This is the Math Kangaroo Solutions video library presenting solution suggestions for levels 11 and 12 from the year 2018. These solutions are presented by Lukasz Nalewskowski. The purpose of the Math Kangaroo Solutions library video 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 problem on your own. Question 1. The picture shows the calendar for a certain month of the year. Unfortunately, some ink spilled on the calendar and most of it cannot be seen. Which day of the week was 27th of that month? Draw out the calendar, we can have Monday through Sunday, and we can see that the second and third days are Thursdays and Fridays. So the first week will be starting off Wednesday, and the fifth will be a Sunday. We continue this on, Monday is the 6th, 13th, 20th, and also the 27th. Therefore, the 27th day of the month is a Monday, so the correct answer is A. Question 2. Which of the following numerical expressions has the highest value? To solve this, we must just take each possible solution and find out what it is, and compare them. First one is 2 minus 0 times 1 plus 8. We use the order of operations, we can simplify this to 2 minus 0 plus 8, which gives us 10. So A will be 10. B is 2 plus 0 times 1 times 8. Use the order of operations, we simplify to 2 plus 0, which is 2. For C, 2 times 0 plus 1 times 8 will give us 0 plus 8, giving us 8. D is 2 times 0 plus 1 plus 8, which gives us 2 times 9, which will give us 18. Finally, E is 2 times 0 plus 1 plus 8, which is 2 times 0 plus 9, which gives us 9. So the correct answer is D, 2 times 0 plus 1 plus 8. Question 3. The figure shows the floor plan to Renee's house. Renee enters her house from the porch and walks through each door exactly once. In which room does she end up? Take a look at the floor plan, and start off on the porch, we have to go through every door at least once. Follow the blue path, we end up in room 2. There is one other way to do this, following the red path, but we still end up in room 2. So the correct answer is B, 2. Question 4. This man has 7 stones and a hammer. Every time he hits a stone with the hammer, the stone breaks into exactly 5 smaller stones. He does this several times, so the following numbers can be the number of stones he may end with. We take this, and we do this every time. We will break off 1 stone into 5, take away 1 and break it into 5 more. We have a net gain of 4 stones every time. We can end up with 17 stones, however, we know that Thor had 7 stones at first, so we must add 6 more stones that were never broken. This will give us the answer D, 23 stones. Question 5. The figure shown is made of 10 cubes glued together. The figure is dipped into a bucket of paint, which covers the surface entirely. How many of the cubes will be painted on exactly 4 of their faces? We look at the shape, we notice that the 2 end cubes are only adjacent to 1 other cube, so only 1 face is covered. They will not only have 4 faces painted. The other 8 cubes are adjacent to a cube on at least 2 sides. Therefore, only 4 of their 6 sides will be painted. The correct answer is 10-2, so it's C, 8. Question 6. The following 2 statements are true. Some aliens are green, the others are purple, and green aliens live only on Mars, therefore it logically follows that blank. Since we know that Venus and Mars are different planets from different places, and there are only green aliens living on Mars, so there are no green aliens on Venus, all the other statements except for the correct one will be false. For example, the first statement, all aliens live on Mars, can be false, since purple aliens can live outside Mars. For the second option, only green alien lives on Mars, that can also be false, since there can be purple aliens living on Mars. For the third option, some purple aliens live on Venus, it can happen that all purple aliens live outside of Venus, and they all live on Mars. For D, all purple aliens live on Venus, it is not necessarily true that all purple aliens will live in Venus, they could be outside of Venus. And E, no green aliens live on Venus is the last option, so the correct answer is E. Question 7. 