This is the Math Kangaroo Solutions video library, presenting solution suggestions for levels 7 and 8 from the year 2016. These solutions are presented by Lucas Naleskowski. The purpose of the Math Kangaroo Solutions video library is to help you learn how to solve math problems such as those presented in the Math Kangaroo competition. It is important that you make sure to read the problem as well as listen as I read the problem. After reading and listening to the question, pause the video and try to solve the problem on your own. Question 1. How many whole numbers are there between 20.16 and 3.17? To get to a whole number, we always have to round up. Since we cannot use 3 as a number, since it is below 3.7, we have to go to the next highest whole number, which is 4. From then on, we can just start counting 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20. Now we can't go any further, since if we went to 21, the next whole number after 20, we would be above the value of 20.16. So if we count up all the numbers together, then we get our solution, which is C, 17. Question 2. Which of the following traffic signs has the largest number of axes of symmetry? An axis of symmetry means that if there is a bend in the shape, it can be folded over itself and be exactly the same on either side. If we examine E, it can be done so in one spot, if it is bent along this red line that we created, that is an axis of symmetry, both sides are exactly the same. So E has one axis of symmetry. D has no axis of symmetry, since there is no place we could put a line that would divide the shape into two exact copies of itself. If we make C a triangle, we can have three axes of symmetry, since any one of them bent would make the shape fold over itself perfectly. Shape B will have one axis of symmetry plus another one, however, no more. And finally, shape A will have one, two, three, four axes of symmetry. Shape A has the most amount of axes of symmetry, so the correct answer is A. Question 3. What is the sum of the two marked angles in the figure on the right? If we take a closer look at the figure, it is a triangle with the hypotenuse extended. We can label the inside angles of this triangle X and Y. We know if we add together the three angles of a triangle, we get 180. This is clearly a right triangle, since it is marked with a box, so X plus Y plus 90 will give us 180. If we subtract 90 from both sides, we get X plus Y equals 90. Now we can move on to labeling the angles we are interested. Let's call them A and B for this problem. We know that A plus X will give us 180, since it is on a straight line. Similarly, Y and B will also give us 180. So if we put it all together, X plus A plus Y plus B will give us 360. Now as we've previously determined, X and Y are equal to 90. So it's really A plus B plus 90 equals 360. If we subtract 90 from both sides, we get A plus B is 270. The question asks us for the sum of the two marked angles, which is 270. So this gives us the answer, C, 270. Question 4. Jenny had to add 26 to a certain number. Instead, she subtracted 26 and obtained negative 14. What number should she have obtained? So we know that Jenny wants to add 26 to a number. So we can call this X plus 26 equals an unknown. But instead, she subtracted 26. So we get X minus 26. And we know that she obtained negative 14 from this. So we can say X minus 26 equals negative 14. With this, we can figure out what X is equal to. X is equal to 12, since we add 26 to both sides to cancel them out. And we get 12. Knowing that X is 12, we can go back to the first problem, which is adding 26 to a certain number, which we now know is 12. So 12 plus 26 will give us the number that Jenny should have obtained, which gives us the answer, D, 38. Question 5. Joanna turns a card over around its lower edge and then about its right edge, as shown. What does she get? Take a closer look at this. All we have to do is turn the shape as well as moving it. So if we turn it on its lower edge, on its right edge, then we get this. We know that she does this again, and the shape will move like so. And now it is in a position that is identical to our answer, which is B. Question 6. Kenga combines 555 groups of nine stones in a single pile. She then splits the resulting pile into groups of five stones. How many groups does she get? We know that Kenga is starting off with 555 groups of nine stones. The total number of stones is 555 times 9. Now, we don't have to take the multiplication and get the solution of this, since this is the total number of stones. And we know that she will split the total number of stones by 5. We can take 555 times 9 and divide it by 5, since 555 can be divided by 5 with no issue. That leaves us with 111 times 9, which gives us our answer, which is A, 999. Question 7. In my school, 60% of the teachers, which is 45 teachers, bike to school. Only 12% of the teachers use their car to get to school. How many teachers use their car to get to