This is the Math Kangaroo Solutions Video Library. These solutions are for problems for Level 3 and 4 from the year 2016. They are presented by Agata Ghazal. The purpose of the Math Kangaroo Solutions Video Library is to help you learn how to solve math problems, such as those presented in the Math Kangaroo Competition. Let me start with a few tips of how to best utilize these solutions. First, make sure that you read the problem, and in this case, listen to it carefully. This will help you understand what kind of information you need to solve the problem. After reading the problem, I will wait for a few seconds. Please use this opportunity to pause the video and try solving the problem on your own. If you get stuck, don't worry. I will start the presentation of the solution with a hint of how to arrive at the correct answer. You can pause the video again at this point and try to finish the problem. My presentation of the solution will follow. Keep in mind that there might be ways of solving the problem other than what I present. I hope that you will learn something new from each problem you attempt, and, most importantly, that you will have fun. This is the Math Kangaroo Solutions video library. These solutions are for problems for Level 3 and 4 from the year 2016. They are presented by Agata Gazal. Problem number 1. Amy, Bert, Carl, Doris, and Ernst each rolled two dice and added the number of dots. Who rolled the largest total? To find the answer, we need to add the number of dots on both dice for each person. Let's start by adding the total on Amy's dice. On her first die, she rolled 1, 2, 3, 4, 5, 6. And on her second die, she just rolled 1. 6 plus 1 is 7. So that's Amy's total. For Bert, the total will be 1, 2, 3 plus 1, 2, 3. 3 plus 3 is 6. So we see that Amy rolled more than Bert. So Bert will not be the answer. For Carl, the total is 1, 2 plus 1, 2, 3. 2 plus 3 is 5. So Amy is still in the lead. Doris rolled 1, 2, 3, 4 on both of her dice. 4 plus 4 is 8. That's already more than Amy. And then the last sum will be Ernst. He rolled 1, 2, 3, 4, 5 on the first die. And 4 on the second, which is 9. So we see that Ernst rolled 9, which was the highest total. So the answer is E, Ernst. Problem number 2. Little Kanga is 7 weeks and 2 days old. In how many days will Little Kanga be 8 weeks old? Hint, since a week is 7 days long, 7 days need to pass from a time Kanga is 7 weeks old to a time she is 8 weeks old. When Kanga was exactly 7 weeks old, she needed to wait 7 days to be 8 weeks old. We can break down the 7 days into the 2 days that have passed and 7 minus 2 equals 5. 5 more days. So we can say that 7 weeks plus 2 days, which have already passed, plus 5 days more will make her 8 weeks old. So it is 5 more days until Kanga will be 8 weeks old. The answer is E, 5. Problem number 3. Which number should be placed in the box of the question mark? The arrows leading from each operation sign suggest that the number in the box should be the result of the operations on the numbers above it. Let's first do the operations in the top line, starting with 17 plus 3, which will go in this box. 17 plus 3 will give us, let's add the 1's together, 7 plus 3 is 10, so 0. Carry the 1, and 1 plus 1 is 2. Gives us 20. Now, let's solve the other problem in the top row, 20 minus 16. Put the answer in the box below. 20 minus 16. We can subtract 6 from 0, so we're going to borrow 1 here, make this 10. 10 minus 6 is 4, and 1 minus 1 is 0. So we get 4 for this one. Now our problem becomes 20 plus 4. 20 plus 4 is 24. We can put that in the box of the question mark. So our answer is A. 24 is the number we should put in the box of the question mark. Problem number 4. What does Peeple the Clown see when he looks at himself in the mirror? In the mirror reflection, the things that were on the right will be on the left, and the things that were on the left will be on the right. I will draw a mirror right here, and here is Peeple. When he looks in the mirror, the things that are closest to the mirror on this side will be closest to the mirror also on the other side, and the things that are furthest away will be furthest away on this side. So we know that his flower will now be on the opposite side, on the left side, which means that answer C and answer E do not work. His face will have the bigger eye closer to the mirror, again on the left side, and the smaller eye on the right side, which means that answer B is not going to work. His hair, the smaller side, is here, and here is the larger side, which is the same in both answers A and D. The bow tie has the section with three dots on the left, one, two, three, and the section with the four dots on the right, one, two, three, four. The face and the hat, and we have the clown that is shown in answer A. Problem number five. Jeff goes with his father to a circus. Their seats are numbered 71 and 72. Which way should they go? We need to figure out in which of the ranges of numbers listed in the chart, numbers 71 and 72 fall. Think of the ranges given in each of the rows as showing us where our numbers should be located. The first row would be for numbers which are greater than or equal to 1 and less than or equal to 20. The second row would have the numbers from 21 to 40 inclusive, or 21 is less than or equal to our number, and our number is less than or equal to 40. So, if our numbers are 71 and 72, we can see that they would fit in the row from 61 to 80. 