4 identical rhombuses and 2 squares are put together to make a regular octagon. What is the measure of the larger angle of each rhombus? To find the answer, all we must do is find the angle of a regular octagon. To do this, all we have to do is 8-2 times 180 divided by 8. Simplified gives us 6 times 45 divided by 2, or 3 times 45, this gives us 135. This will be the angle of one angle of the octagon. If we look at the size in the rhombus, then we will notice that one angle of the rhombus is also equal to one angle of the octagon. So the correct answer is A, 135 degrees. Question 8. There are 65 balls in a box, 8 of the balls are white, the rest are black. In one move, at most 5 balls can be taken out of the box, no balls may be put back in the box. What is the smallest number of moves needed to ensure that at least one white ball will be taken out? We take out 5 balls at most every time, then we will see that we take out 5, 10, 15, 20, and so on balls. The most we can take out, while still only taking out black balls, is 55 after 11 moves. After that, there will be 10 balls left in the box, 8 of which can be white. So the 12th move will ensure that we have at least one white ball taken out. So the correct answer is B, 12. Question 9. The faces of a rectangular brick have areas A, B, and C as shown. What is the volume of the brick? To find the volume of a 3-dimensional shape, we must do length times width times height. Since we know that the area of the faces are A, B, and C, we can represent the length, width, and height by lowercase a, b, and c for every side of the rectangular brick. We just do V equals the root of A times the root of B times the root of C. We can simplify this to get our correct answer, B, root A times B times C. Question 10. In how many ways can the number 1001 be written as the sum of two primes? Since the number 1001 is an odd number, we know that if we add two even or two odd numbers, we will get an even answer. So we will have to add together an odd and an even number to get the sum of 1001. The only even prime number is 2, and if we do 1001 minus 2, we get 999, which is not a prime number. Therefore, it is impossible to get the number 1001 as the sum of two primes, so the correct answer is A, none. Question 11. Two cubes of volumes V and W intersect. The part of the cube of volume V, which is not common to the two cubes, is 90% of its volume. The part of the cube of volume W, which is not common to the two cubes, is 85% of its volume. What is the relationship between V and W? We know that if we look at just the part where the cubes intersect, it is equal to 10% of or 15% of W, because we have the 90 and the 85% as stated in the problems question. If we set these equal to each other, because they are both talking about the same piece, we know that 0.1 V is equal to 0.15 W. This can be simplified to V equals 0.15 divided by 0.1 W, and this can be further simplified to 3 halves. So the correct answer is B, V equals 3 halves W. Question 12. A vase is filled up to the top with water at a constant rate. The graph shows the height h of the water as a function of time t. Which of the following can be the shape of the vase? Look at the graph of height and time. We notice that in the beginning, the height increases rapidly, and then as time goes on, it slows down. If we look at the first option, A, a cylinder, we know this cannot be the vase, since the cylinder's height would increase at a constant rate over time. If we look at the next shape, B, then we notice that it narrows, which means the height would be slowly increasing and then speed up, and then as it opens back up, it would slow down. So this cannot be the graph either, since it does not look like that. C is a sphere. It would slow down in filling up, and once it reached the midpoint, it would speed back up. So this cannot be the shape. E is just a cylinder and a sphere put together. The only answer left is D, a cone. Since it is narrow at the base, it would quickly fill up, and as it filled up more, it would widen and it would take longer to fill up, so it would slow down like the graph shows. So the correct answer is D, the cone. Question 13. Absolute value of root 17 minus 5 plus absolute value of root 17 plus 5 equals blank. We must first recognize that the root of 17 is a smaller number than 5. Since the root of 25 is 5, and the root of 16 is 4, therefore root 17 is smaller than 5. This allows us to replace root 17 with an X. The X becomes insignificant, as the 5 is the larger of the two numbers. Simplify this, you can get to 5 plus 5. This gives us the answer of A, 10. Question 14. An octahedron is