This is the Math Kangaroo Solutions video library, presenting solution suggestions for levels 5 and 6 for the year 2020. These solutions are presented by Patryk Lipski. The purpose of the Math Kangaroo Solutions video library is to help you learn how to solve math problems, such as those featured in the Math Kangaroo competition. It is important that you make sure to read each problem as well as listen as I read the problem. After reading and listening to a problem, pause the video and attempt to solve the problem on your own before viewing the solution. In the top left corner, we have a quarter circle missing. Next to it are two long, thin triangles. And finally, a quarter diamond or inverted circle. We see that the same shapes occur in the last answer. Therefore, that answer is E. Problem 2. As Amira is walking from A town to B town, she passes the five signposts shown. One of them is incorrect. Which one? So we can draw the road from A to B town and a signpost on it. This is signpost D. Now, it tells us that the distance to A town is 8 kilometers and the distance to B town is 3 kilometers. Therefore, the distance from A town to B town, according to sign D, must be 8 plus 3, or 11. Similarly, sign A gives us the distance as 2 plus 9, or also 11. B, 3 plus 8, also 11. C, 5 plus 6, also 11. D we have covered. And E says 9 plus 4, which makes 13. That's the odd one out, so E is incorrect. Problem 3. A large square is made up of smaller white and gray squares. What does the larger square look like if the colors of the white and gray squares are switched? We can take the square and reverse the colors row by row to get the correct answer. The first row is gray, gray, white, gray, so we will make it white, white, gray, white. The second row is white, gray, gray, white, so we will make it gray, white, white, gray. The third row is white, white, gray, gray, so we will make it gray, gray, white, white. And the fourth row is gray, gray, white, white, so we will make it white, white, gray, gray. Now we can see that the correct answer must be D. Problem 4. Mikos wants to bake 24 muffins for his birthday party. To bake 6 muffins, he needs 2 eggs. Eggs are sold in boxes of 6. How many boxes does Mikos need to buy? To bake 6 muffins, Mikos needs 2 eggs. Since he wants to make 24 muffins, we need to find out how many sets of 6 muffins Mikos is making. 24 divided into sets of 6 is 4 sets. So Mikos is making 4 sets of 6 muffins. Then each set needs 2 eggs. So Mikos needs 4 times 2 eggs, which is 8. Since eggs come in boxes of 6, 1 box of eggs will not be enough. But 2 boxes of eggs is 12 eggs, which is enough for Mikos. Therefore, the answer is 2, or B. Problem 5. Flora reflects the letter F over the two lines shown. What will the reflections look like? We need to reflect the letter F. First, let's start with the line at the bottom. When we reflect, we will get this image. The F is still pointing the same way, but now it is upside down. Next, let's reflect over the side. We get this image. The F is now backwards, but still right side up. These images match the answer, E. Problem 6. Problem 6. Kim has several chains of length 5 and of length 7. By joining chains one after the other, Kim can create different lengths. Which of these lengths is impossible to make? Kim can make 10 by combining two chains of length 5. She can likewise make 15 by combining three chains of length 5. 14 is made out of two chains of length 7. And if she combines a chain of length 5 and of 7, she will get a chain of length 12. 13, however she does it, is not possible to make. Problem 7. Maria starts out with 10 sheets. Now, she cuts a sheet into 5 pieces. We can count and see that she has 14 pieces. How did this happen? If she cuts another sheet, we can count again and see that she has 18 pieces. Every time she cuts one sheet into 5, she is adding 4 sheets. 5 minus 1 is 4. We can see this highlighted. Then, if we divide the difference of 22 and 10, which is 12, by 4, we should know how many sheets she has to cut in order to make 12 more pieces. That is 3, and we can confirm she needs 3 sheets by cutting the third sheet. Therefore, the answer is 3, or A. Problem 8. Cindy colors each region of the pattern shown to the right red, blue, or yellow. She colors regions that touch each other with different colors. She colors the outer region blue. How many regions of the completed pattern are colored blue? The outer piece is blue. Since we don't know which piece is yellow or which is red, we can start with 2 shades of orange. At this point, there are 3 pieces that touch. Here, we have both shades of orange already, so we need a blue piece. Then, at this point, there are also 3 regions that touch, and since we have blue and light orange already, we can do dark orange. Then, at these 3 points, we will have a similar situation, and so we can finish the figure by following the same rules. Now we see that we have 3 blue regions. We can even flip our 2 shades of orange, or flip red and yellow, to see that we will still have 3 blue regions. Therefore, the answer is 3, or B. Problem 9. 