school? Since we know that 60% of the teachers use their bikes to get to school, and we know that 60% is equal to 45 teachers, we can move on with the problem. We want to figure out what 12% of the teachers is. All we need to do is simply take 60% equals 45, and we can divide both sides by 5. Since when we divide 60 by 5, we will get 12. So 60% divided by 5 is 12%, and 45 divided by 5 is 9. So therefore, the answer is C, 9 teachers use their car to get to school. Question 8. What is the size of the shaded area? We take a closer look at the shape. It is a rectangle that is 20 by 10. There are some circles and lines inscribed in it. We take one of these circles, then we can turn it around. It is still the same shape. This is just to help us visualize it. Looking at this now, we notice that the half of the rectangle, the 10 by 10 area, with the circle in it, is the same as the other half of the circle, except inversed. So this means that half of the area is shaded, or 50%. And the entire area of the rectangle is 20 times 10, which gives us 200. And since we know that 50% of the area is shaded, or half, then we just have to multiply 200, which is the area of the whole rectangle, by 50%, or just divide by 2 to get half. And that gives us the answer C, 100. Question 9. One piece of rope has a length of 1 meter, and another piece of rope has a length of 2 meters. Alex cuts the two pieces of rope into several parts. All the parts have equal lengths. Which of the following cannot be the total number of parts he obtains? So we have the 1 meter rope, and the 2 meter rope. So all together, we will have 3 meters of rope, since 1 plus 2 is 3. Now if we take a look at all the possible answers, 6, 8, 9, 12, and 15, all we have to do is just compare them, and see which of these answers is not divisible by 3. Now all the answers, 6, 9, 12, and 15, can be divided into 3, or by 3, into equal parts. And only 8 cannot be divided into 3 parts evenly. So that means our correct answer will be B, 8. Question 10. Four towns, P, Q, R, and S, are connected by roads, as shown. A race uses each road exactly once. The race starts at S, and finishes at Q. How many possible routes are there for the race? Look at the possible roads, and all the towns more closely, you notice it makes a four-sided shape. Now there are multiple ways to get from S to Q, using each road once. First way is to go from S to P, P to Q, Q to S, S to R, and R to Q. This way, we used each road only once, and got from S to Q. This is one way to do this. Another way is to follow the same beginning, going from S to P, P to Q, however this time taking a different turn, and going Q to R, then R to S, and S to Q. This still results in reaching Q from S using each road only once, so this is a second method. Next method is to start off from S to R, R to Q, Q to S, S to P, and P to Q. This is a third possible method. Similarly, we can go S to R, R to Q, this time making a different turn, going Q to P, P to S, and S to Q. A fourth method of using each road only once, getting from S to Q. Next method is to go straight to Q from S, but then go back to R, R to S, S to P, and P to Q. This way we have still used each road only once, and ended up in Q starting from S. Similarly, we can go from S to Q, take a different turn, and go from Q to P, P to S, S to R, and finally R to Q, reaching the destination. This is the sixth method, and the final one. Therefore, there are six different ways to get from S to Q using each road once. So our answer is going to be C, 6. Question 11. The diagram shows four identical rectangles placed inside a square. The perimeter of each rectangle is 16 centimeters. What is the perimeter of the big square? We look at the big square with the four rectangles in it, and we know that each rectangle's perimeter is equal to 16 centimeters. Then we can start looking for the sides of each rectangle. Now a rectangle is composed of two sets of two equal sides. Let's name these sides X and Y, and their opposing are also equal to X and Y. So therefore we know that 2X plus 2Y is 16, since the perimeter of a rectangle is 16 centimeters. With this, we can look at the square, which is composed of these rectangles. We see that one side of the square is composed of an X and a Y. Each side, since it is a square, is composed of the same. And if we add all these together, then we get the perimeter of the square is 4X plus 4Y. We can simplify this and say 2 times the sum of 2X plus 2Y, which is the same as 4X plus 4Y. Just taking a 2 away and placing it in a previous part. Now we know that 2X plus 2Y equals 16, so we can replace that with a 16, getting 2 times 16 is the perimeter of the square. So with this, we just have to multiply 2 times 16 and get our answer, E, 32 centimeters. Question 12. Petra has 49 blue beads and one red bead. How many beads must Petra remove so that 90% of her beads are blue? Since we want 90%, we know that 90% is equal to 90 divided by 100, which can be simplified