Hi, this is the Math Kangaroo Solutions Video Library. This set of videos is for levels or grades one and two from the year 2013. My name is Agata. George and I prepared these solutions. If you do have any questions, please email me at agata at mathkangaroo.org. To use these solutions, here are a few hints. First of all, have some paper and a pencil ready because we would like you to try to solve the problem as well. So if you listen to the problem and you haven't done this one before, try solving it by yourself first. Then as you're listening to the video, if you hear a hint or something that helps you understand the problem better, pause the video and finish the problem yourself. Then check the answer. We really hope you enjoy this set of solutions and that you will learn a lot of math through it. Problem number one. Which digits are missing on the right? A, three and five. B, four and eight. C, two and zero. D, six and nine. E, seven and one. We're going to look at the answers and see which one is missing. So we'll start with three and five. So here's a three. And here's a five. So that one we have. Next we'll look at four and eight. Here's a four. And here's an eight. So we also have that one. Now we have two and zero. And we have a two. And we have a zero. So we have two and zero. Now six and nine, we do not have. And seven and one. Here is a seven. And here's a one. So we have that one. So six and nine is missing. Sorry, answer is D. Six and nine. Problem number two. There are 12 books on the shelf and four children in a room. How many books will be left on the shelf if each child takes one book? A, 12, B, eight, C, four, D, two or E, zero. In order to solve this problem, we are going to do 12 minus four. The reason we are doing this is because if there are four children and they each take one book, that is four books taken. So 12 minus four is eight. Sorry, answer is B, eight. Problem number three. Which of the dresses has less than seven dots but more than five dots? A, B, C, D or E? The only number that is less than seven but more than five is six. So we are going to count the dots on each dress to see which one has six dots. So A has one, two, three, four, five, six. One, two, three, four, five, six. So A has six dots. So B has one, two, three, four dots. So this could not be our answer. C has one, two, three, four, five, six, seven, eight. So this could also not be our answer because it is too much. D is one, two, three, four, five, six. One, two, three, four, five. This could not be our answer because it is too little. And then we have E, which is one, two, three, four, five, six and seven. So this could not be our answer because it is too much. Sorry, answer is A. Problem number four. There are white, gray and black kangaroos. Which picture has more black kangaroos than white kangaroos? Picture A, picture B, picture C, picture D or picture E. We are going to look at each picture and figure out which one has more black kangaroos than white kangaroos. We're also gonna count the kangaroos on each picture. Starting with A, these are white kangaroos, these are gray kangaroos and these are black kangaroos. We're going to see how many of each we have. So we have one black. We are going to ignore the gray ones because they are not needed in this problem. And then there are two white ones. So this cannot be our answer because it has more white kangaroos than black kangaroos. Then B has two black kangaroos and two white kangaroos. And this cannot be our answer because it is even and there has to be more black kangaroos than white kangaroos. So here, there is one black kangaroo and three white kangaroos. This cannot be our answer because there are three white and only one black. Then D, we have three black kangaroos and two white kangaroos. This should be our answer because there are more black kangaroos than white kangaroos. And then E has three black kangaroos and three white kangaroos. And this cannot be our answer because they are even and the black is supposed to outnumber the white. So our answer is D. Problem number five. How many more bricks are in the larger stack? A, four, B, five, C, six, D, seven, E, 10. Now I'm going to count the number of bricks in each pile. So the smaller pile has one, two, three, four, five, six, seven, eight, nine, 10. This one has 10 bricks. The larger stack has one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15. So the taller stack has 15 bricks. Now 15 minus 10 is five. The larger stack has five more bricks than the smaller one. So our answer is B, five. Problem number six. The picture shows a path made of square tiles. How many tiles fit in the area inside? A, five, B, six, C, seven, D, eight, E, nine. So I'm going to draw them in first and then count them. All right, so we have 1, 2, 3, 4, 5, 6, 7, 8, 9. So our answer is E, nine tiles. Problem number seven. Lana cuts a big piece out of a cake. Which one? So if you look here, we have the before picture and the after picture. So the piece that is cut will look like this. So there's three nuts, a star, and the outline of some frosting and the piece of an arm. So if we look at A, it has three