4 baskets contain 1, 4, 6, and 9 apples, respectively. How many apples should we move between the baskets to have the same number of apples in each basket? We have our groups of apples organized here by row, so they're easier to work with. To find out how many apples should be in each basket, we can add them up, and 1 plus 4 plus 6 plus 9 makes 20, and divide by the number of baskets, or divide by 4. So we get that there should be 5 apples per basket. So now we have our target for each basket. 2 baskets are missing apples, which definitely need to be brought in from the bigger baskets. An apple from the 6 basket can fit nicely into the basket with 4 apples. This works because 6 minus 1 and 4 plus 1 are both 5. Similarly, 4 apples from the 9 basket can go into a 1 basket, and since 9 minus 4 and 1 plus 4 are both 5. We have moved 5 apples between the baskets, and that was the minimum needed to fill the 1 basket and the 4 basket, so the answer is C. Problem 10. A dog and a cat are walking in the park along the path marked by the thick black line. The dog starts from P at the same time as the cat starts from Q. The dog walks 3 times as fast as the cat. At which point do they meet? So the dog takes 3 steps for every 1 step the cat makes. So at first, the dog covers 3 segments while the cat covers 1. Then the dog covers 6 while the cat covers 2. Continuing this way, eventually the dog will have walked 12 pieces and the cat has walked 4. And now, they are separated by 4 equal diagonal pieces. So since the dog is moving 3 times as fast as the cat, the dog will cover 3 while the cat covers 1. They meet at point E. Problem 11. The numbers from 1 to 10 have to be placed in the small circles, one in each circle. Numbers in any two neighboring circles must have the same sum as the numbers in the two diametrically opposite circles. Some of the numbers are already placed. What number should be placed in the circle with the question mark? We have a circle, and let's also keep track of which numbers are already on it. We can already clear some of the numbers. The problem is saying that pairs across the circle, like this one, or like this one, should add up to equal sums. In the current pair, we can find the missing number by equating 9 plus 5 and 10 plus the missing number. Since 9 plus 5 is 14, the missing number must be 4. We can fill this in on our circle and cross it off our list. This next pair we can also solve by setting 1 plus 10 equal to 5 and the missing quantity. 10 plus 1 is 11, so the missing number is 6. Fill it in and cross it out. With that 6, we can move on to the next pair. This time, 6 plus 2 equals 1 and the missing number. 6 plus 2 is 8, so we're missing a 7. After we cross it out, we can move on to the last pair. 7 and the unknown equals 2 and a blank. We only have two numbers left to use, a 3 and an 8. Since 7 is bigger than 2, it makes sense that the 3 should be with the 7 and the 8 with the 2. Checking our math, we get 11 equals 11, so the answer is 3 or 8. Problem 12. When Elisa Bat leaves her cave, a digital clock shows 2020. When she returns and is hanging upside down, she sees 2020 on the clock again. How long has she been away from her cave? If Elisa sees 2020 on her clock, then flipping it upside down, we'll get 202. To see how much time has passed from 2020 until 202, we're going to need to get through midnight. So first, let's add 3 hours to 2020 and we get 2320, or just before midnight. And then we need to add 40 more minutes to actually hit midnight. At that point, all we need is 2 hours and 2 minutes to get to 202. Now, summing up all the additions, we see that Elisa was gone from her cave for 5 hours and 42 minutes. Therefore, the answer is E. Problem 13. An elf and a troll meet. The troll always lies while the elf always tells the truth. They both say exactly one of the following sentences. Which one? We have an elf and we have a troll. The elf always tells the truth, the troll always lies, which is another way of saying that the elf cannot tell lies and the troll cannot tell truths. So let's start with E. One and only one of us is telling the truth. Now, this is a general truth, so the elf has to say it, but the troll can't. Let's look at D. I always lie. For the elf, this is a lie, and so the elf will not say it. For the troll, this is the truth, and so the troll actually also cannot say it. Let's look at C. We are both telling the truth. This is a lie, so the elf cannot say it, while the troll has to. In B, you are telling the truth. For the elf, this is a lie, since the troll always lies, so the elf cannot say it. For the troll, this is the truth, and actually the troll also cannot say it. Now let's look at A. I am telling the truth. For the elf, this is the truth, so the elf can say it. For the troll, well the troll never tells the truth, so this is a lie, and that means that the troll can say it. Therefore, the answer is A. Problem 14. Mary has exactly 10 white cubes, 9 light gray cubes, and 8 dark gray cubes, all of the same size. She glues all of these cubes together to build a big cube. One of the cubes below is the one she builds. Which one? So 4 of these cubes must be impossible to build in some way. But