to 9 over 10. So with this, we want Petra to have 10 beads in total, 9 of which will be blue, so that 90% of her beads will be blue. She currently has 49 blue beads and one red bead, or 50 beads in total, and we want to get to 10, so we must do 50 minus a number of beads to give us 10. We subtract 50 from both sides, we can simplify this and get 40. Petra must remove 40 blue beads so that she will have 9 blue beads and one red bead, or 90% blue beads. So our correct answer is going to be E, 40. Question 13. Which of the following fractions has a value closest to one half? One half can be rewritten as 0.5. So we can use long division to find the answers in decimals of each of the questions. If we take 25 divided by 79, we can find out that it is approximately 0.31. We take a look at B, which is 27 divided by 51, we say that it is approximately 0.45. Next up, if we look at 29 divided by 57, we get an approximate answer of 0.5. D, 52 divided by 79, is approximately 0.65. And E, 57 divided by 92, will give us approximately 0.61. Out of the five possible solutions, C is the closest to one half. So the correct answer is C, 29 divided by 57. Question 14. Ivor writes down the results of the quarterfinals, the semifinals, and the final of a knockout tournament. The results, not necessarily in this order. Bart beat Anthony, Carl beat Damien, Glenn beat Henry, Glenn beat Carl, Carl beat Bart, Ed beat Fred, and Glenn beat Ed. Which pair played in the final? So to solve this, we must know which people fought and were eliminated in the quarterfinals. Since it is a knockout tournament, once they lose, they will not play again, and there will be less people playing in the semifinals and the finals. So the people that are in the quarterfinals, that are only mentioned once, will be there, and will have lost. So for example, Ed vs. Fred. Since we only see Fred's name once in the list of the results, we know that this is a fight that took place in the quarterfinals. And since Fred's name is only once, and it is paired with Ed, we know that Ed beat Fred, as his name is highlighted. Glenn vs. Henry, the same principle applies. Henry's name only shows up once when he is fighting Glenn, and Glenn beats Henry. Then we have Carl vs. Damien, and Damien only showed up once, with Carl beating him. And then we have Bart vs. Anthony, Anthony losing, because he only shows up once, so he only took place in one fight, and Bart moved on. So now that we know that Ed, Glenn, Carl, and Bart have moved on to the semifinals, we must just find the results of the people who fought twice, and we will know that they were knocked out, since the people whose name shows up three times will make it and fight in the finals. So Glenn and Ed will have fought. Since we know that Glenn beat Ed, Ed's name only shows up twice in the results, so he is out, and Glenn will move on. And Carl vs. Bart, we know that these two fought, Carl won, and Bart will have lost, since his name is only mentioned twice as well, when it is talked about. Carl and Bart fighting, and Bart and Anthony fighting. And so we have our two winners, Glenn and Carl. They beat the semifinals, and they move on to the final round, giving us answer B, Glenn and Carl. Question 15. Anne has glued some cubes together, as shown to the right. She rotates the solid to look at it from different angles. Which of the following can she not see? We look at the solid to our right, which Anne has glued together. We can take our left hand, put your thumb up in the direction of the shorter column of the cube, and then your open fingers vertically up along the adjacent column. When you curl your fingers, they will point in the direction of the third column of the cubes. Thus, this solid has a left-hand orientation. Rotations don't change the orientation. Among the five options shown, only answer B will have a right-hand orientation, since we have to use our right hand. We put our thumb in the direction of the shorter column of the cubes, and our open fingers along the adjacent column. We will see that when we curl our fingers, they point in the direction of the third column of cubes. So, solid B has a right-hand orientation, and therefore, since we've established that the way that Anne glued the cubes together was in a left-hand orientation, it cannot be a right-hand orientation shape. So, she cannot see shape B. Question 16. Tim, Tom, and Jim are triplets, three brothers born on the same day. Their twin brothers, John and James, are three years younger. Which of the following numbers could be the sum of the ages of the five brothers? We write out T as the age of a triplet, and D as the age of a twin. We can write out T plus T plus T plus D plus D as the sum of the ages. We add three and three, since the twins are three years younger, then we will have a number that is divisible by five, like so. If we look at all the solutions, and we add six of them, and we take out the ones that aren't divisible by five, we should be left with a one answer. 