nuts and a star and a piece of an arm but has no frosting outline. B, it has three nuts, the frosting and an arm, but a heart instead of the star. C has three nuts, an arm, the outline of frosting, and a star. So this could be our answer. D has only two nuts, and a star, and an arm, and the outline of some frosting. But it cannot be our answer because it only has two nuts when the slice you're looking for has three. And E has three nuts, outline of frosting, and an arm, but it has no star in the corner. So that means our answer is C. Problem number eight. Anne has earrings, Barb gave Eve a necklace, Jim has a hat, and Bob has glasses. Who is Barb? So we are going to look at all the different characters and see who has what. The earrings are for Anne, and this girl must be Anne because she has earrings. This necklace is for Eve, and this girl down here is wearing a necklace, so that must be Eve. Jim has a hat, this boy must be Jim, and Bob has glasses, so this must be Bob. So because Barb is the person who does not have anything from these four items, the character on D has no items from these four, so that must be our answer, D. D is Barb. Problem number nine. Father gives five apples to each of his three children. Anna gives three apples to Sanya, and then Sanya gives half of her apples to Mikhail. How many apples does Mikhail have now? How many apples does Mikhail have now? A, four, B, five, C, seven, D, eight, or E, nine. You can solve this problem by using numbers or by drawing pictures. There are three children. There's Anna, Sanya, and Mikhail. Anna has five apples, Sanya has five apples, and Mikhail has five apples. Let's draw them in. One, two, three, four, five. Now we're ready to look at who gives how many apples to whom. Anna gives three apples to Sanya. So instead of five apples, Anna will have two, and Sanya will have eight. Take these three apples and move them. The second step is that Sanya gives half of her apples to Mikhail. Now Sanya has eight apples at this point, and half of eight is four. So she'll take four of her apples, leaving her with four, and give them to Mikhail. Mikhail will now have five plus four, that's nine apples. That was the last step. So Mikhail has nine apples. Answer is E, nine. Problem number 10. George has two cats of the same weight. What is the weight of one cat if George weighs 30 kilograms? A, one kilogram. B, two kilograms. C, three kilograms. D, four kilograms. Or E, five kilograms. If we look on the scale that George and the cats are standing on, George weighs 30 kilograms. And this is 36. So 36 minus 30 is six. So the weight of two cats is six kilograms. Now since there are two cats, we are going to divide six by two. So six divided by two is three. So our answer is C. Three kilograms. Which kind of square appears the most often? A, B, C, D, or E. All are equal. We are going to look at each square individually and see which one has the most. Starting with the gray square, we have one, two, three, four, five, six, seven of the gray square. We have seven of those. Moving on to this one, we have one, two, three, four, five, six. Six of those. Moving on to this one, we have one, two, three, four, five, six of those. Then this last one, we have one, two, three, four, five, six of those. So A, seven, and the other three have six. So our answer is A with the gray square. Problem number 12. How many cats can the rabbit eat walking freely in this maze? A, seven, B, eight, C, nine, D, 15, or E, 16. Here's the rabbit, and he can walk anywhere in the maze where there is not a wall stopping him. So we're going to try to go as far as we can in all directions and count how many carrots we can get to. If the rabbit goes here, there are no carrots, and the wall here stops him from going over to the other section. The wall here stops him from going over to the other section. He can go up here, he can go in this cubby, pick up this carrot. If he goes up here, he can go to the carrot here. That's our second one. There are walls around this carrot, so he won't be able to pick it up. He can go in here, get this carrot. Here there's more walls, so have to back up. So far he has three carrots. Going this direction, there's a fourth carrot. And keep going through here. If he goes in here, there's nothing. In here, there's the fifth carrot. And this is as far as he can go this way. So he can back up and go over here. Sixth carrot. Keep going. Here's another one. That's seven. I have to go the other direction. Eight. And down here, there's no more carrots. So it looks like he got eight carrots. Let me just double check my counting. One, two, three, four, five, six, seven, eight. The answer is B. The rabbit can eat eight carrots. Problem number 13. Cat and mouse are moving to the right. When mouse jumps one tile, cat jumps two tiles at the same time. On which tile does cat catch mouse? A, one, B, two, C, three, D, four, or E, five. One way to solve this problem is to track each move of the cat and the mouse and then figure out where they meet. We need to remember that they're moving at the same time. In the first move, the cat moves from here to here. This is