before we start, let's note that each cube has 8 cubes inside it that we cannot see. Okay, cube A has 9 dark gray cubes showing. That's impossible for Mary to build. Cube C has 11 white cubes showing, which is also impossible for Mary to build. Cube D, like cube A, has 9 dark gray cubes, so again, D cannot be the answer. And E has 10 light gray cubes, so E also cannot be the answer. B is the only cube left, and we see that it has 9 light gray cubes in the front and 10 white cubes behind them. The 8 dark gray cubes are just that little cube that we pointed out earlier, so the answer is B. Problem 15. The diagrams show 5 paths from X to Y marked with a thick line. Which path is the shortest? Let's start by noting that every path has 5 straight segments. Likewise, every path has 2 segments on the middle circle. Therefore, it's just the large circle and the small circle that will make the difference. A has 2 large segments and 2 small segments. B has 2 large segments and 1 small segment. C has 1 large segment and 2 small segments. D has 2 large segments and 2 small segments. And E has 2 large segments and 1 small segment. A and D both have 4 segments, which makes them bigger than B, C, or E. B and E have the same size, which is suspicious. But also, 2 large and a small is bigger than a large and 2 smalls, so B and E can't be right. Therefore, the shortest path is C. Problem 16. A father kangaroo lives with his 3 children. They decide on all matters by vote, and each member of the family gets as many votes as his or her age. The father is 36 years old, and the children are 13, 6, and 4 years old, so the father always wins. How many years will it take for the children to win all the votes if they all agree? The father now has 36 votes, while the children have the sum of their age, or 23. Each year, the father gains a vote, while the children each gain a vote, so they gain 3 total. The next year, the father will have 37 votes, while the children will now have 26. The year after that will have 38 and 29, respectively. Since the children's vote grows faster than the father's, eventually the children will have more votes than the father. We can see when this happens, and it happens in seven years. Therefore, the answer is C. Problem 17. Giorgio has two identical pieces of wire of this shape. Which of the following shapes can it be made by putting together these two pieces? To make our search easier, we can focus on this long part of the piece. Let's try to find it in each of these shapes. In A, we see two long segments right here. In B, there are two long segments on either side. Same with C. And in D, there's two long segments right here. E only has one long segment, which immediately disqualifies it as a possible shape. Just for good measure, let's show the separators in A, B, C, and D. Problem 18. Amy glues the six stickers shown onto the faces of a cube. The pictures show the cube in two different positions. Which sticker is on the face opposite the face with the mouse? From the first cube, we see that the mouse is just behind the ladybug. On the second picture, we actually also see the ladybug. And since it's behind is on the back of the cube, we know that the mouse must be there. Therefore, opposite the mouse is the dog. So, the answer is D. Problem 19. The picture shows the friendships of the six girls. Anne, Beatrice, Chloe, Diana, Elizabeth, and Fiona. Each number represents one of the girls, and each line joining two numbers represents a friendship between those two girls. Chloe, Diana, and Fiona each have four friends. Beatrice is friends with only Chloe and Diana. Which number represents Fiona? Since Chloe, Diana, and Fiona each have four friends, their numbers must be the three that have four lines coming out. Therefore, they must be one, three, and four. Beatrice only has two friends, Chloe and Diana. So, Beatrice must be the only point with two lines coming out of it. Therefore, five is Beatrice, and one and four must be Beatrice's friends, Chloe and Diana. This leaves three as the only possible friends. Chloe and Diana. This leaves three as the only possible choice for Fiona. Therefore, the answer is B. Problem 20. Mary put the same amount of liquid in three rectangular vessels. Viewed from the front, they seem to have the same size, but the liquid has risen to different levels in the three vessels. Which of the following images represents the three vessels when viewed from above? We have three volumes of water in the shape of rectangular prisms. Though they have different heights, they have the same widths. Since the volume of a rectangular prism is width times length times height, the lengths, or the view from above, will have to be balanced with different heights. The middle vessel is the highest, the first vessel is next, and the third vessel has the lowest volume of water. As the heights increase, the lengths have to decrease in order to maintain the same volume. Therefore, the lengths should be in the reverse order of the heights. We see that this means that two should be the shortest. Only answers A and B have vessel two as the shortest. In A, vessel one is next, while