inscribed in a cube with a side length of 1. The vertices of the octahedron are at the centers of the faces of the cube. What is the volume of the octahedron? We know that the centers of the faces of the cube are perpendicular to the top face of the cube, or vertices of the square parallel to the top face. The area of the square, we call it B, is half the area of any face of the cube. The distance from the center of the top face to the square, called H for height, is half of any edge of the cube. The octahedron consists of two pyramids, with a square as the common base. So if we use the formula for finding out the volume of a pyramid, then we can get that the volume of the octahedron is 1 third times B times H, and since there are two of them, multiply by 2. Which, simplified down, is 2 times 1 third times 1 half times 1 half, further simplifying into 1 third times 1 half, giving us the answer D, 1 sixth. Question 15. The vertices of a triangle are A, P, Q, B, point RS, and C, T, U, as shown. The midpoints of the sides of the triangle are the points M, negative 2, 1, N, 2, negative 1, and P, 3, 2. What is the value of P plus Q plus R plus S plus T plus U? Use the midpoint formula, and we get M is equal to P plus R divided by 2, which is negative 2, and Q plus S divided by 2 is negative 1. N will be R plus 2 divided by 2, 2, as well as S plus U divided by 2, negative 1. T will give us T plus P divided by 2, and that'll be 3, as well as U plus Q divided by 2, which gives us 2. If the totals, we can get 3 and 2. If we add those together, then our solution is D, 5. Question 16. Five predictions were made before the soccer game between Real Madrid and Manchester United. First prediction, the game will not end in a tie. Second prediction, Real Madrid will score. Third prediction, Real Madrid will win. Fourth prediction, Real Madrid will not lose. Fifth prediction, three goals will be scored. What was the final score of the game between Real Madrid and Manchester United if exactly three of the predictions came true? Suppose that Real Madrid will win, which is the third prediction, then predictions 1, 2, and 4 must also be true. So in total, those are four predictions. So we know that Real Madrid cannot win. So that is one of the false predictions. Now that we know that, we can suppose that Real Madrid will not lose. The only possibility then is a tie, but that contradicts the first prediction. So the number of goals will have to be even, but since it is three goals scored, this cannot be the case. At this moment, we know that Real Madrid will lose. So the first prediction is true. The third prediction is false, as well as the fourth prediction. The only triple of the true predictions is that the game will not end up in a tie. Real Madrid will score, and three goals will be scored. So the game must end with one goal for Real Madrid and two goals for Manchester United. So the correct answer is D, 1 to 2. Question 17. We cut out a regular pentagon from a lined piece of paper. In each step, we rotate the pentagon counterclockwise around its center by 21 degrees. The position after the first step is shown. What will we see when the pentagon first fits back in the opening? So for the pentagon to return back into an opening, it has to be rotated by a variable of 72 degrees, since the central angle of a regular pentagon is 360 degrees divided by 5, or 72 degrees. And since it is only rotated by 21 degrees every turn, then it has to be a number of steps in which 72 and 21 line up. We let k be the smallest number of steps after which the pentagon returns to the opening, and k is the smallest natural number of k times 21 degrees, and it has to be a multiple of 72 so that they line up. Now that 21 is 3 times 7, and 72 is 3 times 24. Since 7 and 24 are relatively prime, k will be 24. Multiply 24 by 21, we get 7 times 72. We use 24 steps, or 7 rotations of 72. Each rotation by 72 degrees keeps the pentagon in the opening, but the lines inside the pentagon are changing direction. After 5 rotations, the pentagon will be in its original position. After the 6th rotation, sides in the 1st and 2nd position will be in the 2nd and 3rd position. After the last, the 7th rotation, then they will be two spots displaced. So the correct answer will be b. Question 18. Which of these 5 numbers does not divide 18 to the power of 2017 plus 18 to the power of 2018? If we look at these, we can divide them into 18 to the power of 2017 times 1 plus 18. We can also separate it into 2 to the power of 2017 plus 3 to the power of 2 times 2017 times 19, since it is a prime factorization of the sum. So it's a multiple of 2 to the power of 3, which is 8, 2 times 3 to the power of 2, which is 18, 2 times 19, which is 38, 2 to the power of 4 times 3, which is 48. It's not a multiple of 28. It's not a multiple of 7. So the correct answer is c, 28. Question 19. 