanted you to have a nice big picture of the scale there. When I tried to put it in a pole that got so tiny, it was hard to see the cubes and the cylinders. Okay, hopefully that worked. I did have several students say they needed the time, so that's why I just gave you some more quiet. All right, so if I look at this balance, I notice they even kind of made it a bit easy by putting the two little cubes and the little cylinders together in this group and this group. So if I took both of those off the balance, the balance would still remain level, right? So what I know is that one of the cubes, I'm going to do a square for the cube. One cube is the same mass as the 20 gram weight plus one of the cylinders. Remember I said I don't like to use O's or circles. I'm going to use a cylinder, a dot inside it to show that it's a cylinder. Okay, so I know that that's the relationship between the cubes and the cylinders. And they gave me this about the total weight of all of them. So if I count up how many cubes do I have? I have one, two, three, four, five cubes, five cubes. And how many cylinders? One, two, three cylinders. And all together that is 500. But instead of thinking about cubes, I could do each cylinder and 20 grams. Right, so this is a substitution. That's what we call it in algebra. We're going to substitute this for everywhere I see a cube. So I have five and then it's 20 grams plus a cylinder for every cube. And then I still have three cylinders. So now I, this is the distributive property. Five times 20 is 100 plus five cylinders plus these three cylinders equals 500. I can combine three cylinders and five cylinders, that's eight cylinders. And if I take away 100 from both sides, subtract it, I get 400. So then I know that each cylinder by division, if eight of them is 400, if I divide by eight, each one is 50 grams. And then I can use this first equation that I used. If I put a 50 in for here, then I know that one of the cubes is 70 grams. Now, some of you may have done this with a slightly different substitution, and that is okay as well. You could have said that the one cylinder equals one square minus 20. And that would work as well. Then you would get your cube answer right away without having to do the second little calculation that I did over here. So either substitution will work. Now, I know that that's kind of heavy on the algebra and using those crazy symbols and stuff, but I think you can do it. Jane bought some books. For the first book, she paid half of her money and $1 more. For the second book, she paid half of the remaining money and $2 more. Finally, for the third book, she paid half of the remaining money and $3 more, thus spending all of her money. How much did the three books cost all together? So three all together is the money that she started with because she spent all of her money. So there's a little bit of trickiness in the wording, but if she spends all of her money, that tells you the price of the books. It's the same, her money and the price all together. I do have a poll for this one, but let me give you a few moments to try to get started before I launch that. This type of problem, some students might work as a working backwards problem. So you might start with the ending and then try to go to the beginning. This is also a problem where you might just take the answer choices and test them. So if all else fails, try one of the numbers and see if it works. That might tell you if you need to go higher or lower. Okay, we have some fun, still have some fun problems ahead of us. So let's see if we can end the poll and go through this one and that way we'll get to finish all the cool problems today. So we have quite a few of the answers, you know, A through, I guess 100, everyone decided 100 doesn't work, but all the other choices were selected with 34 being the most popular answer. So let's take a look. I'm gonna do the working backwards method and see what happens. So we know for that, for the third book, she paid the remaining money and three, she paid half of the remaining money and $3 more. So that's a really big clue. If she paid half of her money and then $3, then we know that $3 is equal to the other half, isn't it? Right? Because if the amount of money she has, if we have half and three, then the whole thing for the third book, the third book must have cost $6. Now let's take a look at the second book. The second book, we know that she had half of her money and $2 more. So if we add $2 to the six, if we add $2 back, that gives us eight, and then we have to double it. So that is $16 that she had. And then if we go back for the first book, we know that she spent half and $1 more. So if we do... Yeah. I'm having trouble following my own notes. So if we know that at this point she had $16, if we add one back, that's $17, and then we have to double it and we get $34 at the start. Now let's try it forwards just to make sure. If we start with $34, then half of that would be 17. And we know it costs one more dollar than that, right? So that's $18 for the first book. And then there are $16 left over. If we take half plus two, that's $10 for the second book, which means only $6 left over for the third book. So backwards or forwards, it works if we use 34. You can set this up with algebraic variables as well. You could use N or X as your initial amount, and you'd be subtracting and dividing by two several times. And eventually you would be able to set that up with an algebraic equation. But I think that for many fifth graders, that's gonna be pretty tricky. Maybe some of my sixth graders have more algebraic experience and would be able to do it that way. Okay. I think Jacob, who's been quiet for so long, wanted to lead you through the dragons problem. Jacob, we don't hear you. Can you check your mic? I'm sorry, can you hear me now? Yes, thank you. Sorry about that. So red and green dragons lived in a cave. Each red dragon has six heads, eight legs, and two tails, but each green dragon has eight heads, six legs, and four tails. If there were 44 tails altogether and there were six fewer green legs and red heads, how many red dragons lived in the cave? So I'll give you guys some time to think about this question, and then we'll go over it. No, I don't have a poll for this problem, but there will be for the next one. The polls take some time away from solving problems sometimes. All right, so hopefully that was enough time, but I'm going to go over it. So I think the best way to envision this problem is to kind of have a table. So Sorry, can I Okay, so the best way I think we can envision this is a table. So we can write Like this. So we have red dragons here and green dragons here and then we have Heads, tails, and legs. So if you have, let's say, our And legs. So if you have, let's say, our red dragons and let's say g green dragons, then we have 6R heads, 2R tails, red tails, and 8R red legs. And for the green dragons, we have 8G heads. We have 6G legs and 4G tails. And since we have 44 tails altogether. That means that we have this number, this, this column should total to 44. So that means that 2R plus 4G equals 44 And we also know that there are six fewer green legs than red heads. So if there are six fewer green legs than red heads. That means that we have the equation. 