in B, vessel three is next. Since the lengths need to go two, then one, then three, A has the correct lengths shown. Problem 21. What does the object in the picture look like when viewed from above? Let's start by examining the quadrilateral made of two black diagonal pieces, a black inner piece and a gray outer piece. We see the same quadrilateral in B, D, and E. However, in A, we almost have this quadrilateral, but with a black outer piece. Therefore, A cannot be the right answer. In C, we also almost have this quadrilateral, but it has a gray inner piece, so C cannot be the right answer. Next, let's move clockwise from the gray outer piece and examine the two black pieces. B has these pieces. D, however, only has one black piece, so D is wrong. And E also only has one black piece, so E cannot be right either. That leaves us with the answer B. Problem 22. Three small squares are drawn inside a larger square as shown. What is the length of the line marked with a question mark? It's important to note that every shape in this problem is a square. So we have our square, and we know that its height must be the same as its base, or 28 centimeters. Now we have this 15 centimeter height, which can give us the height of the square above it. We know that together they must make 28, so we subtract 28 minus 15 to get that the height of that square is 13. And its base, because it is a square, is also 13. We can do the same thing for the length 22 and find the height of the square above it. Doing 28 minus 22 gives us that the height must be 6, and the base is 6 as well. Now we can do this process backwards to find the height of the middle square, and then the question mark height. Using the fact that the base is 28 minus 6 minus 13 gives us that the height must be 9. Therefore, the missing height must be 28 minus 9, which is 19. So our answer must be E. Problem 23. Nine tokens are black on one side and white on the other. Initially, four tokens have the black side up. In each churn, you have to flip three tokens. What is the least number of churns you need to have all tokens showing the same color? To start, let's note that one flip is not possible. If we flip all three black tokens, we will have one remaining. And if we flip all three white tokens, we'll still have two white tokens remaining, so one is not possible. Two, however, is possible. First, let's flip two black tokens and one white token. That way, we have three black tokens remaining, and now we can flip all three to get an all-white set. Therefore, the answer is two flips, or B. Problem 24. Which of the following options will definitely balance the third scale? We see that we have an arrow on the first scale, as well as two boxes on the second scale. Then we could maybe combine these scales so that we at least have what we need on the third scale. Now notice that we have two extra boxes and two extra triangles on our scale, so we can take off the boxes and take off the triangles. Now we have recovered the third scale. In that case, what we see on the right of our new scale should be what's on the question mark. Therefore, we need three circles and a triangle, so that's answer C. Problem 25. Ten people each order one scoop of ice cream. They order four scoops of vanilla, three scoops of chocolate, two scoops of lemon, and one scoop of mango. They top the ice cream scoops with four umbrellas, three cherries, two wafers, and one chocolate chip. They use one topping on each scoop in such a way that no two ice cream scoops are alike. Which of the following combinations is not possible? So we have our ice cream flavors. We have four vanillas, three chocolates, two lemons, and one mango. Now we need four umbrellas, but for that to happen, we need a different flavor under each umbrella, so we can assign the umbrellas like this. Now we need three cherries and a different flavor under each one, so we could do this. Same, we need two wafers, so that's a vanilla and a chocolate wafer, and then we have one chocolate chip on the vanilla. Looking now, we have a chocolate with a cherry, we have a mango with an umbrella, we have a vanilla with an umbrella, and we have vanilla with a chocolate chip. What we do not have is the lemon with a wafer. Problem 26. We call a three-digit number nice if its middle digit is greater than the sum of its first and last digits. What is the largest possible number of consecutive nice three-digit numbers? Let's say that we started with a sequence like this, where we go into two different decades, 379, 380, 381. This is impossible. Looking at 379, we see that it is not nice. In fact, any number of a ones digit of 9 won't be nice, because that 9 will be greater than or equal to the middle digit, so this sequence is impossible. To get the most nice numbers possible, our middle digit should be as big as possible, so it should be 9. Our hundreds digit should be as small as possible, since it won't change, so we should put a 1 there. Now, let's start with 0 as the ones digit, and continue up. If we try and do 198, we'll find that 1 plus 8 equals 9, so 198 is not nice. All of these numbers, however, are nice. Since there are 