36 plus 6 gives us 42. Since this number is not divisible by 5, this cannot be the answer. 53 plus 6 gives us 59, also not the answer. 76 plus 6, 82. 89 plus 6, 95. And 92 plus 6, 98. Only one of these is divisible by 5, 95. So, the correct answer will be D, 89. Question 17. A three centimeter wide rectangular strip of paper is gray on one side and white on the other. Maria folds the strip as shown. The gray trapezoids are identical. What is the length of the original strip? Take a closer look at the strip. Notice that the folded strip is 27 centimeters long. Since all these shapes are identical, we can divide it up into nine equal sections. Once we have nine sections, we can do 27 centimeters divided by nine, which gives us three. So, we know each of these nine sections is three centimeters long. We can do the same in the other direction. All that's left is to count the amount of sections we've drawn out, such as this. We count them out. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19. There are 19 total sections. So, we just have to do three centimeters times 19 to give us the correct answer, which is D, 57 centimeters. Question 18. Two kangaroos, Jum and Pur, start to jump at the same time from the same point in the same direction. They make one jump per second. Each of Jum's jumps is six meters in length. Pur's first jump is one meter in length. The second is two meters in length. The third is three meters in length, and so on. After how many jumps does Pur catch up to Jum? We make a table like so. With the first second already passed, we have Jum at six meters and Pur at one. Jum will jump six more, going to 12, and Pur will jump two more, going to three. Next, after three seconds, Jum will have jumped six more, so 18, and Pur will have jumped three more, so six. We keep going. We have 24 and 10, since they jumped six and four meters, then six and five meters each. They both jumped six meters, then six and seven meters, and slowly, as Pur jumps more every time, we notice he slowly gets closer and closer, until they're at 66 and 68 after 10 seconds, and on the 11th second, Pur beats Jum. And if we count out all those grids, we get that this took 11 seconds, so the correct answer is going to be B, 11. Question 19. Seven standard dice are glued together to make the solid shown. The faces of the dice that are glued together have the same number of dots on them. How many dots are on the surface of the solid? We draw a flattened die like so. We can label all the different values of dots. Now, if we add together all the dots on one of these die, we get a total of 21. Since there are seven die together, we can multiply 21 by 7 to get a total of 147. We know that one of the dice will be completely enveloped, and none of its dots will be showing. Then each other die touching it will have one face covered, and that face will be a different face for each die. So it will also be a value of 21, since there will be six different dice, each with a different face being covered. So in total, we get 147 minus the sum of two dice worth of dots, or 21 and 21. So 147 minus 21 minus 21 gives us our answer, which is D, 105. Question 20. There are 20 students in a class. They sit in pairs so that exactly one-third of the boys sit with a girl, and exactly one-half of the girls sit with a boy. How many boys are there in the class? We write out P as mixed pairs. We know that 3P will be boys, since only one-third of the boys will sit with a girl, and 2P will be girls, since half of the girls sit with a boy, or in a mixed pair. So 5P will give us all 20 students. This means that there will be four mixed pairs, since 20 divided by 5 gives us four. So those will be the four mixed pairs looking like so, blue being for boys, red being for girls. Now we also know that there are two more pairs of girls, and there are also eight more boys, or four more pairs. We count up all of the blue rectangles for boys, then we get a total of B, 12. Question 21. Inside a square with an area of 36, there are shaded regions as shown. The total shaded area is 27. What is P plus Q plus R plus S? We take a closer look at the square. Each of the sides is a total of 6, since the area is 36. Now we know that the area of a triangle is base times height times half, that is the area of a triangle. We draw a line through the middle, or through the diagonal of the square, then we can get four different shaded triangles. Shaded triangle S, shaded triangle R, shaded triangle Q, and shaded triangle P. Now we know that the total area of the shaded part will be 27. So if we add together the areas of the four triangles we just made, the shaded triangles S, R, Q, and P, we'll get 27. So we can do four times base times height times half. Now the height for each of these will be 6, since that is the side of one of the one of the sides of the square. And the B will be either S, Q, R, or P, depending on each triangle. But since we are only