3 of the 5 cards shown are given to Nadia and the rest to Reini. Nadia multiplies the 3 values of her cards and Reini multiplies the 2 values of his cards. Turns out that the sum of the two resulting products is a prime number. What is the sum of the values of Nadia's cards? If 6 is among 2 of Reini's numbers, then Nadia must have a 3 or a 4, because Reini cannot have both of them. First case, Nadia's product and Reini's product are multiples of 3. So the sum of these products is a multiple of 3, but the sum is greater than 3, so it is not a prime number. Second case, the products are multiples of 2. The sum greater than 2, also not a prime number. So Nadia has to have a 6. And Reini cannot have a 3 or a 4, so those will go to Nadia. So we can give Reini the 5 and the 7. Their products are 72 and 35. The sum is 107. The number 107 cannot be divided, so it is a prime number. It is not divisible by 2, 3, 5, or 7. This fulfills a requirement. Now we just add Nadia's numbers. 6 plus 3 plus 4. That gives us the correct answer of B, 13. Question 20. Two rectangles are inclined to the vertical line at 40 degrees and 30 degree angles, as shown. What is the measure of the angle theta? We draw out the rectangles in the line. We label each point accordingly as A, B, C, D, E, and F. F being theta. And we also have 90 and 30 degrees. We can figure out the angle CAB will be equal to 180 degrees minus 90 minus 40. We get the 90 because we have a quadrilateral, which is a rectangle. So that's 90 degrees, and the 40 is from the inscribed 40 degrees we know. 180 is because it is a straight line. So angle CAB will be 50 degrees. We get the measure of CBA in the same way, except replacing the 40 with the 30, giving us 180 degrees minus 90 minus 30, which is 60. BCA will give us 180 minus 50 minus 60, since we found out the other two angle measures. That gives us 70 degrees. At 70 degrees, we can also figure out angle DCE is also 70 degrees, since it crosses over. With that, we can use angle DFE, or just angle theta, since they are the same values, as 110 degrees, since 180 minus 70 is 110. The options are 105, 120, 130, and 135 degrees. None of those are 110 degrees. So the correct answer is E. None of these. Question 21. Take a closer look at the prism. We label each as ABC. Let's start getting formulas for one side of it as a square. 1 plus X plus 5 plus C will equal S. 1 plus A plus B plus C is S. A plus X plus B plus 5 equals S. With that, we can get the formula 2 times this sum of 1 plus 5 plus X plus A plus B plus C equals 3S. This can be further elaborated upon by taking what's inside the parentheses and setting it equal to 1 plus 2 plus 3 plus 4 plus 5 plus 6, since the 6 vertices are numbered from 1 through 6. We know that 1 plus 5 plus X plus A plus B plus C is 21. 21 equals 3S, which simplified will give us S equals 14. Once we know that S equals 14, we can plug it back into the formulas, which has the equation giving us 14 minus 1 minus 5, which will give us 8. With that problem, we have the solution of 14 being S. We can work backwards with the three original equations. Once we go through them, we find out that the correct answer is A2. Question 22. M and N are the roots of the equation X squared minus X minus 2018 equals 0. What is the value of N squared plus M? We know that N is a root, so N squared equals N plus 2018. N and M are roots, so the formula N plus M equals 1, as well as N times M equals negative 2018. If we do N squared plus M, that'll give us N plus 2018 plus M, which gives us 2019. The correct answer is D, 2019. Question 23. Four brothers named A, B, C, and D have different heights. They state the following. A. I am neither the tallest nor the shortest. B. I am not the shortest. C. I am the tallest. D. I am the shortest. Exactly one of them is lying. Who is the tallest? If we assume that A is the only one lying, then he is either the tallest, which