8 numbers here, the answer is D, or 8. Problem 27. Magnus has to play 15 games in a chess tournament. At a certain point during the tournament, he has won half of the games he has played. He has lost one third of the games he has played, and two have ended in a draw. How many games does Magnus still have to play? We can analyze all the games he has played so far, the wins, the losses, and the draws. Half of his games were wins, and a third were losses. Let's find out what fraction of the games he has played so far are draws. To do this, we can subtract a half and a third from 1. By finding common denominators, this is equal to 6 6ths minus 3 6ths minus 2 6ths, which is equal to 1 6th. Therefore, 1 6th of the games that Magnus has played so far are draws, which is equivalent to 2 games. If 2 draws makes 1 6th of Magnus's games, then multiplying both sides by 6 tells us that he has played 12 games so far. In that case, since he has 15 games to play, we can subtract 15 minus 12 to get 3, and that will be the answer, B. Problem 28. Vadim has a square piece of paper divided into nine cells. He folds the paper as shown, overlapping horizontally and then vertically so that the gray square ends on top. Vadim wants to write the numbers from 1 to 9 into the cells so that once the paper is folded, the numbers will be in increasing order with number 1 in the top layer. What numbers should he write instead of A, B, and C? Let's color code our columns so we can examine them better. We see here that all of the gray pieces should be on top of the blue pieces, and both of those should be on top of the yellow pieces. Let's unfold. Since all the grays are on top, those should correspond to 1, 2, and 3. Now let's look how the grays are connected to the blues. Each time we fold, we switch the direction of the column. That is to say, the bottom of the gray column touches the top of the blue column. Here. So, the blue column should be in reverse order and go 4, 5, 6 from the bottom. Similarly, the top of the blue column touches the bottom of the yellow column. Here. So, the yellow column should have switched back to normal order and go down 7, 8, 9. Now, we can simply look and see that A is 6, B is 4, and C is 8. That corresponds to answer A. Problem 29. Don builds a pyramid using balls. The square base consists of 3x3 balls. The middle layer has 2x2 balls, and there is one ball at the top. There is glue at each contact point between the 2 balls. How many glue points are there? Here are the 3 layers of balls. Any ball that is not in the bottom layer rests on top of 4 other balls. That creates 4 glue points, as shown here. Therefore, all the balls in the top 2 layers will have 4 contact points. Since there are 5 balls, we can multiply 4 times 5 to get 20 points total. There are also contact points between balls within layers, as shown here. Then we have another row of 3 right below this, and then more contact points going vertically. This creates 2 sets of 2x3, or 6 contact points. In the middle layer, we have more contact points. This creates 2 sets of 2x1, or 2 contact points. We can add these up. That makes 2 sets of 2 plus 2 sets of 6, or 16 contact points. And finally, we can add together all our contact points to get a total of 36. That gives us the answer E. Problem 30. The figure shows a map of some islands and how they are connected by bridges. A postman has to visit each island exactly once. He starts on the island marked Start and would like to finish on the island marked Finish. He has just reached the black island in the center of the map. In which direction should he move to be able to complete his route? To show the correct direction, we have to examine which paths through the islands are possible. The best way to start is to work backwards from Finish. Let's see if it's possible to get to Finish from the island to the left of Finish. If this were the case, then what happens when the postman gets to the island below Finish? The postman gets stuck. He cannot move forward since he'll eventually visit Finish from the left, so he shouldn't come from the left, but from below. Then we know we have to end up at this fork two islands below Finish. Can we get here from the left? Again, no, because if we go to the island below this fork island, we will be stuck, unable to advance to the fork, so we must come from the bottom corner. Moving backwards, we come to another fork, and another similar situation. We must avoid being stuck again. We continue on around the bend. Here, if we come in from the left, our path will be much too short, skipping many islands, so we come from the right. Now, can we get to this spot by leaving south from the black island? No, because again, we'd have a dead-end island. We must continue around to the black island. Now it is evident that if a path through the islands is possible for the postman, that he must go east from the black island. Therefore, the answer is either B or E, but a path is in fact possible, and it's shown here, so the answer must be B.

Math Kangaroo

problem-solving

logical reasoning

arithmetic calculations

geometry

probability

pattern recognition