61 is less than or equal to 71 and 72, and both these numbers are less than or equal to 80. This is the arrow Jeff and his father would need to follow. So, this is answer D. Problem number six. Anna shares some apples between herself and five friends. Everyone gets half an apple. How many apples does she share? First, figure out how many people there are eating the apples. We have six people eating the apples. Anna plus five friends means that there are six people eating. If each person gets half an apple, then there are six halves of apples, and six halves make three whole apples. So, the answer is B, three apples. Problem number seven. A rectangle is partly hidden behind a curtain. What shape is the hidden part? Our choices are A, a triangle, B, a square, C, a hexagon, D, a circle, and E, a rectangle. Draw in the parts of this rectangle which you cannot see. We will lengthen this edge through here, and this through here. Color in the part we drew in. We see that it has one, two, three sides, which means that it is a triangle. The answer is A, triangle. Problem number eight. Which of the following sentences correctly describes the picture on the left? A, there are as many circles as squares. B, there are fewer circles than triangles. C, there are twice as many circles as triangles. D, there are more squares than triangles. E, there are two triangles more than circles. Reading through the answers, we see that we need to know the number of squares, circles, and triangles in the picture. So, let's start by counting them. First, let's count the circles. There are one, two, three, four of them. So, for circles, we can put down four. Next, let's count the triangles. One, two. We have two triangles. And finally, let's look at the number of squares. One, two. Now, let's go through the sentences and see which ones are true. Sentence A says there are as many circles as squares. We know there are four circles and two squares. So, A is not true. There are not as many circles as squares. There are more circles than squares. Choice B says there are fewer circles than triangles. Well, we have four circles and two triangles. That means that B is false, because there are, in fact, more circles than triangles. Choice C, there are twice as many circles as triangles. We do have four circles and two triangles. And four is twice as many as two. Two times two is four. That means that C is correct. Let's check the other statements as well. D, there are more squares than triangles. There are two triangles and two squares, which means that this is false. There's the same number of squares and triangles. And E, there are two triangles more than circles. Well, actually, there are four circles and two triangles. So, there are two triangles less than circles. So, this is not true. We see that the only statement that's true is statement C. There are twice as many circles as triangles. Problem number 9. The sum of the digits of the year 2016 is equal to 9. What is the next year after 2016 when the sum of the digits of the year is equal to 9 again? Hint, if the digits add up to 9, what does that tell you about the number? A number whose digits add up to 9, or a multiple of 9, is itself divisible by 9. So, 2016 is divisible by 9 because its digits 2, 0, 1, and 6 add up to 9. We will check the next multiple of 9 to see if its digits add up to 9. Let's add 9 to our original number, 2016. So, 2016 plus 9 will be... 6 plus 9 is 15. I've carried a 1. This one is 2, and 0, and 2. And we get 2025 as the next number as a multiple of 9. Now, let's see what its digits add up to. 2 plus 0 plus 2 plus 5 is... 2 plus 2 is 4, plus 5, it is 9. So, because the digits add up to 9, and we know this is the next number that can possibly have digits that add up to 9, we know that B is the answer. Problem number 10. The mouse wants to escape from the maze. How many different paths can the mouse take without passing through the same gate more than once? Hint, there are a lot of gates, but there aren't as many choices. Let's start by looking at the top gate. If the mouse goes through this gate here, then it has to go through this gate. So, this constitutes only one choice. If it goes through the bottom gate here, it will need to go through this gate. There's the other choice. In this room, it can either go here, and go through this gate, it goes here, and then it can go out, or it can go downward, and has to go through this gate, and then go out. So, in reality, the mouse can only go through the orange gate, and then through the green, through the orange, and then through the red, or through the pink, and through the green, and through the pink, and then through the red, which makes for only four choices of the ways that the mouse can go to get out of the maze without backtracking. The answer is B, four. Problem number 11. Here are the numbers on each side of both cards. The sum of the two numbers on the first card is equal to the sum of the two numbers on the second card. The sum of the four numbers is 32. What could be the two numbers on the sides that we cannot see? We know the sum of