This is the Math Kangaroo Solutions Video Library, presenting solution suggestions for Levels 1 and 2 from the year 2018. These solutions are presented by Lucas Naleskowski. 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. It is important that you make sure to read the problem, as well as listen as I read the problem. After reading or listening to the question, pause it, and try to solve the problem on your own. Question 1. What do you get when you switch the colors? If we look more closely at the initial picture, the first step is to count up all the dots. We have 1 dot, 2, 3, 4, 5, 6, 7. If we look at all possible solutions, we notice that only solution C and solution E have 7 dots. If we look at dot number 7 of the initial photo and look at its placement, and look at the same spot on circle C and E, we notice that circle C does not have a dot in that spot, while circle E has a dot in the same spot as the initial picture, except with the reverse color. Therefore, the correct solution is E. Question 2. Alice draws a figure connecting all the ladybugs in the order of increasing number of dots. She starts with the ladybug with 1 dot. Which figure will she get? If we look at the 5 ladybugs, and we start counting from the one with 1, 2, 3, 4, 5 dots, we must connect the ladybugs in ascending order, meaning we go from 1 to 2, from 2 to 3, from 3 to 4, from 4 to 5. This is the shape we are given, therefore the answer is D. Question 3. Mary glued 4 ray stars together like so. At least how many stars did she use? To do this, we must continue stacking stars on top of each other. We stack 1, 2, 3, 4, 5, 6, 7, 8 stars, then we get a picture that is the same as the one in the question. Therefore, the answer is D. Question 4. This pizza was divided into equal parts. How many parts have been taken? If we divide the pizza onto an axis, like so, in the demonstrated image, then we can count out that there are 2 slices on one side, 2 slices on the other side. Therefore, if we count out 1, 2, 3, 4 slices upon this side, we can assume and know that there will be 4 on the other side. Therefore, the answer is D. Question 5. How many kangaroos must be moved from one park to the other in order to get the same number of kangaroos in each park? If we look at the initial photo, we have 14 kangaroos in the left park and 4 kangaroos in the right park. If we continue to move the kangaroos from the left to the right park, we go from 12 to 6, 11 to 7, 10 and 8 until we have 9 in each park. We initially started off with 14 kangaroos in the left park. We now have 9. Therefore, the correct answer is 5 because 14 minus 9 is 5. We have moved 5 kangaroos from the left to the right park in order to have an equal number of kangaroos in each park. Question 6. Which of these ladybugs has to fly away so that the rest of them have 20 dots total? To solve this problem, we must take a closer look at all 5 ladybugs and count up all the dots that each one has. The first ladybug has 5 dots. The next ladybug has 7, then 5, 6, and the final ladybug has 4 dots. The next step is to add all the dots together so we know the total. When we do this, we get a total of 27 dots. The question asked us to find which ladybug has to fly away so that there is a total of 20 dots. So we must figure out 27 minus 1 will give us 20. There is only one ladybug that will give us this solution, the ladybug with 7 dots. Since 27 minus 7 gives us 20, the correct answer is B, the ladybug with 7 dots. Question 7. Emily builds towers in the following pattern. What will the 16th tower look like? If we take the initial pattern that Emily has given us, then we notice that after 4 tries, the towers start repeating. If we add more and more, we can continue to see how the pattern repeats itself. And upon reaching the 16th tower, we notice that the answer is E. Question 8. Little Theodore assembled a stacking toy as in the picture. How many rings will he see when he is looking at it from above? For a ring to be visible looking from the top down, it must be larger than all the rings above it. There are exactly 3 rings like this. All other rings are hidden beneath these larger rings. The topmost ring will always be visible, therefore it is one of these rings. The ring below it is larger than all the rings above it, therefore it will also be visible from the top. The two rings below that are smaller, therefore will not be seen. However the ring under them is larger than all rings above it and will be seen. And the ring below that is smaller and will not be seen, therefore the correct answer is C, 3 rings. Question 9. Juana the Friendly Witch has 5 broomsticks in her garage. Each broomstick is marked with a letter at the end of its handle. Juana removes the broomsticks one by one without moving the others. Which broomstick will she remove last? To remove a broomstick without moving the others, we can only remove the topmost broomstick. Therefore we must start off with broomstick D, remove it, then broomstick A, followed by broomstick E, finally broomstick C. Then we are left with the white broomstick, broomstick B, therefore the answer is B. Question 10. Which of the following figures can be made by placing these two transparent squares on top of each other? You can rotate both squares. We take a look at both squares, we put them on top of each other, and then rotate the second square 90 degrees clockwise. We see we get a