would make C lying as well, or he is the shortest, which would mean D is also lying. So A cannot be lying. If B is the only one lying, then he is the shortest, which means D is also lying. So B is telling the truth. If D is the only one who is lying, then D is not the shortest. The other three brothers are telling the truth. So none of them is the shortest. That's a contradiction, so D must be lying. The only lying brother is C, saying that he is the shortest, the tallest. C cannot be the tallest. D is not the tallest. He is the shortest. A is not the tallest. So therefore B must be the tallest. The correct answer is B. Question 24. Let f be a function such that f of x plus y equals f of x times f of y for all integers x and y. If f of 1 equals 1 half, find the value of f of 0 plus f of 1 plus f of 2 plus f of 3. To solve this, we must use the fact that f of x plus y equals f of x times f of y. We know that f of 1 is equal to 1 half, and we can do f of 1 plus 0. Using the equation, we know that f of 0 is equal to f of x times f of y. So with that, we can justify f of 0 being 1. Following this logic, f of 2 will equal f of 1 plus 1, or f of 1 times f of 1, giving us 1 fourth, since f of 1 is 1 half, and if we multiply it, it's 1 fourth. f of 3 will be f of 2 plus f of 1, or f of 2 times f of 1, which is 1 fourth times 1 half, giving us 1 eighth. Now we must add them all together. With that, we get 1 plus 1 half plus 1 fourth plus 1 eighth. With that, we get the solution d 15 eighths. Question 25. A quadratic function f of x equals x squared plus px plus q is such that its graph intersects the x-axis and the y-axis in three different points. The circle through these three points intersects the graph of f at a fourth point. What are the coordinates of this fourth point? If we look at the three points, let's label the first one q at point 0,q,u at u,0 and v at v,0. They intersect the x and the y-axis. Set up the equations. u plus v will be negative p, and u times v will be q. Knowing this, we can do x equals u plus v divided by 2, as well as negative p divided by 2. That is the axis of symmetry of the parabola. There is only one circle passing through the points q, u, and v. The axis of symmetry of the parabola is the bisector of the segment u,v. So it's also the axis of symmetry of the circle. So point p will be symmetric to point q at 0,q, which belongs in the parabola and the circle. The x-coordinate of p is negative p, because 0 plus negative p divided by 2 is negative p halves. So the y-coordinate is q. So the correct answer will lead to c, negative p, comma, q. Question 26. We are given a rectangular billboard table with side length 3 meters and 2 meters. A ball is shot from the point m on one of the longer sides. It bounces off on each of the other sides as shown. What is the distance from point a? Will it hit the initial side if bm equals 1.2 meters and bn equals 0.8 meters? Look at the table more closely. We can notice that starting point m makes a triangle with points m, b, and n. This triangle is the same triangle as the one in the opposite corner. Since we know that the lengths of the legs of this triangle are 1.2 on m, b, and 0.8 on bn, we can assume the same thing is true for the shorter side from d almost halfway through to a. That point is also 0.8 meters, making the point from the point where it hits off the wall and a be 1.2 meters. Since we know that the triangles will be the same, the larger triangles will also be the same. The legs of the larger triangle, such as the one that is made up of c, n, and the point on the other wall, are 1.2 as well as 1.8. So we know that the length of am is 3 minus 1.2, which gives us the answer e, 1.8 meters. Question 27. How many real solutions does the equation absolute value of 4 to the power of x minus 3 minus 2, all absolute valued, equal 1 half? We set 4 to the power of x minus 3 as a, then we can use it as a variable and set it up as absolute value of a minus 2 equals 1. There are two possible options. We can either have 1 or 3. If a is 1, then the answer will be negative 1, which turns into 1 thanks to the absolute value. If it is 3, then 3 minus 2 gives us 1 as well. Those are two possibilities. Now if we set a as 4 to the power of x minus 3, there are two possibilities. 