all four numbers, and we know that the two sums for each card are equal. So we can find the sum for each card. If the sum for the four numbers is 32, and the sum on each of the two cards is the same, then if we divide 32 by 2, we will have the sum for each card. So 16 is the sum of the two numbers on this first card, and of the two numbers on the second card. Now all we need to do is to subtract the number on the side of the card shown from 16 to find what the number on the other side is. 16 minus 5 is 11. So the first number will be 11, and then 16 minus 12 is 4. So our two numbers are 11 and 4. This is answer C. Problem number 12. Which tile should go in the middle of the pattern in the picture? One way to solve this problem would be to draw in the missing arcs connecting the paths that we see. I will start by drawing in the blue arc. We'll connect a short distance here. Then let me draw in the red. Again, just a simple curve, and likewise with the yellow. To make things easier to see, I will just erase around this piece. We see that the tile in the middle has a blue arc in the upper left-hand corner, a yellow line going across the bottom half, and a red one going from the middle bottom to the upper right. This looks exactly like tile B, so our answer is B. Problem number 13. Each of five children had a paper square, a paper triangle, and a paper circle. Each child placed their own papers in a pile, as shown in the pictures. How many children placed the triangle above the square? Think about what we would see if the triangle is on top of the square, as opposed to the square being on top of the triangle. Let's ignore the circle and see what happens when we put the triangle on top. That is, first we will put down the square, and then we will put down the triangle. We can see that the bottom corners of the square are not showing. Now, if we were to take the square and put it on top of the triangle, now we clearly see the whole square. Let's put the triangle back on top. We are looking for pictures where the bottom corners of the square are not showing, like over here. This picture fits. This one doesn't, because we can see the bottom corners of the square. This one also doesn't fit. Here the square is clearly on top. The fourth one fits, because we cannot see the bottom corners of the square. And the fifth one looks exactly like our example. We also cannot see the bottom corners of the square. So, there are three pictures, or three children, who place the triangle above the square. The answer is D, 3. Problem number 14. Which three of the five jigsaw pieces shown on the left can be joined together to form a square? The square and the rectangle as pieces 2 and 4 don't look like they can be put together to form a square, unless some of the triangles also make a square. So, let's start with the triangles. Piece number 3 looks like a good piece to start with, because it is the largest and most irregular shape. Let's bring it down here. Now, let's look at piece number 1. looks like piece number 1 has the same side length as piece number 3. So we can try to put it on the opposite edge, and then flip it upside down. And we can move it down right here, and it fits. Now it looks like we just need a triangle to finish making a square. Let's bring piece number 5 down, and we'll need to rotate it. And right there we have made a square. We have used pieces numbers 1, 3, and 5, so the answer is A. Pieces 1, 3, and 5 can be joined together to form a square. Problem number 15. Lois has started to write some numbers in the table. She decides that each row and column will contain the numbers 1, 2, and 3 exactly once. What is the sum of the numbers that she will write in the two shaded squares? Let's first look at the first row and figure out where the numbers 1, 2, and 3 belong there. We have the numbers 1, 2, and 3 to put in each row and each column. Looking at row 1, we already have the number 1 written in. The number 2 cannot go right next to the 1, because then it would be in the same column as this 2. We have two 2's in the same column. Instead, we will need to put the 2 over here. This already gives us the answer as to what the sum of the two shaded numbers is. Since the numbers in the last column are 1, 2, and 3, and the number 2 is not on the shaded square, then the sum of the remaining numbers in this column will be 1 plus 3, which is 4. Our answer is C, 4. Do you want to see the further solution? In row 1, we can put the 3 right here. In the second row, the 3 can go here. The shaded square would have the 1 here, which leaves us a 3 for this shaded square, a 2 for the first square in that row, and a 1 for the last square. Problem number 16. John has a board with 11 squares as shown in the picture. He puts a coin in each of 8 neighboring squares without leaving any empty squares between the coins. What is the maximum number of squares in which one can be sure there is a coin? To solve this problem, let's find all the different ways in which the coins can be placed, and then compare those ways to find in which square there is always a coin. Here is one board. Let's start by putting the 8 coins on the furthest left. They have to be next to each other, so 1, 2, 3, 4, 5. 1, 2, 3, 4, 5, 6, 7, 8. Let's draw another board and shift all the coins over 1. 