interested in the sum of all four, S, R, Q, and P, we don't need to make that distinction. We can simplify this problem, 4 times B times 3 equals 27. And then if we divide out 3, we get 4B equals 9. And as stated previously, the B, which is the base of a triangle, can be explained as being S, R, Q, or P. But since we did this problem as the sum of all four of the triangles, 4B will be S, R, Q, and P all together. So that means 4B gives us our answer, 9. So the correct answer is D, 9. Question 22. Theo's watch is 10 minutes slow, but he believes that it is 5 minutes fast. Leo's watch is 5 minutes fast, but he believes it is 10 minutes slow. At the same moment, each of them looks at his own watch. Theo thinks it is 12. What time does Leo think it is? So starting off with Theo, he thinks it is 12, but he believes that his watch is 5 minutes fast. So we can see and add 5 minutes to whatever time Theo sees. Since Theo will see his watch, see 12.05, believe it is fast, and subtract the 5 to get to 12. But the watch actually displays 12.05. But since his watch is displaying 12.05, but the watch is 10 minutes slow, we can add 10 more minutes. So we know that it is 12.05. The actual time is 12.05. Now if we take Leo, we know that Leo's watch is 5 minutes fast, but he believes it is 10 minutes slow. We can add 5 minutes since it is 5 minutes fast, giving us 12.20. But since he believes that it is 10 minutes slow, he will be adding an additional 10 minutes to give us whatever time Leo thinks it is, 12.20 plus 10 minutes, which gives us our answer, 12.30. So the correct answer is D, 12.30. Question 23. 12 girls met in a cafe. On average, they ate 1.5 cupcakes each. None of them ate more than two cupcakes, and two of them had only mineral water. How many girls ate two cupcakes? Now to find the amount of cupcakes they ate, we have to multiply the number of girls times the average of cupcakes eaten. So 12 times 1.5 will give us 18 cupcakes. So we know that 18 cupcakes were eaten in total. Since we know two of the girls only had mineral water, only 10 girls had cupcakes. Now if we take 8 times 2, meaning 8 girls times 2 cupcakes each, we get a total of 16 cupcakes eaten. This leaves 2 more cupcakes left to be eaten. So that means 2 girls ate 1 cupcake each. So that means that in total, 8 girls ate 2 cupcakes each. So the correct answer is going to be E, 8. Question 24. Little Red Riding Hood is delivering waffles to three grannies. She starts with a basket full of waffles. Just before she enters each of the grannies' houses, the big bad wolf eats half of those waffles in her basket. When she leaves the third granny's house, she has no waffles left. She delivers the same number of waffles to each granny. Which of the following numbers definitely divides the number of waffles she started with? To solve this problem you have to go backwards. Since we know that Little Red Riding Hood will end with zero waffles and she gives the same amount of waffles to each granny, we can label the number of waffles each granny receives as X. And so she gave X waffles to the third granny. But the Big Bad Wolf eats half of her waffles each time. So if she gave X waffles to the third granny, the Big Bad Wolf must have eaten half of those before she gave them to her. So that means we multiply by two. Since if she had two X waffles, the Big Bad Wolf eats half, or X. She's left with X. We can repeat this. With the second granny, she would have given her X waffles. So we can make this total 2X plus X, meaning before she gave the waffles to the second granny, she would have had 3X waffles. And before the Big Bad Wolf would have eaten those half, we can multiply by two. Means before the Big Bad Wolf got to her at the second granny, she would have had 6X waffles. Then the first granny, right before she gave her waffles, X waffles, she would have had 7X waffles. And before that, the Big Bad Wolf would have eaten half of those. So if 7X is half of what she started with, we multiply by 2 and get 14X. So we get 14. Now 14 can be divided into 2 and 7. So the correct answer is going to be D, 7. Question 25. The cube on the right is divided into 64 smaller cubes. Exactly one of the small cubes is gray. On the first day, the gray cube changes all of its neighboring small cubes to gray. Two cubes are neighbors if they have a common face. The second day, all the gray cubes do the same thing. How many gray cubes are there at the end of the second day? Since it is a 4 by 4 grid, we can draw this out, knowing that the first layer on the first day will have one gray cube. We have no other gray cubes, so this is what the three layers look like. On the first day, after the shift, all the neighbors of the one gray cube will turn gray, like so, since all of these have a neighboring face. On the last day, or the second day, then they will do the same thing and spread out like so. Now we just have to count out the number of gray cubes and we get our answer, which is E, 17. