Hello and welcome to the Math Kangaroo 2017 levels 1 and 2 solutions video series. First of all I'd like to congratulate you on participating in the Math Kangaroo. I used to take the exams myself every year from my first grade in elementary school to my senior year in high school, so I took a lot of the kangaroo exams and I am definitely familiar with how challenging they can be, so you should really be proud of yourself for taking on such a great mathematical task. Now the purpose of this series is to show you a method to get to the correct answer for each individual question on the exam. Now keep in mind some of the methods that you see in the videos might not be the only methods to get the correct answers, so sometimes you'll get the same correct answer but by doing the problem in a completely different way. In this case sometimes your method is equally good, but there are also situations where your method might only work for that one particular problem, or maybe you just kind of got lucky and the method doesn't actually make sense, but you somehow got the correct answer anyway just by luck. If you have any questions on whether your method makes sense compared to the one shown in the video, please feel free to email me at thomas at mathkangaroo.org, the email listed below, or if you have any other questions about the exam please feel free to get in contact with me. Now before you actually move on to the videos, I'd like to offer a couple of pieces of advice. First of all, make sure that before you actually watch a video for a problem that you try solving that problem on your own first, because very often you can just look at the video for a problem without trying to solve it and you'll see how it's solved and you'll think, oh that's super easy I could totally do that, and then you won't have actually learned anything, and if you get a similar problem in the future you might not actually be able to do it, because you didn't struggle with the problem beforehand. It's very important that you try doing the problem first on your own, so that way you understand why it's difficult, why it's tricky, and what you really need to be thinking about, and only if you can't get it after actually trying should you use the video, because then the video will help you figure out what your mistake was. The other piece of advice I have to offer is to make sure that you write down as much of your work as possible, because a lot of times when you're good at math you try doing a whole lot of the math in your head and keeping everything in your head, and that's just a way to make silly mistakes. Everyone who's good at math makes that silly mistake at some point in their lives. Try to avoid it. Plus it's easier to check where your mistake is if you write down everything that you're thinking, so there are only benefits to writing down your work. It actually also goes slightly faster than doing mental math a lot of the time, because you don't have to remind yourself, oh what was this number, oh what was this number, because you wrote them down somewhere, so you can just look and that'll be much faster. Now without further ado, I hope you find all of the videos useful and helpful, and I hope you have lots of fun taking this exam. So good luck, have fun, and thanks for watching. Problem number one, who caught the fish? So for this problem, the quickest way to do it, of course you can check each individual person and try going along their lines and getting to the end of their hooks, but the easiest way is actually to just work backwards from the fish, and if you trace the line going from the fish and follow the path, it should look something like this, and you see that the line ends up going back to David, so our final answer is D. Problem number two, in the picture there are stars with five points, stars with six points, and stars with seven points. How many stars that have only five points are there? So we can go ahead and just check which stars have five points on them, and the ones that are circled are the only ones that have five, so we see that four of them are circled, and therefore our final answer is C, four. Problem number three, the entire pie seen in the picture is divided among several children. Each child receives a piece of pie with three cherries on top. How many children are there? So because the cherries are set up on the pie in a pretty symmetrical way, we can very easily just cut the pie up so there are three cherries on top of each slice, and if we do so it would look something like this, and we see that that actually cuts the pie into four slices, so there are four children, and therefore our final answer is B. Problem number four, into how many parts do the scissors cut the rope in the picture? So if I actually got this problem on the exam, I would go ahead and just write a little mark on each individual string of rope, and then cut the number of marks. So I would mark one roughly here, roughly here, anywhere on this long end, maybe somewhere just right here, one here, and one here, but because this is sketchbook, I can go ahead and color in each of the individual marks, and that makes it a lot more visible. So all we have to do is count how many different pieces there are, and we have one, two, three, this is a long one, four, and then five pieces, so our final answer is going to be A. Problem number five, Ellen wants to decorate the butterfly with these stickers, which butterfly can she make? So we see that there are two blue circles, two pink circles, and two black ovals, so for instance we can't use this butterfly because there's two black dots, and there are no stickers like that, and each of the