1, 2, 3, 4, 5, 6, 7, 8. Here is another one. Shifting one more, 1, 2, 3, 4, 5, 6, 7, 8. And we can do one more. 1, 2, 3, 4, 5, 6, 7, 8. We cannot shift our coins any further, so these are the only 4 choices. If we look at our picture, we see that these 5 squares always have a coin in them, no matter which layout we use. There are 1, 2, 3, 4, 5 such squares. So the answer is D, 5 squares. Problem number 17. When a transparent card is turned on its right edge, the result can be seen in the picture to the right. What will we see when this card is turned over its upper edge? Solve this problem in two steps. First, figure out what the original card looks like, and then figure out what it will look like turned over its upper edge. When the card is flipped, we can think of it as being reflected over an axis. If we think of the first axis being here, the card was only flipped horizontally, so it will be switched left to right, but not top to bottom. The original card looked like this. Now, when we flip it over the top edge, we are flipping it over this axis. So when it's flipped, it will look like this. It's transposed top to bottom, but not left to right. So, this is our answer, which is card D. Problem number 18. Tim, Tom, and Jim are triplets, three brothers born on the same day. Their brother Paul is exactly three years older. Which of the following numbers can be the sum of the ages of the four brothers? What would the sum be if all four were born on the same day? How is that different from the sum right now? Since Tim, Tom, and Jim are triplets, their ages are all the same. So, Tim's age is equal to Tom's age to Jim's age. Now, Paul's age is three years more than any one of those ages. So, let's use Tim's age. Paul's age is equal to Tim's age plus three. So, the sum of all their ages would be like having Tim's age plus Tom's age, which is the same as Tim's age, plus Jim's age, which is also the same as Tim's age, plus Tim's age plus three, which is Paul's age. Which means that the sum of the ages is four times Tim's age plus three. So, to get the answer, we will first subtract three from each one of the answers from A to E, and then we will see which one is divisible by four, because an age needs to be expressed in a whole number here. So, let's see. Twenty-five minus three is twenty-two, which is not divisible by four. So, that cannot be the answer. Twenty-seven minus three is twenty-four. Twenty-four is divisible by four. Twenty-four divided by four is six. So, B should be the answer. Let's check the other ones. Twenty-nine minus three is twenty-six. That is not divisible by four. Thirty minus three is twenty-seven. That is not divisible by four. And sixty minus three is fifty-seven, which also is not divisible by four. So, the only number that works is twenty-seven. That's answer B. Problem number nineteen. Magic trees grow in a magic garden. Each tree has either six pears and three apples, or eight pears and four apples. There are twenty-five apples in the garden. How many pears are there in the garden? Compare the number of pears to the number of apples on each tree. We have pictures of the two types of trees. On the first tree, there are six pears and three apples. On the second tree, there are eight pears and four apples. Notice that on both kinds of trees, the number of pears is twice the number of apples. On the first tree, two times three apples is six, and there are six pears. On the second tree, two times four is eight, and there are eight pears. That means that no matter which kind of trees we have in the garden, or what mixture of them, there will always be twice as many pears as apples in the garden. So, if we have trees that together have twenty-five apples, there will be twice as many pears. Twenty-five times two is fifty. So, our answer is D. There are fifty pears in the garden with twenty-five apples. Problem number twenty. Problem number twenty. My dogs have eighteen more legs than noses. How many dogs do I have? Hint. First, figure out how many more legs than noses one dog has. Let me first draw a picture of a dog. Here's the head, the snout, the ear, neck, body, two paws in front, two in back, and a tail. And here's his nose. So, this dog has one nose, and one, two, three, four legs. Four minus one is three. So, one dog has three more legs than noses. So, if there's one dog, there will be three more legs. If there are two dogs, and each dog has three more legs than noses, there will be six more legs than noses. If there are three dogs, there will be nine more legs. Nine more legs. So, let's find the number that multiplied by three is eighteen. Eighteen divided by three is six. So, six times three is eighteen. So, six dogs will have eighteen more legs than noses altogether. So, our answer is six, which is C. Problem number 21. Karen wants to place five bowls on a table in order of their weight. Shorty placed bowls Q, R, S, and T in order. Bowl T weighs the most. Where must she place bowl C? We can solve this problem by figuring out which of the shapes weighs the most and which one weighs the least. Let's find two bowls in the first row that differ only by one shape to figure out which shape weighs more or which one weighs less. If we look at S