figure identical to that in the answer A, therefore the correct answer is A. Question 11. Peter drew a pattern twice as in the picture. Which point will he reach when he draws the third pattern? If we look at the pattern and we repeat it one more time, following the same path as before, we notice that there is only one point that the pattern will cross through. That is point D. Therefore the correct answer is letter D. Question 12. Lisa has four puzzle pieces, but only needs three for her puzzle frame. Which one will be left over? If we look at the four puzzle pieces and the puzzle frame, the easiest way to figure out which three pieces will be used is by seeing which one piece won't be used. If we try to put in puzzle piece A, there is only one spot it will fit in. Then it requires two puzzle pieces just like B, but there is only one. Therefore puzzle piece A cannot fit, and the answer is A. Question 13. On her first turn, Diana got six points with three arrows on the target, as shown in the left part of the picture. On her second turn, she got eight points, as shown in the middle picture. How many points did she get on her third turn? If we take the amount of points, six, and the amount of arrows, three, that Diana shot, and divide them, we get two. Therefore each arrow that makes it into the outer circle is worth two points. Following this logic, in the middle picture, where we have one arrow in the middle circle and two arrows in the center circle, we must do eight, the total points, minus two and two, the two arrows that are in the outer circle to get the value of the middle circle, and this gives us four. Therefore an arrow in the middle circle gives us four points. When there are three arrows in the center circle, we must add up their values, which is four each, therefore four plus four plus four, and we get the answer of 12. Therefore the correct answer is C. Question 14. The dog went to its food following a path as shown. Along its journey, it had to make a total of three right turns and two left turns. Which path did the dog follow? If we inspect each path individually, we can count up the answer. In the first path, the dog makes three left turns and two right turns. In the preceding path, the dog makes three left turns and two right turns as well. In the third path, the dog makes two left turns and three right turns. In path D, the dog makes four left turns and two right turns, and in the final path, the dog makes four left turns and one right turn. The correct answer, therefore, is letter C, two left turns and three right turns. Question 15. How many times does a right hand appear in the picture? To solve this problem, you must just follow along with your own hands. First picture is a left hand, then a right hand, a left hand again, another right hand, Then we must total up all the right hands. There are five, therefore, the correct answer is C. Question 16. Charles cut a rope in three equal pieces and then made some identical knots on the pieces. Which figure correctly shows the three pieces with the knots? Since we know all three pieces started off at the same length, and each time a knot is made, the length is shortened, we know that the piece with one knot will be the longest and the piece with three knots will be the shortest, while the piece with two knots will be a medium length in between the one with one and three knots. There is only one image that satisfies these conditions, therefore, the answer is letter B. Question 17. The number of dwarves that can fit under a mushroom is equal to the number of dots on the mushroom cap. The picture below shows one side of each mushroom. The number of dots on the other side is the same. If 30 dwarves are seeking shelter from the rain, how many dwarves will get wet? And look at all the mushrooms, we are given four, but we also know that from the back they will have the same amount of dots, so we have to redraw them. Now we must just count up the number of dots on each mushroom to know how many dwarves will be able to hide under there. First mushroom can hide 3, plus 3, so 6. The next mushroom will be able to hide 4 plus 4, 2 and 2, and the last mushroom will be able to hide 5 plus 5 dwarves. We simply add all these numbers together, 3 plus 3, plus 4 plus 4, plus 2 plus 2, plus 5 plus 5, taking into account all the dots on the mushrooms, we get 28. Now since we know that the number of dots is the number of dwarves that can hide under the mushroom, we know that 28 dwarves will be able to hide, and there are 30 dwarves. So if we simply do 30 minus 28, we get 2, which is our answer, A, 2. One ice cream cone costs $1. There is a sale so you can buy 6 ice cream cones for $5. How many ice cream cones at most can you buy with $36? To get the most ice cream, we have to use the promotion of $5 equals 6 ice cream cones as much as possible. The most we can get into 36 with 5 is 7 times. So if we do the simple equation of 5 times 7 plus 1, we get 36. Therefore we will be able to use the promotion 7 times, so we will get 6 ice cream cones 7 times, for a total of 42 ice cream cones. With the remaining dollar, we can buy an additional ice cream cone to purchase 43 ice cream cones. Therefore the correct answer is D, 43 ice cream cones. Question 19. How many different numbers greater than 10 and smaller than 25 with all different digits can we make by using the digits 2, 0, 1, and 8? We know that all these numbers have to be greater than 10 and smaller than 25. So first let's count off all the numbers that are between 10 and 25. We have 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, and 24. These are all the numbers between 10 and 25. Now we can start crossing off the numbers. We can cross off 11 since it does not use all different digits. 