6G Plus 6 is equal to 6R. Right, so And we want to find the value of R now. So now we have two equations and we can just solve these two equations. So when we divide the second equation by 6, we get G plus 1 equals R. We can substitute R into the first equation. So then we get 2G plus 2 plus 4G equals 42. So then G, so then 6G equals, oh, sorry. This 42 should be a 44. So Then we can solve for 6G is 42. So that means that G equals 7. Remember that we want to find for, we want to find R. So since G is 7, then we can use this equation here. So then we get R equals 7 plus 1, which is just 8. So there are eight red dragons. So answer should be C. Yeah, for students who have not done a lot of algebra, Jacob is relying on the fact that if we divide everything, both sides by 6, then we can get rid of like 6 divided by 6 is 1, 6 divided by 6 is 1, 6 divided by 6 is 1. And then the substitution I showed you earlier how we could do a substitution. All right, there is one more problem today that I think we're going to get to do. And I think that Jacob wanted to lead this one as well. Yeah, so Magnus has to play 15 games in a chess tournament. At a certain point during the tournament, he has won half the games he has played. He lost one third of the games he has played, and two have ended in a draw. So we want to figure out how many games does Magnus still have to play. So I'll let you guys think about this for a little bit, and then we'll go over it. For anybody who's not familiar with the draw means it was a tie. So no one win and no one lost. Hey Jacob, I think we have the majority of them answering the poll. Anyone else want to put your your best guess into the poll before we close it and share and Jacob shows you a good way to solve this? All right, that's most of them. Here are the results. Look at that, Jacob. We thought it was tricky, but almost 80 percent of them have it correct. They're doing really well today. Yeah, definitely. So, okay. So, for this question, what we're going to do is we're going to try to set a variable, and that variable is going to represent the total amount of games that Magnus has played so far. So, let's call this variable G. So, G is the amount of games that Magnus has played so far, and so that means that, and remember the amount of games that we have so far, each game ends up as either a win, a loss, or a tie, if you know chess. So, we can say that there are G over 2 wins, G over 3 losses, and 2 draws, and since these are all the different outcomes that we can end up with, that means that this has a total to G. So, now we just have a simple equation, or not really simple, but it looks a little bit simple. We have an equation that we can try to solve, right? And so, when we have G over 3 plus G over 2, that comes out to being 5G over 6 plus 2, and then we can subtract the 5G over 6 onto the left side, so we get 1 over 6 times G equals 2, so G equals 12. But remember, G represents the amount of games that we have played so far, and you want to find the amount of games that Magnus still has to play, and Magnus played a total of 15 games, so our answer should be 15 minus G, which is 3, so our answer should be B. That's a really nice solution, Jacob. For our students who haven't done as much algebra here, I like the way you set it up here on the left-hand side, that we know the games that he's played can be divisible by 2 and divisible by 3. So, if it's divisible by 2 and by 3, we know it's a multiple of 6, so we can start to look, how can we have, if it's a multiple of 6 and we have 2 draws, it's not going to quite work, so you're going to look for multiples of 6, that also gives you 12 as a multiple of 6. So, you can do that as well, looking at multiples. Okay, so hopefully that gives you another option. There's a couple of different ways to set up the algebra, but I think Jacob did it in a really nicely effective way. All right, so remember, in algebraic thinking, we had some patterns. We were doing representative relationships, trying to write things down, sometimes with variables helps, and you have to use some reasoning as well. So, that last one where I said, okay, it has to be multiples of 6. We had the other problems where we had, what are we doing with these chests? How do we open the chest? How much do we have to do? We had problems that required the sum of all the numbers was 39, but we had factors of 80 we had to add. We had the balance. Remember, if you have the same things on the two sides of the scale, you can pretend that you take off the matching blocks and that makes your scale easier. Okay, so that is where we're going to have to end today's lesson. I'm glad that you are participating so well in the webinar. I'm really happy to be able to chat with some of you and give you some smiley faces and happy faces. I know that Jacob is doing the same. So, I hope you'll come back next week and we will solve more math kangaroo problems. Remember, we have access codes in your registration for this course and also in your registration for the contest where you can get some practice contests. Contests from past years are going to be pretty good ideas about what you might expect to see this year. We tend to use similar types of thinking and tools, so if you can solve a past contest, that's going to prepare you for solving the contest coming up in March. All right, everybody, thank you very much. Bye-bye.

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