and T, the only difference is that bowl S has a triangle. The only difference is that bowl S has a triangle, whereas bowl T has a circle. So, since all of bowl S weighs less than bowl T, and because the only difference is the triangle instead of a circle, we know that the triangle weighs less than the circle. So, we'll put our triangle on the left and our circle on its right. Now, let's figure out where the square will fit. If we compare bowls R and T, we know that T is heavier, and the only difference is that one of the circles in bowl R was replaced with a square in bowl T. That means that the square is heavier than the circle, therefore it's the heaviest of the three objects. Bowl Z has a square as well as a triangle and a circle. If we compare with bowl Q, it will be heavier because this circle is heavier than the triangle that it has, and the other two shapes are the same. If we look at bowl R, bowl Z will be lighter because its triangle is lighter than the circle in bowl R, and the other two shapes are the same. Therefore, we need to put bowl Z in between bowls Q and R. This is answer B. Problem number 22. Rachel adds seven numbers and gets 2,016. One of the numbers she adds is 201. She replaces the number 201 with 102. What sum does she get? We can solve this problem by finding the sum of the six numbers apart from 201 and then adding to it 102. We start with the number 2,016 and we will subtract 201. 6 minus 1 is 5. 1 minus 0 is 1. 20 minus 2 is 18. We get 1,815. This is the sum of the six numbers when 201 is omitted. Now, we will take the sum of the six numbers, 18, 15, and add to it 102. 5 plus 2 is 7. 1 plus 0 is 1. 8 plus 1 is 9. The number we get after 201 is replaced with 102 is 1,815. The number we get after 201 is replaced with 102 is 1,917. This is answer C. Problem number 23. Malte built a bar of 27 bricks. He breaks the bar into two bars in such a way that one of them is twice the length of the other. Then he takes one of the new bars and breaks it the same way. He continues in this way. Which of the following bars will he not be able to get? There is a pattern to the way that he is breaking up the bar. Here is the bar that Malte starts with. It is 27 bricks long. When he breaks it down in such a way that one part is twice the length of the other, he breaks it down into lengths of 9 blocks and 18 blocks. In other words, he is dividing the block into thirds and keeping two of the pieces together. Let's see what we can do with the bar on the left, which is 9 pieces long. 9 divided by 3 is 3. The other part will be 6 blocks long. We see that we can get a piece that is 6 blocks long. It is this one here. C will not be the answer of what we cannot get. If we break down the piece that is 3 blocks long, 3 divided by 3 is 1. The other part will be 2 pieces long. Now we have made a piece that is 2 blocks long. A is not something that we cannot obtain. We cannot break these two down any further. Let's see what we can do with the block that is 6 pieces long. 6 divided by 3 is 2. The other piece is 4 blocks long. Now we have a piece that is 4 blocks long. B will not be the answer. We cannot break down the pieces 2 blocks long or 4 blocks long any further because neither one of those is divisible by 3. Let's see if we can get either a piece that is 8 blocks long or 10 blocks long from this piece which is 18 blocks long. 18 divided by 3 is 6. The other part is 12 blocks long. We already had made a piece 6 blocks long. Let's see what we can do with the one that is 12 pieces long. 12 divided by 3 is 4. The other part will be 8 pieces long. Now we have made a piece that is 8 blocks long. We can make the piece with an answer D so it's not going to be the right answer. We have not made a piece that is 10 blocks long and we have broken our bar down already in such a way that there is no way to make it. So E will be the right answer. Malte cannot make a piece that is 10 blocks long. Problem number 24. Five sparrows sit on a branch as shown in the figure. Each sparrow chirps the same number of times as the number of sparrows it sees. For example, David chirps three times. Then one sparrow turns to look in the opposite direction. Again, each of the sparrows chirps the same number of times as the number of sparrows it sees. This time the total number of chirps is more than the first time. Which of the sparrows turn to look in the opposite direction? To find the answer, we need to figure out which sparrow will see more sparrows if it turns. Because if it sees more sparrows, then it will chirp more times. Let's start with Angel. Angel sees four sparrows right now. She chirps four times. If she turned the other direction, she would see zero. So the number did not increase. It went down. Bertha chirps once. If she were to turn, she would chirp three times. Charlie chirps two times. If he turns, he will chirp two times again. David chirps three times. If he turns, he will chirp three times again. If he turns, he would chirp once, so the number would go down. And finally, Eglio chirps four times. If he turned, he would chirp zero times. So we see that Bertha is the only sparrow that would chirp more times if she turned. So the answer is B, Bertha.

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Agata Ghazal