This is the Math Kangaroo Solutions Video Library, presenting solution suggestions for Levels 7 and 8 from the year 2019. These solutions are presented by Lucas Neliskowski. 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 and listening to the question, pause the video and try to solve the question on your own. Question 1. Which cloud contains four even numbers? First, we must find out what an even number is. It is any number that is divisible by 2, such as 2, 4, 6, 8, 10, and so on. So we can start looking at the different clouds. Cloud A has 3, 9, 6, and 5. Only 6 is an even number, so it is not A. B has 33, 3, 13, and 23. None of those are even numbers, as none of them can be divided by 2 to get a whole number. C has 3, 30, 27, and 9. Here, the only even number is 30, so it cannot be C. Cloud D has 1, 9, 6, and 3. Again, there is only one even number, which is 6, and the others cannot be divided by 2 by getting a whole number. When we look at cloud E, which has 10, 2, 34, and 58, this cloud has 4 even numbers, so the answer is E. Question 2. How many hours are there in 10 quarters of an hour? So we know that 1 hour is equal to 60 minutes. Then 1 quarter of an hour will be 60 minutes divided by 4, which gives us 15. So 1 quarter of an hour is 15. However, we want 10 quarters of an hour, so we multiply by 10. With that, we get 150 minutes. So now we have to convert 150 minutes to hours. We can do this, and we find out that there are 2 intervals of 60 and 1 of 30. So that's 1 hour plus 1 hour plus 30 minutes, or 2 hours and 30 minutes, or answer E, 2 and a half hours. Question 3. A 3x3x3 cube is built from 1x1x1 cubes. Then, some cubes are removed from front to back, from left to right, and from top to bottom, as shown. How many 1x1x1 cubes are left? So a 3x3x3 cube will be 3x3x3, or 27. So this cube is made up of 27 1x1x1 cubes. Then, we know that we remove 6 cubes, 1 for each side of the cube. Now we also end up removing the cube in the very center, so we can subtract 1 more. So we have 27 minus 1 times 6 minus 1, which gives us 27 minus 7, which will lead us to our answer, which is 20. See? Question 4. 3 rings are linked, as shown in the diagram. Which of the following diagrams also shows the 3 rings linked in the same way? If we look at our given figure, we can see that the white ring passes through the gray ring. So that means answers B, C, and E are invalid. Now, we also see that the black and the gray rings do not connect to each other, so that means answer A cannot be correct. This only leaves us with answer D. Question 5. Which of the diagrams below cannot be drawn without lifting your pencil off the page, or drawing along the same line twice? Now, if we look at a regular box, we can easily draw it without lifting our pencil. Adding different lines, such as C or E, will still make the task just as easy. Same with A. We will have to just draw a circle around the box. Now, if we look at shape D, we try to draw it out like so. Then we notice, if we are following the line, we get to a point where we have to choose between one or the second direction. And this means this shape cannot be drawn without lifting our pencil or passing over a line a second time. So our answer is going to be D. Question 6. Five friends met. Each of them gave a cupcake to each of the others. They then ate all the cupcakes they had been given. As a result, the total number of cupcakes they had decreased by half. How many cupcakes did the five friends have at the start? So if there are five friends, and each one gives a cupcake to each one of the others, that means one friend will give out four cupcakes. Since there are five friends, we will do 5 times 4, which means there is a total of 20 cupcakes distributed in this way. Now, we know that half of the cupcakes have been eaten. So if we just multiply 20 by 2, we will get our answer. Our answer is 40. D. Question 7. In a race, Lothar finished before Manfred. Victor finished after Jan. Manfred finished before Jan. And Eddie finished before Victor. Who finished last of these five runners? Let's label them. Lothar will be L. Manfred M. Victor V. Jan J. Eddie E. Now, since we know Lothar finished before Manfred, we can put L in front of M. We also know that Victor finished after Jan. We also know that Manfred finished before Jan. So if we put V before J and set that aside, and now we deal with Manfred finishing before Jan, we can move the J before the M. And since we know that Victor is before Jan or J, then we can put the V over here. Finally, we know that Eddie finished before Victor. So Eddie will be in front of Victor. So now we know who finished last. Lothar was first, and last was A. Victor. Question 8. The pages of the book Julia is reading are all numbered. The numbers used on the pages contain the digit 0 exactly 5 times and the digit 8 exactly 6 times. What is the number on the final page? Our first number or page will be page 8, as it contains the digit 8. Next, we will have 10 as it contains the 0. 