12 can stay since it uses the digits 1 and 2. 13 cannot stay since we are only allowed to use digits 2, 0, 1, and 8. There is no 3. The same reason is for 14. There is no 4 we can use, or a 5, 6, or 7. 18 does use these digits, 1 and 8. 19 does not. 20 is allowed to stay since it uses the digits 2 and 0. 21 also uses digits 2 and 1, so they can both stay. 22 has the same problem as 11. It does not use all different digits. 23 and 24 can be crossed out since we're not allowed to use digits 3 or 4. We are only left with four numbers. Therefore, the correct answer is A. Question 20. A pirate has two chests. There are 10 coins in the chest on the left, and the other chest is empty. Starting tomorrow, the pirate will put one coin in the chest on the left and two coins in the chest on the right every day. In how many days will the two chests have the same number of coins? We know that the chest on the left has 10 coins, while the chest on the right has 0. After one day, the chest on the left will have 11 coins, and the chest on the right will have 2 coins. If we continue this reasoning, then the chest will slowly start to even out. On day 8, for example, the chest on the left will have 18 coins, while the chest on the right has 16 coins. If we continue this for two more days, on day 10, both chests will have 20 coins. Therefore, the correct answer is C. 10 days. Question 21. Alice has three white, two black, and two gray pieces of paper. She cuts every non-black piece of paper in half. Then she cuts every non-white piece of paper in half. How many pieces of paper will she have? First step is to visualize the problem. So we should draw out three white, two black, and two gray pieces of paper. The first step is to cut every non-black piece of paper. You can do this. Then you must cut every non-white piece of paper in half, like so. And now all that is left to do is to count up all the pieces of paper. You can see that there are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18 pieces of paper. Therefore, the correct answer is D. Question 22. A student had some sticks with a length of 5 centimeters and a width of 1 centimeter. With the sticks, he constructed the fence below. What is the length of the fence? To do this problem, we must divide the fence into segments. See that the first part is just a width of a stick. Therefore, it is 1 centimeter. Now if we take two widths, which is 2 centimeters, and subtract it from the length of a stick, which is 3, we get the next part, 3. Then we can take the width, 1, and so on, until we have the entire length of the fence. If we add all these together, we get an answer of 21. So the correct answer is B. Question 23. The road from Anna's house to Mary's house is 16 kilometers long. The road from Mary's house to John's house is 20 kilometers long. And the road from the crossroads to Mary's house is 9 kilometers long. How long is the road from Anna's house to John's house? We take the road from Anna's to Mary's house, we get a length of 16 kilometers. John to Mary, 20 kilometers. And from Mary's house to the crossroads is 9 kilometers. This, we can reason that from the crossroads to John's house, it is 11 kilometers. Since from John's to Mary's, it is 20. And if we take the 9 kilometers away from the crossroads, we get 11. We can do the same if we measure from the crossroads to Anna's house, we get 7 kilometers. Now to get the length of the road from Anna's to John's house, we just have to take the distance from Anna's to the crossroads and from the crossroads to John's house and add them together. 11 plus 7 is 18. Therefore, the correct answer is E, 18 kilometers. Question 24. Nellie bought four toys in the store. Their costs satisfy their qualities. Three cones equals a rabbit. One cone and a rabbit equals a doll. A bear and a rabbit is equal to a cone and a doll. What is the cheapest and most expensive toy? So if we take a closer look at all three equalities, we notice that if the rabbit is equal to three cones and we have a cone plus a rabbit equal to a doll, then we can look at this equation and substitute one rabbit for three cones and get the new equation of four cones equals one doll. We can do the same thing with the next equation and substitute the rabbit with three cones and the doll with four cones, and with this we get one bear plus three cones is equal to five cones. If we eliminate one cone from each side, keeping the equation equal, a bear and two cones equals four cones, a bear and a cone equals three cones, and finally, a bear equals two cones. With this, we can now find the solution. All out of the four toys, the cone is the cheapest since two cones is equal to one bear, three cones is one rabbit, and four cones is a doll. There is nothing that is worth less than the cone, and the toy that is worth the most cones since we have all three toys set to cones is the doll since the doll is worth four cones. Therefore, the correct answer is E.

Math Kangaroo

problem-solving

Lucas Naleskowski

pattern recognition

arithmetic operations

mathematical thinking