shown already by the arrows. I'm like, the cat's here. The mouse moves from here to here. In the next move, the cat will move from this space, jumping over one so that the jump covers two spaces and will land here. The mouse will move just one space ahead to this one. Then the cat will move from this space to the space where the mouse had started. The mouse will move from the space where it is to the space with the one. Next, the cat will move. I'll mark it on the top so it's easier to see. The cat moves to space, goes over here. The mouse was on one, now it's on two. In the next move, the cat will move over two, land on two. But the mouse has time to go to the space with the three. Then the cat moves from two to four. And the mouse moves from three to four. So this is where they finally meet at four. So the cat will catch the mouse on the tile with the four. Answer is D, four. Problem number 14. Peter built a podium, as in the picture. How many cubes did he use? A, 12. B, 18. C, 19. D, 22. Or E, 24. So we are going to count how many blocks there are in this structure. And then we are going to multiply it by two since there are two layers. One, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve. So twelve times two is 24. So that means our answer is E, 24 because the structure consists of 24 bricks. So our answer is E, 24. Problem number 15. There are five children in a family. Kitty is two years older than Betty, but two years younger than Danny. Teddy is three years older than Annie. Betty and Annie are twins. Who of them is the eldest or the oldest? A, Annie. B, Betty. C, Danny. D, Kitty. E, Teddy. With this kind of problem, I find it helpful to draw some kind of a diagram or a picture. You can start with the first statement. Kitty is two years older than Betty. So if we put the older on this side and younger on this side, we can put Kitty here and Betty here. And there's a two-year difference between them. We also know that Kitty is two years younger than Danny. That means that Danny is two years older than Kitty. So we'll put him here. The difference of two years. Years. Next, we know that Teddy is three years older than Annie. We don't have Annie on here yet, so we'll come back to this one. Finally, Betty and Annie are twins. So Annie will go right here. And we can come back to the statement about Teddy. So far, we have Kitty, who is two years older than Betty and Annie. And Danny, who is two plus two, or four years older than Betty and Annie. So Teddy, who is three years older than Annie, and also three years older than Betty, will be over here. So we have Teddy. This shows us that Danny is the oldest of the five children. He's one year older than Teddy, three years older than Kitty. And four years older than Betty and Annie. The answer is C, Danny. Problem number 16. Which is next? A, B, C, D, or E? We are given a pattern of the letter A being rotated. And the problem is asking us what's going to come next in this pattern. The first thing to do is to figure out how the letter A is being rotated. It starts upright, and it is tilted. It is turned clockwise. Each time by half of a quarter turn, or an eighth of a turn. Here's an animation that shows this movement. Ending with the letter A upside down, like here. And when the letter is upside down, when it rotates again clockwise, the tip is being moved to the left. So the next turn will be the next turn will be, it will not be a quarter turn, it will be half a quarter turn. So the letter will be at a slant. It looks something like this. And this is answer A. Problem number 17. Kasha has three brothers and three sisters. How many brothers and how many sisters does her brother Mike have? A. Three brothers and three sisters. B. Three brothers and four sisters. C. Two brothers and three sisters. D. Three brothers and two sisters. And E. Two brothers and four sisters. There are seven children in total. There's Kasha and her three sisters, which makes four. Then there are three brothers, and that adds up to seven. And that adds up to seven, because four plus three equals seven. Mike is one of the brothers, which means from his perspective, he has two brothers and four sisters. And that is our answer. So two brothers and four sisters means E. Problem number 18. In a certain game, it is possible to make the following exchanges. One pair for two apples, an apple for three plums, and a strawberry for two plums. Adam has six pairs. How many strawberries will Adam have after he trades all his pairs for just strawberries? A-12, B-36, C-18, D-24, or E-6. There's no exchange going straight from pairs to strawberries, so Adam will have to take several steps to trade all his pairs for strawberries. First, he can trade the 6 pairs for apples. He gets 2 apples for each pair, so he will have 6 x 2 or 12 apples. Then, he can trade each apple for 3 plums, so he will get 12 x 3 or 36 plums. And finally, he can trade the plums for strawberries. Now each strawberry is worth 2 plums, so he will need to divide the number of plums by 2 to get the number of strawberries he can have. 36 divided by 2 is 18. So after all the exchanges, Adam will end up with 18 