18, 20, 28, 30, 38, 40, 48, 50, and 58. For this, we have had 5 0's and 6 8's. So we know that our answer will be B, 58. Question 9. A large square is divided into smaller squares. What fraction of the large square is colored gray? If we take a closer look at the square, we notice it is initially divided into fourths. So the large bottom right square that is colored gray is 1 fourth of the area. Now, the top left square is also divided into 9 parts, 7 out of 9 of which are colored gray. So that has an area of 1 fourth times 7 ninths. Do this, and we get 7 over 36, which is the product of 1 fourth and 7 ninths, plus 1 fourth, which is the other gray square. This can be simplified to 7 thirty-sixths plus 9 thirty-sixths, and then we get 16 thirty-sixths. If we simplify this fraction, we get 8 over 18, and then 4 over 9. And with this, we get our solution, which is the 4 over 9. Question 10. Andrew divided some apples into 6 equal piles. Boris divided the same number of apples into 5 equal piles. Boris noticed that each of his piles contains 2 more apples than each of Andrew's piles. How many apples does Andrew have? So first, we can label apples as A. Since we know the same amount of apples will be divided into 6 as well as 5 piles, the number has to be divisible by both 6 and 5. Now, A, or apples, divided by 6 will give us X, and A divided by 5 will give us X plus 2. We just start plugging in the different answers we have, we'll get our answer. Now, if we say apples are 60, A equals 60, we do 60 divided by 6 gives us 10, and 60 divided by 5 gives us 12, which satisfies our initial thought. The 6 piles will have 10 apples each, and if there are 5 piles, they'll have 12 apples each, meaning 2 apples more than the other set of piles. So our answer will be A, 60. Question 11. A different 4-digit number was written on each of three pieces of paper. These pieces of paper are arranged so that three of the digits are covered. It's shown the sum of the three 4-digit numbers is 10,126. Which are the covered digits? If we set up an equation like so, leaving the digits unknown as blank, and writing out the sum on the bottom, we can get our answer. First, we'll add together all the digits in the ones place, 3, 7, and 6, which gives us 16. So the 6 and 10,126 is correct, as we would think, and we move a 1 over to the 10 spot. Next, we add all the digits in the 10 spot. We get 1 plus 4 plus 2, which gives us 7. Now we know this answer has to end with a 2, so the only way to do that with a 1-digit number is to do 7 plus 5 equals 12. So that way the 2 will be on the bottom, and we'll carry a 1 to the hundreds place. Here we do the same thing, 1 plus 2 plus 1 plus something, so 4 plus something has to give us a number that ends with a 1. 7 plus 4 will give us 11, so 7 will go here, and 1 will be carried over. And again, 1 plus 1 plus 2 plus something will give us something ending with a 0. 4 plus 6 gives us 10, and this lets us carry on the last 1 digit to the 10,000 spot. So with this, the 3 digits we filled in are 6, 7, and 5, or if we order them in a different way, we get our answer a 5, 6, and 7. Question 12. In the diagram, PQ is equal to PR, which is equal to QS, and the angle QPR is equal to 20 degrees. What is the size of angle RQS? Let's take a closer look at our triangle. Now we see the segment PR and PQ as well as QS are all equal to each other. Now since PQ and PR are equal length, we know that triangle PQR is an isosceles triangle, meaning that angles PQR and PRQ will be the same. So with this information, angles PQR plus angle PRQ plus angle QPR, all the angles in triangle PQR total together 180 degrees, which is true of any triangle. Now since we've established that it is an isosceles triangle, angle PQR is equal to angle PRQ. So we can rewrite our formula and get 2 times angle PRQ plus 20 degrees is equal to 180 degrees. Subtract 20 degrees from both sides and get 2 times angle PRQ is equal to 160 degrees. So angle PRQ is equal to 80 degrees, and that in turn also tells us that angle PQR is also 80 degrees. Next up, we know that angle QRS plus 80 degrees will be 180 degrees, since angle PRQ as well as angle SRQ will together be 180 degrees, as that is true of any straight line. So this gives us angle QRS is equal to 100 degrees. Next, with our information, we know that angle QPR is equal to angle QSR, which means this will be 20 degrees. We know this since segment PQ and segment QS are the same length, so triangle PQS is also an isosceles triangle. With this information, we can now go on to our final step with RQS plus angle SRQ plus angle QSR equal to 180 degrees. Since we know what angle SRQ and angle QSR are, we get our answer of angle RQS plus 100 degrees plus 20 degrees is equal to 180 degrees. When we subtract these, we get angle RQS is 60 degrees, so our answer is B, 60 degrees. Question 13. Which of the