remaining butterflies also have sticker setups that just aren't possible given the stickers that she has, so this one has four black stickers when there are only two black stickers, this one has four blues when there are only two blues to work with, and this one has two small blue circle stickers, and these are clearly much bigger, so that just leaves us with A, and we see we have the two pink ones which match with these, we have the two blue ones which match with these, and then we have two black ovals, so that means our final answer is A. Problem number six, how many bricks like this are missing from the igloo? So we can go ahead and try just filling in the white space with bricks, but we need to find out how many bricks can fit into a certain space. Well one way to think about it is that the line from the gap between two bricks in the row above or below should be right in the middle of the brick in this row, so for instance you know we have a brick here which means there's a midpoint right here which hits the middle of this brick. Now with this in mind we can go ahead and fill in the bricks, we have just one brick here because there's one line going down the middle, and then one line going down the middle, and then so in the second row we're going to have two bricks because there's going to be the one line from the gap in between them cutting this brick in half, so that means that we can fit exactly two bricks here. And then here we can fit three bricks just following the same thing, we have a middle point and a middle point, we have a middle point and a middle point for this brick, and we have a middle point and a middle point for this brick since there are two bricks up here that creates a middle point here. And if we were to go ahead and count up all of these bricks that would be six total, so our final answer is A. Problem number seven, in the drawing we see a string with four beads, which of the strings below is the same string? So this problem is pretty simple to realize once you notice something that four of the answer choices have in common, four of them are actually all the same bead, or the same string I should say, and since you can't have multiple correct answers in multiple choice we can eliminate all four of those answers since they're just saying the same thing. For example, here we have a black, a white, a white, and a black, but if we just go ahead and rotate this black bead around until it goes to the end here, then it'll be two whites followed by two blacks, which is answer choice D. So these two answer choices are actually the same thing, the beads are just rotated around, and looking at this we don't know how the beads will be rotated when we actually unfold the necklace, so it doesn't make sense to ask oh is it going to look like this when they unfold, or is it going to look like this when they unfold, because you just don't know from what side you're going to be looking at the necklace from. So in that case E is the only unique one because it's the only one where you can't rotate the bead so that it's two blacks and two whites, and in fact if you try following this string around we see we have a white bead, and then we keep on going, and we hit a black bead, and then we keep on going, and we're on the back string now, and we hit a white bead, and then we hit a black bead, so it is white black white black, so it is alternating just to make sure. So we can eliminate the first four answer choices, and that leaves us with answer choice E, final answer. Problem number eight, four of the numbers 1 3 4 5 and 7 are used, one in each square, so that the equality is correct. Which of the numbers is not used? So in this case you don't really have any sort of shortcuts that you can take, you just have to try pairing numbers. A good idea is generally to try taking two of the outer numbers, like 1 and 7, and try balancing them with two of the inner numbers closer to the middle, because if you add two small numbers and two big numbers you're gonna get a big sum and a small sum, so probably your answer isn't gonna look something like that. So that's why you want to pair like the numbers that are on the extremes, the really small with the really big, and the kind of small with the kind of big. So a natural guess is to try adding 1 and 7 and 3 and 5, and in both of those cases they add up to 8, so the equality is actually true. And that means the one number we don't use is 4, so our final answer is C. Problem number nine, in the country of jewelries you can trade three sapphires for one ruby, picture one. For one sapphire you can get two flowers, picture two. How many flowers can you get for two rubies? So we start off with two rubies, and we see that for one ruby we get three sapphires, so this ruby will give us three sapphires, and one ruby will give us three sapphires, and we're just drawing all this out to make it as clear as possible. Now we have six sapphires here, which means that we should be able to get twice as many flowers based on this picture. So really all we're doing is 2 times 6 equals 12, or if we were to write all of it out it would look like this. For this sapphire we get two flowers, for this sapphire we get two flowers, and so on. So drawing it out is really not necessary at all. The quicker way is to do multiplication if you know it, but if you don't then drawing it out is kind of the only way that makes sense. But if you know multiplication it's as simple as, oh, one of these is equal to three of these, well then two rubies times three sapphires per ruby is equal to six, and then six sapphires times two flowers per sapphire is equal to twelve. In either case you get