strawberries. The answer is C, 18. Problem number 19. Anne has a square sheet of paper. She cuts these pieces out of the sheet as many as possible. How many pieces does she get? A-1, B-2, C-3, D-4, E-5. The square sheet of paper is made up of 4 rows and 4 columns, which means it has 16 small squares in it. The figure is made up of 4 squares. There's no way that Anne can get 5 pieces, because 4 x 5 is 20, which is more squares that the sheet has. Anytime we're trying to fit as many things as possible into a shape, it's a good idea to try starting at a corner. So maybe the first piece that Anne cuts out could go like this. The problem doesn't say that it matters whether the piece is flipped or not, so she could fit another one in here. And one more right here, getting 3 pieces. So she gets 3 pieces, so it's more than 1 or 2. Now the question is, can she somehow get 4 pieces out of the square? Turns out that she cannot. Notice that when I place the piece here, this corner was covered, but there's no way to place a piece here. It would have to stick out something like this, maybe, or even another way. If I set the first piece at a different corner, say over here, then there's no way to put a piece on this corner. Again, it would have to be sticking out or overlapping, which doesn't work. We can try that for all the corners, and it shows us that there's no way that Anne can cut the square into 4 pieces like those shown. So the largest number of pieces that Anne can get is 3. Problem number 20. Sophie makes a row of 10 houses of matchsticks. In the picture, you can see the beginning of the row. How many matchsticks does Sophie need altogether? A, 50. B, 51. C, 55. D, 60. E, 62. The houses in the row are connected, but there should be a pattern of how many matchsticks Sophie needs. Looking at the first house, she used 1, 2, 3, 4, 5, 6 matchsticks. Now let's look at the second house. Here, she used 1, 2, 3, 4, 5 matchsticks to add it on. Now let's look at the third house. She used 1, 2, 3, 4, 5 more matchsticks to add the third house on. And then she used 1, 2, 3, 4, 5 to add the fourth house. So she used 6 matchsticks for the first house, and then 5 for each of the other houses, all the way to the end. You can see that because if these weren't here, you would still be using this matchstick on the far right. So for 10 houses, she needs 6 matchsticks for the first house, and then she's making 9 more houses out of 5 matchsticks each. So she will need 9 times 5 matchsticks for those remaining houses. So 6 for the first one, and 9 times 5 is 45 for the other 9 houses, which means that for 10 houses, she needs 51 matchsticks. So the answer is B, 51. Problem number 21. A tile falls off the wall. Caroline has 3 extra tiles. See the picture below. Which tiles fit the pattern? So our choices are choice A, only tile B. Choice B, both tiles A and B. Choice C is tiles B and C. Choice D, only tile C. Or E, all 3 of them fit. First, let's figure out what the tile should look like. The pattern is a zigzag. So this tile will have lines like this, with these parts being dark. So it slants from the upper left to the lower right. Given where these tiles are, if it's rotated 180 degrees, it will again look the same. Now, let's look at the tiles that we have. Tile A, if we rotate it just a tiny bit clockwise, we see that it's slanting from the lower left to the upper right. So this tile will not fit. Tile B, if you consider it rotated a little bit counterclockwise, we do have the stripe going from the upper left to the lower right. So this is a good tile. Tile C, if we rotate it a little bit counterclockwise, we see it's exactly the same as tile A, with the stripe going from the bottom on the left to the top on the right. So tile C also will not fit in the pattern. So the only tile that fits is tile B. The answer is A, only B. Problem number 22. Anna has one 5 cent coin, one 10 cent coin, one 20 cent coin, and one 50 cent coin. How many different values can she make with these coins? A, 4, B, 7, C, 10, D, 15, or E, 20. Each coin has a different value, and Anna can put the coins together in different ways, too. She can put two coins together, or three coins together, or even four coins together to get different values. So we'll have to look at all the combinations and figure out how many different values there are. First of all, we have the four values of the coins. 