following 4x4 tiles cannot be formed by combining the two given pieces? We look at our 4x4 grid that is given by E, and we look at our long right angled shape, this one, which has gray piece in the corner and then a gray piece separated by a white piece, and we see that shape E has no way of fitting this sort of piece since it has two gray pieces in its corners, but then the other gray pieces do not fit in the same spots, so the correct answer is E. Question 14. Alan, Bella, Claire, Dora, and Eric met at a party and shook hands exactly once with everyone they already knew. Alan shook hands once, Bella shook hands twice, Claire shook hands three times, and Dora shook hands four times. How many times did Eric shake hands? We write out our five friends, Alan, Bella, Claire, Dora, and Eric. We can look. Now, since we know that Dora shook hands four times, it means that Dora shook hands with every one of the people here, including Eric. Now, we also know that Alan shook hands once, so that means Alan only shook hands with Dora. Next, if we look at Claire, we shook hands three times. She'll shake hands with Eric. She already shook hands with Dora, which we've established. That's the second, and third, she shook hands with Bella since Alan will no longer be shaking hands. With this, we know that Bella has shaken the hands of Dora as well as Claire. We know that Claire has shaken hands with Bella, Dora, and Eric, and Dora has shaken hands with everyone. This means that Eric shook hands twice, so the answer will be B, 2. Question 15. Jane is playing basketball. In a series of 20 shots, Jane scored 55% of the time. After five more shots, her scoring rate increased to 56%. On how many of the last five shots did she score? We do 20 shots times 0.55, the 55%, we get 11 times she scored. We do 20 shots and then plus five more extra shots. That gives us 25 shots in total. We multiply by 0.56, the 56%, we get 14 times scored. Now, if we want to find out how many of those were scored in the final five shots, we just have to take the number of scores she got in 25 shots, subtracted from the number of scores she got in 20 shots, so 14 minus 11, which gives us our answer, which is 3C. Question 16. Kathy folded a square sheet of paper exactly in half twice, and then cut it in the middle twice, as shown in the diagram. How many of the pieces that she obtains are squares? Now, if we look at a square sheet of paper that has been folded in half exactly twice, it'll look like this. And when it is cut down the middle twice, like Kathy did, then the cuts will be made in these spots. Now, the pieces that are squares will be highlighted in blue. So there's 1, 2, 3, 4, and the final one will be in the middle, the large one. So in total, there are five. So our answer will be 5C. Question 17. Michael has dogs, cows, cats, and kangaroos as pets. He tells Helen that he has 24 pets in total. That one-eighth of them are dogs, three-fourths are not cows, two-thirds are not cats. How many kangaroos does Michael have? And we know there are 24 animals in total. If we take 24 animals, and we know that one-eighth of them are dogs, we can multiply 24 by one-eighth, which gives us three dogs. We know there are 24 animals, and three-fourths are not cows. That means 18 are not cows. If we do 24 minus 18, we get 6 cows. If we do 24 animals times 2 divided by 3, which are not cats, we get 16 animals are not cats. If we do 24 minus 16, we get 8 cats. Now if we take these three totals, then we can add them together and get 17 dogs, cows, and cats. If we take the number of animals, 24, and subtract them by 17, we'll get the number of kangaroos. And we get our answer, 7D. Question 18. Some identical rectangles are drawn on the floor. A triangle with a base of 10 centimeters and a height of 6 centimeters is drawn over them, as shown. And the region inside the rectangles and outside the triangle is shaded. What is the area of the shaded region? So you get closer look at our rectangles and triangle. Now looking at the diagram closely, we can see that there are five rectangles along the length of the triangle of 10 centimeters. So if we do 10 centimeters divided by five, we get two centimeters, which is one side of each of the rectangles. We do the same thing for the height of the triangle, six centimeters. We notice there are four rectangles. So we do six divided by four, gives us 1.5 centimeters for each of the rectangles. So each one rectangle is two centimeters by 1.5 centimeters. Now, to get the area of one rectangle, we just multiply the sides and get three centimeters squared. Next, we take three centimeters squared and we multiply it by 14, the total number of rectangles we have. That we get 42 centimeters squared. Next, we do six centimeters times 10 centimeters divided by two to find the area of the triangle. That we get 30 centimeters squared. Now, to find the area of the shaded region, we do 42 centimeters squared, which