1 equals 4 to the power of x minus 3, as well as 3 equals the absolute value of 4x minus 3. We start off with 1 equals absolute value of 4x minus 3, then we either want positive 1, which will lead us to 4 equals 4 to the power of x, which gives us x plus 1, or if we want negative 1, we have 2 equals 4 to the power of x, which gives us x equals 0.5. To start off on the other side, we can do 4 to the power of x equals 6, and thus we have our three solutions, x equals 1, x equals 0.5, and the solution of 4 to the power of x equals 6. So the correct answer is B, 3. Question 28. ABCDEF is a regular hexagon. G is the midpoint of AB. H and I are the points of intersection of the segments GD and GE, with FC respectively. What is the ratio between the area of the triangle GIF and the area of the trapezoid IHDE? We know that segment IH is equal to half of ED, as well as segment FC is equal to two segments ED. With this, we know that segment FI is equal to the difference of FC and IH divided by 2, which can also be written out as 2ED minus half ED divided by 2, which gives us segment FI is equal to three-fourths ED. With this, we can start trying to figure out the areas. FGI is equal to 0.5 times FI times H. And so with that, we can figure out that the triangle is half the area of the trapezoid, so the ratio will be one-half, giving us answer A, one-half. Question 29. There are 40% more girls than boys in a class. How many pupils are in this class if the probability that a two-person delegation selected at random consists of a boy and a girl is equal to one and a half? So the total number of students will be girls plus boys. We can start off with the solution of the sum of G and B times G plus B minus 1 divided by 2, which you can write out as 4 times G times B, or the sum of G plus B times G plus B minus 1. Then we can simplify into G plus B equals G minus B squared. Since we know that G equals 1.4B, we can reason that 2.4B will be equal to 0.4 B squared. And with that, we can write out the equation B equals 2.4 divided by 0.4 squared, or 240 divided by 16, which gives us a total of 15. We know there are 15, and we can do 15 plus 1.4 times 15, which will give us 15 plus 21 for the total number of girls and boys. We do 15 plus 21, and we get the correct answer, C, 36. Question 30. Archimedes calculated a factorial of 15. The result is written on the board. Unfortunately, two of the digits, the second and the tenth, are not visible. What are these two digits? To do a factorial, we must write down all values from 1 to 15 and multiply all the values in between. We divide up the values that are prime, such as 4, 6, 8, 10, 12, and 14. We can get a string of prime numbers. We group them all together and we get 2 to the 11th times 3 to the 6th times 5 cubed times 7 squared times 11 times 13. Since the last three digits of our number are zeros, then we can take away the 2 cubed and the 5 cubed and simplify to 10 cubed so that the factorial of 15 will end with the three zeros and then the unknown tenth digit, since it is not visible. We can reason that that digit will not be a zero, since 2 to the 8th times 3 to the 6th times 7 squared times 11 times 13. It's not a multiple of 5. So that way we can cross out the answer A. Now, if we say that we try to do a multiple of 8, since 2 to the power of 8 is a multiple of 8, so that way that tenth and covered digit after the 3, 6, and then blank must be a multiple of 8, since 360 and 368 are the only multiples of 8 in the range of 0 to 9, but knowing that X cannot be 0, it must be 8. So we can narrow our answers down to either B or E. Now, if we assume that it is 8, we can look at the rest of the numbers. And if we do, the sum of all the values of the digits, 1 plus a blank, plus 0, plus 7, plus 6, plus 7, plus 4, plus 3, plus 6, plus 8, we get the value of 42 plus a variable, and that must be a multiple of 9. Since 42 is already a multiple of 3, the unknown other number must also be a multiple of 3. Only possibilities 0, 3, 6, and 9. Only if we have a 3 do we get the total of 45, which is a multiple of 9. So this can be reasoned that the answer will be 3 in the second digits area. This problem can also be solved just by multiplying out all the numbers and finding the value, but knowing the second digit will be a 3, and the tenth digit will be an 8. We get the correct answer, E, 3, and 8.

Math Kangaroo

2018 competition

solutions

problem-solving

geometry

algebra

logical reasoning

Lukasz Nalewskowski