5 cents, 10 cents, 5 cents, 10 cents, 20 cents, and 50 cents. If we put two coins together, we can have 5 plus 10, which is 15, 5 plus 20, which is 25, 5 plus 50, which is 55. So that's if we put the 5 cent coin with one of the other coins. If we put the 10 cent coin together with one of the other coins, well, we already did 5 plus 10, so we now duplicate that. We can do 10 plus 20, which is 30, 20, which is 30, and 10 plus 50, which is 60. The only other two coins we haven't put together are 20 and 50, which is 70. When we put three coins together, we can do all of them except the 50 cent coin. So that would be 5 plus 10 plus 20, which is 35, or all of them except the 10 cent coin. So 5 plus 20 plus 50, which is 75. All of them except the 20 cent coin, 5 plus 10 plus 50 is 65, or all of them together except for the 5 cent coin, 10 plus 20 plus 50 is 80. If we put all four coins together, the value is 5 plus 10 plus 20 plus 50, which is 85. Look at these values. There are no repetitions. So we have four values for just one coin, six values, or six different ways to put two coins together, four values for putting three coins together, and one more value when we put all the coins together. So the total will be 4 plus 6, that's 10, 4 more is 14, plus 1 more is 15. The answer is D, 15. The answer is D, 15. Problem number 23. Anya makes a large cube from 27 small white cubes. She paints all the faces of the large cube green. Then Anya removes a small cube from four corners, as shown. While the paint is still wet, she stamps each of the new faces onto a piece of paper. How many of the following stamps can Anya make? A, 1, B, 2, C, 3, D, 4, or E, 5. Here are the five different stamps that are given as the options for answers. We just see how many of these can be made from the cube that's given. So the number will correspond to how many of the stamps she can make, not which stamp she can make. The nice thing about this problem is that we're told that Anya removed four corners. All four of these corners are shown. Here's one, here's another, here's the third back here, and the fourth. So the one corner we cannot see, the one in the very back on the bottom, still has a cube on it. That will help us determine which stamps she can make. Let me erase this so we can see the faces more clearly. There are three faces here that we can see. There's one like this, which we can tell right away corresponds to this stamp. So this is one that she can make. There's a face that's missing three of the corners, almost making a cross, but not quite. It actually looks like this stamp. The top face is very similar to the first face that we outlined. It can make the same stamp, so I'll just put another check mark. Now let's think about the faces that we cannot see. We already know that the corner in the lower back that we cannot see is filled in. So this face is only missing one square. It will look something like this. If we rotate it, we can make this image. The face on the back is missing two squares on corners next to each other, so it will look like this. It can make the very first stamp, this one. The face on the bottom is also missing two corners, so it will look the same as the one over here. Again, we can make a stamp like this from it. So the one we have not seen is the last one. To make this stamp, the four corners would have to be removed from the same face, and we see that that's not the case. There's no face that has all four corners removed. So this last step is one that Anya cannot make. So she can make four out of the five stamps. So the answer is D, four. Problem number 24. A square box is filled with two layers of identical square pieces of chocolate. Carol has eaten all 20 pieces in the upper layer along the walls of the box. How many pieces of chocolate are left in the box? A, 16, B, 30, C, 50, D, 52, or E, 70. This problem will have several steps. First, we're going to use the information that there were 20 pieces of chocolate in a layer along the walls to find out how many pieces of chocolate are in a whole layer. Then, remembering that there are two layers, we can find out how many pieces of chocolate were in the box at first. And finally, we just subtract 20, the pieces that Carol has eaten, to find out how many pieces are left in the box. So let's start by finding how many pieces are in one layer. We know that the box is a square. It looks something like this. And the pieces are arranged something like this. These outer pieces along the walls are what Carol ate. And obviously, there are more pieces in the box than I've drawn here, because he ate 20 pieces along the walls. So for any square like this, four of the pieces will be along the corners. If there are 20 pieces along the sides, you can subtract the four pieces from 20 to get 16, and these will be the pieces that go here, here, here, and here. It's going to be the same number of pieces in each of these regions, and there are four regions. So 16 divided by 4 gives us four pieces in each of these regions. So we know there are four pieces there, four pieces here, four pieces here, and four pieces here. Oh, we can kind of extend this to show that there are square pieces throughout. The important part is that there's a piece at each corner, and four pieces in between. So there are six pieces along each side, it'll be true on top, and along the side. So if the square has six pieces along one side and six pieces around the other, there are 6 times 6, or 36 pieces of chocolate in one layer. The problem says that there are two layers. In the box, 36 times 2 is 72. So at first, there were 72 pieces of chocolate in the box. The problem told us that Carol ate 20 pieces, so 72 minus 20 gives us 52 pieces left in the box. So the answer is D, 52.

Math Kangaroo

math problems

grades one and two

solution strategies

arithmetic operations

logical reasoning

problem-solving skills

2013 competition