a final answer of D. Problem number 10. At one point Jim and Ben sat on the Ferris wheel as shown in the picture to the left. The Ferris wheel turned, moving Ben to the place where Jim was previously. At that moment where was Jim? So first let's see how many moves it takes for Ben to get to Jim's location. It shouldn't matter whether we do this problem going clockwise or counterclockwise, since if Ben moves counterclockwise then Jim will move counterclockwise, and if Ben moves clockwise then Jim will move clockwise by the same amount of moves. So we can pick either direction. So let's do counterclockwise because it's a little bit quicker. We can see Ben making one, two moves over to get to Jim's location. So Jim would move one, two chairs over to this location. Alternatively if we would go clockwise it would be one, two, three, four, and so Jim would go one, two, three, four, which is actually the exact same spot as when going counterclockwise. So it works out either way and we see that Jim ends up here, so our final answer is going to be A. Problem number 11. How many triangles are there in the picture on the right? So we have this picture blown up a little bit to make it a little easier to see. First things first, let's go ahead and go ahead and highlight the easy triangles to count in red. We have one, two, three, four, four triangles here, five, six, seven. Now these are the obvious small triangles. There are a few bigger ones that we have to consider. So for instance these two highlighted in blue, this big triangle on the roof, that's going to be, let's see, we had one, two, three, four, five, six, seven, so this is going to be eight, and this one is going to be nine, and then we have number ten highlighted like so, and then we have number eleven, this one right here, and that covers all of the triangles. So when we count all of them together we have eleven, so our final answer is D. Problem number 12. Alfred was rotating a shape. The first three turns are shown in the picture. He did six turns in total. How does the shape look at the end? So when we look at these turns we see that they turn 90 degrees each time. One, two, three, each one is 90 degrees, so the fourth turn is actually just going to bring it back to its original position. So after the fourth turn there's going to be two more turns left, the fifth one and the sixth one, and we know after the fourth turn we have this situation here. So then this will bring us to the fifth term, and this will bring us to the sixth. So this is the position that we actually end up with for the shape, and that looks like it matches answer choice E, so that's going to be our final answer. Problem number 13. In which picture are there twice as many apples as carrots and twice as many carrots as pears? So from this problem, if we want twice as many carrots as pears and we want even more apples, that means most of the food should be apples, most of the fruits and vegetables, and the least should be pears. So looking around, the most obvious one to start looking at is the one with only one pear and with mostly apples. We have one pear and two times that is two carrots, and two times that is four apples. So this is actually the correct answer. Other reasonable ones to look at might have been this one, since there's only a couple of pears, but there's the same number of carrots as apples, so that one doesn't match up. And none of the other answer choices match up either. But the key thing is to just look for the one with very few pears and very many apples. And so we see that answer choice D satisfies the conditions in the problem. So that is our final answer. Problem number 14. Brian and William are standing in line. Brian knows that there are seven people in front of him. William knows that there's a total of 11 people in the line. If Brian is just in front of William, how many of the people in the line are behind William? So for this, I would just draw a picture as quickly as possible. And it's gonna look something like this. Assuming this is the front of the line and this is the back of the line, we have Brian, we have William. And Brian has seven people in front of him. One, two, three, four, five, six, seven. And then there's Brian and William who are eight and nine. So we just count backwards for the rest of them. And that's 10 and 11. And we see that there are two people behind William. So that leaves us with answer choice A. Problem number 15. The ancient Romans used Roman numerals. We still use them today. Here are some examples. I equals one, I I equals two, V equals five, I X equals nine, X equals 10, X I equals 11, X X equals 20. This year we celebrate math kangaroo number XX. What year was math kangaroo number XV? So let's note first that this year is 2017, since this is the 2017 exam. And XX is equal to 20. Now XV is equal to 10 plus five or 15. So we wanna go five years back since we're currently celebrating 20 and they want us to go back to 15. So 20 minus 15 is five. We wanna go five years back. And five years back from 2017 is 2017 minus five, which is equal to 2012. So our final answer is going to be C. Problem number 16. Liz is making crowns like this using the following stickers. There are two types of sticker sheets shown on the right. If she wants to make three crowns, what is the smallest number of sheets that she will use? So if she wants to make three crowns, she's gonna need three of the crosses or plus signs or whatever you wanna call them and three of these long lines. So at minimum, she's going to need two of these upper sheets. She's also going to need one, two, three, four times three, which is 12 circular stickers. So giving