is the total area of the rectangles minus 30 centimeters squared, which is the area that the triangle takes away. And when we do 42 minus 30, we get our answer, which is 12 centimeters squared, B. Question 19. Julio has two cylindrical candles with different heights and diameters. The first candle lasts six hours while the second candle lasts eight hours. He lit both candles at the same time and three hours later, both candles were the same height. What is the ratio of their original heights? We draw one candle with height H and the second candle without height I. Then we know that three hours later, the first candle with height H will have a height of H divided by two. Since it is after three hours with a total of six hour length. Second candle will be five I divided by eight. Since it is an eight hour candle, five hours left out of eight since three have passed. We know the ratio will be H divided by two is equal to five I divided by eight. With this, we can tell the ratio of their original heights, which is five to four, answer C. Question 20. Eileen wants to create a path of matches using as few matches as possible. She places each match on the piece of paper like the one shown along one of the dotted lines. Her path returns to the left-hand end of her original match. The numbers shown in some of the cells are equal to the number of matches around that cell. How many matches are in this path? Take a look at our grid. We can start off by placing our first match, which we know will be here since we know the path will return to the left-hand end of the original match. Next, since we know that the numbers represent how many matches are on each side, we can put down three matches in these two spaces since that is the only way they can be put down. Next, we will connect these together. Now we can start placing our new matches. Since we see a grid of zero, we will have to go upwards away from that three. Next, looking back at our original match, since it is on a box of two, and then to the right of that there is a box of one, there will have to be a match like so. And then the fastest way, and only possible way to connect this, is by putting the matches like so. When we count all these matches, we will get our answer. Our solution will be 16C. Question 21. The integers from one to N, inclusive, are equally spaced in order around a circle. The diameter through the position of integer seven also goes through the position of 23 as shown. What is the value of N? Now, if we take a look at a different example of a circle, such as this one, which goes from one to 12, we can look at how this works. When we have a line that goes through as a diameter, such as a line that will go from 12 to six, we see that 12 minus six will give us six. Now, if we do 12 and move forward six positions, we get back to six. And if we do one plus six, we get to seven, which if we look at a diameter from one, we will see we get to seven. So with this, we know that if we take one point and subtract it from the point opposite to it, we get a number that when multiplied by two will give us the total amount of points on that circle. Following the same logic, we can do 23 minus seven, which gives us 16. And then we can multiply by two. And this will give us our answer, which is 32B. Question 22. Liam spent all his money buying 50 soda bottles at the store for $1 each, hoping to make a profit. He raises the price and sells all 50 soda bottles at a new higher price. After selling 40 bottles, he has made $10 more than he started with. He then sells all the remaining bottles. How much money does Liam have now? First off, we know that Liam bought 50 soda bottles for $1 each, which means Liam initially had and spent $50. Now, if we know that 40 bottles will end up with a new price, we'll end up making him $60, $50 plus 10 more, and we can do $60 divided by 40 bottles. With this, we know that he is selling each bottle for $1.50 or $1.50. Next, if we know that each bottle is sold for $1.50, we can do 60 plus $1.50 times 10, which is the remaining bottles. With that, we get $60 plus $15. That will give us our answer, which is $75B. Question 23. Natasha has many sticks of length one. The sticks are colored either blue, red, yellow, or green. She wants to make a three-by-three grid, as shown, so that each one-by-one square in the grid has four sides of different colors. What is the smallest number of green sticks that she could use? If we look at our grid and we start outlining, like so, then we will quickly notice that there will be an outline that makes five distinct boxes, like so. We have the four corner boxes and the one in the middle. Using this information, if we strategically place our green sticks, we will have to have at least five. We put one here, then it will complete one box. We put a green stick here, this will be included in two different boxes. So the top row, each of the three squares has at least one