This is the Math Kangaroo Solutions video library, with problems from Levels 3 and 4 from the year 2018. These solutions are presented by Agata Gazal. 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. Each stamp has one of the digits, 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. She stamps the date of the kangaroo contest, 03-15-2018. How many of the stamps does she use? A, 5, B, 6, C, 7, D, 9, E, 10. Lina uses the same stamp each time she stamps a given digit. The date has 8 digits, but we only need to count how many different digits there are to know how many of the stamps Lina used. Let's go through the date and color the same digits the same colors. The date starts with 0, so let's color this 0 in blue. And we have one more 0 in 2018, so we'll color that one in blue as well. The next digit we see is 3. Let's color that one in. There are no more 3s. We have two 1s. We can color those in yellow. There is a 5, just one. I'll color that one in red. We have 1, 2. Color in purple. And that leaves the 8, which I will color peach. So even though we have 8 digits, we only have 6 different colors. Blue, green, yellow, red, purple, and peach. So the answer is B, 6. Problem number 2. The picture shows three arrows that are flying and nine balloons that can't move. When an arrow hits a balloon, the balloon pops and the arrow keeps flying in the same direction. How many balloons will be hit by the flying arrows? A, 2. B, 3. C, 4. D, 5. E, 6. The easiest way to solve this problem is to trace the path of each arrow. Let me pull up a ruler so I can make straight lines. Let's trace the path of the first arrow. It will go like this. So we see that it will hit the green balloon. So that one pops. And one of the big red balloons. Let's trace the path of the second arrow. The second arrow hit the small yellow balloon and the large blue balloon. Now let's look at the last arrow. This one also hit two balloons. The large yellow one and the small red one. Each arrow hit two balloons. So in total, 2 plus 2 plus 2 balloons were popped. So six balloons were hit. The answer is E, 6. Problem number 3. Susan is six years old. Her sister is one year younger and her brother is one year older. What is the sum of the ages of the three siblings? A, 10. B, 15. C, 18. D, 21. E, 30. The most straightforward way to solve this problem is to find the age of each sibling and then add them together. Here is Susan. We are told that she is six years old. Her sister is one year younger. 6 minus 1, 5. The brother is one year older than Susan. 6 plus 1 is 7. We now have the ages of the three siblings. 6 years old, 5 years old, and 7 years old. We can just add them together. 6 plus 5 plus 7 is... 6 plus 5 is 11, plus 7 more gives us 18. The answer is C, 18. Problem number 4. The picture shows five screws in a block. Four of the screws are the same length. One screw is shorter. Which screw is the short one? A, 1. B, 2. C, 3. D, 4. E, 5. We know that there are four long screws. So, if we find two screws that are the same length, we know that this is the length of all four long screws. Then, we just need to find the one screw that cannot possibly be that long. Let's start by looking at the length of screw 1. We'll mark it from top to bottom, this long. Now, screw 2 is about the same length. We've marked it here. Keeping this in mind, let's look at screw number 3. If it were the same length as screws 1 and 2, it would end about here. Screw number 4, likewise, could be as long, certainly close to that length. If it is that long, it ends over here. We know that screw 4 is not longer than screws 1 and 2 because four screws are the same length and the fifth one is shorter. Finally, we have screw 5. If it were the same length as screw 1, it would end way down here. The type of screw 5 starts at the same place as the type of screw 1, which means that if it were as long as screw 1, it would end in the same place, but it does not. We know that, for sure, screw 5 is somewhat shorter than screw 1. The answer is E, 5. Screw 5 is the short one. Problem number 5. Here is Sophie the ladybug. She turns around. Which of the ladybugs below is not Sophie? We should be able to tell which ladybug is or is not Sophie by looking at the number of dots on her right wing and on her left wing. Here's a close-up of Sophie. We see that on her left wing, looking from the top with her head facing upwards, she has three dots. And on her right wing, she has four dots. To make things easier, I will outline the left wing. We can now compare Sophie with the answers listed. The left wing in answer A is here and has three dots. And we see that the right wing has four dots. In B, the left wing would be this one. It has three dots. And the right wing has four dots. In 3, this is her left wing. Three dots. Right wing has four dots. In D, the left wing is here and it has four dots. Whereas the right wing has three dots. So we see that this one does not match our original ladybug. Just to make sure, let's look at answer choice E. This is the left wing. And there are three dots here and four dots on the right wing. So by counting the number of dots on both wings, we see that the answer is D. D is the ladybug that is not Sophie. Problem number six. Lucy folds a sheet of paper in half. Then she cuts a piece out of it as shown here. What will she see when she unfolds the paper? Here is a close-up of the sheet. We see that the fold line is on the left side of the paper. And here it is unfolding. The image we see when the sheet is unfolded is what is shown in answer D. So the answer is D. Problem number seven. On her first turn, Diana scored 12 points total with three arrows. On her second turn, she scored 15 points. How many points did she score on her third turn? How many points did she score on her third turn? A, 18. B, 19. C, 20. D, 21. E, 22. Hitting an area of a given color with an arrow gives the same number of points for each arrow. Looking at the first target, we see that Diana got 12 points by hitting the light blue area with three arrows. We know that three times four would make 12. So the light blue area must be worth four points for each arrow that makes it there. Looking at the second picture, there are two arrows in the light blue area and one arrow in the dark blue. If each of the arrows in the light blue area is worth four points, then the two arrows in the light blue area are worth plus four or eight points together. So the dark blue area will be worth 15, which is the total, minus eight, the two arrows that are not in the blue area, which gives us seven. And there's only one arrow there. On her third turn, Diana got all three arrows into the dark blue, which means that her score would be seven, which is the value of the dark blue area, times three, which is how many arrows she got into it. So her score is seven times three, 21. The answer is D, 21. Problem number eight. Mike sets the table for eight people. He must set the table correctly for each person sitting at the table. Setting the table correctly means that the fork is on the left of the plate and the knife is on the right of the plate. For how many people did Mike set the table correctly? A, five, B, four, C, six, D, two, E, three. We can solve this problem by finding a place where it's easy to tell whether the fork and the knife are set correctly and then compare it with the other plates. The problem tells us that for the table to be set correctly, the fork needs to be on the left and the knife needs to be on the right. The fork needs to be on the left and the knife needs to be on the right. So our plate would look something like this. Here's the fork and here's the knife. If we look on the bottom part of the table here, it's easy to tell what's on the left and what's on the right. It's just whatever is facing us. So we already see that the plate right in the middle corresponds to the image that we wanted. So this one is correct. We have the fork on the left and the knife on the right. And the plate to the right of it is also correct. However, the plate on the left is not correct because the utensils are switched. Once we start going around the table, we kind of have to rotate. So as we go around, if we want to think about it this way, we should come to the fork first and then to the knife. So this place is correct. Let's go around. This one is correct. We see the fork first. This one is not correct because we have the knife first. The next one is correct. We have to turn again. We are coming to the knife first. So that is not correct. Now we have circled all the ones that are correct and crossed out the ones that are not. We have 1, 2, 3, 4, 5 plates that are set correctly. So Mike set the table correctly for 5 of the people. The answer is A, 5.

Math Kangaroo

problem-solving

logical thinking

pattern recognition

Agata Gazal