two of these sheets so that we have enough lines, we would then have one, two plus signs. We need a third one. So we're gonna need at least one more sticker, one more of these sticker sheets. And then we need to make sure we have enough circles. So one, two, three, four, five, six, seven. Okay, that's not enough. So that's why we need this extra sticker sheet. Eight, nine, 10, 11, 12. So we have 12 circles, three pluses and three lines. And to get that total, we need four sticker sheets total. So our final answer is B. Problem number 17. In the table, the correct additions were performed in the squares according to the pattern shown. What number should replace the question mark? So looking at the square, we see that 10 plus some number over here is equal to 14. Well, then this number has to be four because 10 plus four is equal to 14, which means that we're going to be adding seven and four to get this question mark. And seven plus four is equal to 11. So our final answer is going to be B. Problem number 18. In old McDonald's barn, there's one horse, two cows and three pigs. How many more cows does the barn need so that the number of all the animals is twice the number of cows? So we'll start by just listing out all the animals that are already there and then adding cows. So we have one, two, three, four, five, six animals. And if we add two cows, then there would be eight animals and four cows. So this satisfies the problem. If we were to just add one cow, we would have seven animals, which is an odd number. So it can't be twice the number of cows. But once we add a second one here, then this is the smallest number of cows we can add to satisfy the condition given in the problem. So our final answer is gonna be C. Problem number 19. Sepper has two paper cutouts. He colored one side of each cutout like this. Which shape can you make using both pieces? So looking at this shape, the one cutout that really seems to make sense, most obviously when you look at it, a lot of these just have a bunch of points sticking out for apparently no reason. These two in the center, C and D, look like the ones that kind of look like two of these are just fitting together sideways. So we should probably look at those. And in fact, if we look at answer choice C, we see that we have one, two, three, four. This shape right here is just this sticker rotated 90°. And this shape right here, one, two, three, four, is just an exact copy of it. So we just have two exact copies of this shape clicked right together. So this is exactly what the problem is asking for. And therefore our final answer is going to be C. Notice that D isn't really an answer because if this would be one of the shapes, just like in the previous case, well then this one has the wrong square shaded in with black. So this is not a correct answer. And these other ones just look completely strange. This one has the wrong number of shaded squares. So C is the only one that actually satisfies the problem. Problem number 20. A certain kangaroo makes 10 jumps in one minute and then rests three minutes after. Then he again makes 10 jumps in a minute and then rests three minutes and so on. What is the least amount of minutes he needs to make 30 jumps? So basically to make 30 jumps, he needs to do a set of 10 jumps three times. So he's gonna need three one minute periods to make those jumps. And he's also going to need a couple of resting periods because he's going to jump for one minute, then rest for three minutes, then jump for another minute, then rest for another three minutes. And then he's gonna do one last minute of jumping. So it should look something like this. One minute of jumping, three minutes of rest. One minute of jumping, three minutes of rest. And then one more minute of jumping. If you add all of these ones and threes together, you get nine minutes total. So our final answer is E. Problem number 21. Which stamp was used to get the picture shown to the right? So if you've ever worked with a stamp, you'll know that stamps have a mirror image on them of whatever image you actually want to put down on the paper. And if you wanna see why this is true, I guess one way to think about it is you're pressing the stamp up against the paper kind of like a mirror against whatever image you're trying to create. That's a very poor way to phrase it. So I guess the easiest way to learn about this would be to pick up an actual stamp and look at how it actually presses the image onto the paper. And it should be a little bit more obvious why it's always a mirror image. But with that in mind, we're basically just looking to flip everything. So this tower on the left side should be on the right side. This face of the building should be on the left side and then this door and windows should switch places. So if we look at all of these answer choices, we see, okay, this one and this one both have towers on the right side and faces on the left side. And these are the only two answer choices that have those two qualities. So now we need to look at the door and the windows. They're flipped differently. Well, the original image has the door on the left side, which means the door should be on the right side for the stamp. So that means our final answer is going to be this one, which is answer choice E. Problem number 22. Each of the four keys fits only one of the four locks and the numbers on the keys refer to the letters on the locks. What is written on the last lock? So we wanna try figuring out which number corresponds to which letter. And we'll start with ADA, A-D-A, because this is the only one that has two repeating letters and these two repeating letters will help