green stick. Next, if we put one here, that includes the next two boxes. One here includes another two boxes, and one stick here includes the last two boxes. So now we have a green stick for every one box that is one-by-one in the three-by-three grid. So the smallest number of green sticks will be five, answer C. Question 24. An ant would like to walk along a marked line on the surface of a cube until it returns to its starting point. From which one of the following nets could a cube be made so that such a journey is possible? So to do this problem, we want all the lines on the cube, once it's folded, to connect. We take a look at the net, since each cube is constructed of the same net, we notice that if we folded it, these two would touch. Now, knowing this, we can look at our answers. Answer A is an invalid option, since when we look at those two sides, when folded, there will not be a line that connects them. Same is true for B. When looking at the two sides we marked, they will not connect into a line or a path that the ant can walk on. Same is true for C. Even though there are paths on both the cubes, they're not on the sides and will not connect when they're folded together. The same is true for D. They only connect when we look at our correct answer, which is answer E. Question 25. Elizabeth had a large bag of 60 chocolates. She started by eating one-tenth of them on Monday, then one-ninth of the remainder on Tuesday, then one-eighth of the rest on Wednesday, then one-seventh on Thursday, and so on, until she eats half of the chocolates remaining from the previous day. How many chocolates does she have left? So if we know that we start off with 60 chocolates and Elizabeth eats one-tenth of them, she will have eaten six chocolates, since we multiply 60 by one-tenth. Then to find out how many chocolates we have left, we do 60 minus six, which leaves us with 54 chocolates. The next day, we start off with 54 chocolates. And following our pattern of going one-tenth, one-ninth, one-eighth, one-seventh, and so on, all the way down to one-half. Next, we will multiply by one-ninth. 54 times one-ninth gives us six. That means Elizabeth ate six more chocolates during this day. Then we do 54 minus six, we get 48 chocolates. And we can do this once again. 48 times one-eighth gives us six. 48 minus six will leave us with 42. Then 42 times one-seventh gives us six. 42 minus six will give us 36 chocolates. Now, if we keep doing this, we will see that there are six chocolates being eaten every day, because of the way that the math plays out. Now, we want to get to one-half, since that is what the question is asking, since we want to know how many chocolates she will have left when she has to eat half of the chocolates remaining. Now, we are at 12 chocolates. There are 12 chocolates, she will eat one-half. When she eats one-half, that is six. Then 12 minus six gives us six chocolates. So that means she has six chocolates left. And the answer is 6e. Question 26. Prab painted each of the eight circles in the diagram, red, yellow, or blue, in such a way that no two circles that are joined directly are painted the same color. Which two circles are necessarily painted the same color? If we look at the diagram that is painted by Prab, we can notice we have a triangle made up of points two, eight, six, and five. Now, if we assume that two will be yellow, and six is red, then we notice that five is connected to both two and six, meaning it will have to be blue. However, eight is also connected to two and six, so that one would also have to be blue. This would be true if we switched out two, instead of being yellow, to being blue. Then five and eight would have to be yellow. So no matter what, five and eight will have to be the same color, since two and six will cancel them out. So our answer is five and eight, a. Question 27. When Ria and Flora compared their savings, they found that the ratio of their savings was five to three. Then Ria bought a tablet for $160. The ratio of their savings changed to three to five. How many dollars did Ria have before buying the tablet? So at first, the ratio is five to three, or five times the amount of dollars divided by three times the amount of dollars for both Ria and Flora. Then our ratio switches to three to five, which is also equal to 5D minus 160, all divided by 3D, since the ratio is three to five, after Ria spends $160. We set up our fractions like so. We can cross-multiply to get our answer. We get five times the difference of 5D and 160, equal to three times 3D. This is simplified, we get 25D minus 800, equal to 9D. If we subtract 9D and add 800, we get 16D equals 800. Simplifying further, 2D equal to 100, and D equal to 50. Now, we want to know how much money Ria had before buying the tablet. Since we know that Ria had five times D, we just do five times 50, and this gives us answer, which is 250C. Question 28. Several three-player teams enter a chess tournament. Each player in a team plays exactly once against every player from all other teams. For organizational reasons, no more than 250 games can be played in total. At most, how many teams can enter the tournament? Now, if we say that the total number of teams is N, then we can start setting up our equations. We will have three times the difference of N minus one. This will mean one player will play how many people. Since a player cannot play against the people in his team, it is N minus one. For example, if there are two teams, one player would only play against three people, all three people from the other team. Next, we can also multiply by 3N. Since there will be three players in each one team. So each player will play against the same people, and then we'll multiply this by three. Then we will also divide this entire number by two. Since every game will include two players, and if we did not divide by two, we would double count the games. And in total, this number has to be less than or equal to 250. Since for organizational reasons, there cannot be any more than 250 games played. So we need to find out the number of groups that are in this tournament. We start off with our smallest value, which is seven. If we put in seven, we get the equation three times the difference of seven and one times three times seven divided by two. Simplified, three times six times 21 divided by two, or 18 times 21 divided by two, or nine times 21, which gives us 189. This value is indeed less than 250. So it is a valid option. However, to make sure that there cannot be eight teams, we should put in that value. Since if there can be eight teams and the number we get, for example, is 249, then that would be the most amount of teams. So we put in eight. This we get three times seven times 24 divided by two, 21 times 12, or 252. So since eight teams gives us 252 games, and our maximum number of games is 250, there cannot be eight teams. So our answer will be seven E. Question 29. The diagram shows the square A, B, C, D with P, Q, and R as the midpoints of the sides D, A, B, C, and C, D, respectively. What fraction of the square A, B, C, D is shaded? If we look at our diagram closely, we notice that R is the midpoint of D, C. And since if we go from point A to point R and from point B to point R, we can see that we have a triangle A, R, B. Since R is the midpoint, we know A, R is equal to B, R, making triangle A, R, B an isosceles triangle. Now, if we divide our square into half like so, then we notice an important factor. We filled in triangles like so, triangles A, D, R, and triangles B, C, R, they would be half of the given area. So that means triangle A, R, B is equal to half the entire area of the square. However, this does not mean that is our answer since we still have the smaller triangle at the bottom that is still blank. So our answer will be one half minus that triangle. Now, we also know that P and Q are midpoints of A, B, and B, C. If we redraw this like so, making a line of P, Q, and we turn this sideways, then we can notice one thing. A, Q, and B, P intersect each other, meaning that this shape is divided into four equal parts, meaning this part is one fourth of the area. Now, this is one half of the entire shape. So that means that triangle is one fourth times one half or one eighth. So we take the area of A, R, B, and subtract it from one eighth. So we can do one half minus one eighth, which is equal to four eighths minus one eighth. And that gives us our answer, which is three eighths, E. Question 30, a train is made up of 18 cars. There are 700 passengers traveling on the train. In any block of five adjacent cars, there are 199 passengers in total. How many passengers are there in the middle two cars of the train? First, let us draw out 18 cars. Now, since we know that in every five adjacent cars, there are 199 passengers, we can change this to follow a pattern, A, B, C, D, E, and then A, B, C. And if we went backwards, we would be C, B, A, E, D, C, B, A, E, D, and so on. So that means every five adjacent cars have 199 passengers in total. So that means A, B, C, D, and E, when totaled together, will give us 199. So this first segment of five is 199. Next segment, 199. Next segment, 199. Now we wanna figure out how much A, B, C has. To do this, we will do 700, the total number of passengers on the train, This will give us 103. Now, this means that A plus B plus C is 103. If we look at the mark above D and E, those are the two middle cars. To get D plus E, we must do A plus B plus C plus D plus E minus A plus B plus C, which will give us D plus E. And since we know the sum of A, B, C, D, and E, 199, and the sum of A, B, and C, 103, it is D plus E is equal to 199 minus 103. That gives us our answer, 96D.

Math Kangaroo 2019

Lucas Neliskowski

geometric puzzles

time conversions

problem-solving techniques

mathematical reasoning

numerical sequences

competitive math