us basically look for other keys with a similar structure, with two repeating numbers. And looking around, we have 141 and 717. So that means A either corresponds to one or seven and D corresponds to either four or one. Now, let's suppose that 717 is correct. We have to consider both options. In that case, A is seven, right? So that means we're going to have to have one of the keys with a seven at the end and one of the keys with a seven in the middle. And it looks like we do have these two keys, so that works. Now, we also need the D to be a one. So where we have a seven at the end, we would need a one at the front for this one. So in other words, we need one, probably four, and then seven, 147 would be this number because the D is the one and the A is the seven, so 147. But we don't have that key anywhere here, which means that our initial assumption must have been wrong. It doesn't work if we assume that 717 works because we would have a one here, a seven here, and there's no key that satisfies that. So that means that the key should be 141. That means A is one, D is four. Well, in that case, it works a little bit nicer because here we have a four, here we have a one, and oh, look, we have a key with a four and a one, and in the middle is a seven, which can correspond to the G. So if G is seven and D is four and A is one, then this would be 417. Oh, look at that, it matches this key perfectly. Look at how nicely all of this matches up. In fact, we see that this key matches up to this lock, this key matches up to this one, this key matches up to this one, and so this key is the only one left and it matches up to this one, and we proved that the one corresponds to an A and the seven corresponds to a G. So this should be G-A-G, which means our final answer is D. Problem number 23. Anne put six toys in one shelf with six cubbies like this. When you look at the shelf, you see that the teddy bear is between the blue bear and the duck, I guess. The blue bear is right above one of the trains, and the doll is to the left of the cart and to the right of the train. I really was not sure how I was supposed to phrase some of these things, but let's try taking a look at the information a little bit. In particular, we wanna figure out what's in this top row, and we see that the blue bear is right above something. It doesn't matter what the something is, but the blue bear is right above something, which means that the blue bear is in the top row. Well, that means that the teddy bear is in the top row because it's between the blue bear and the duck, and in fact, that means the duck or chicken or whatever this is, the bird, is also in the top because these three guys all have to be in the same row, and if this one is in the top row, then all three of them must be in the top row. So now, how to figure out what the order is. We know it's definitely not gonna be the teddy bear since the teddy bear has to be in the middle. So is the blue bear here or here, and is the bird here or here? Well, now let's take a look at the third piece of information. The doll is to the left of this cart. In other words, the cart must be here, the doll must be here since the doll has to be in the lower level because the upper level's all taken up. So the doll is going to be in the center and it's to the left of the cart and to the right of the train since going from the train, you would go right to get here, and going from the cart, you would get left to get here. So this is where the cart is, this is where the train is, and again, the blue bear's right above the train. If the train is here, that means the blue bear is here, and that means we can fill in the rest of the top row pretty easily, and so that means that the bird is going to be in the top right cubby. So our final answer is going to be B. Problem number 24. In a stack of three cards with holes, the top of each card is white and the bottom is gray. Basil threaded these cards on a rope. See the picture on the right. Which of the following can he obtain without untying the rope? So the way to think about this is if you go ahead and lay down the rope and the cards on a table so they're flat, like-colored faces should have the loop going out in the same direction. So for instance, if you have three white faces, face up and visible, then the loop should be going through either to the left for all of them or to the right for all of them. So if we look at answer choice D, this is not the case. This one goes out to the left, this one goes out to the left and this one goes out to the right, as you can see. So really, that's what we're looking for. We wanna see that if two faces are of the same color, then the loop goes out the same direction. And if it's of the opposite color, then it goes out in the opposite direction. If you look at all of these answer choices, you see that the only one that satisfies this is E because for the two dark ones, it goes out to the left and for the white one, it goes out to the right. So E is the only one that satisfies the condition that is given in the problem. So that's going to be our final answer. And if it's unclear why the loop should be facing out the same way for same colored sides, simplest way to understand this is to actually just make a physical version of this. Just get three index cards and get a piece of string or something and go ahead and mimic exactly what you saw in the problem. And you'll see it is impossible to flip the cards without changing the orientation of the string. So again, our final answer is E.

Math Kangaroo 2017

levels 1 and 2

problem-solving

mathematical reasoning

exam solutions

arithmetic puzzles

spatial reasoning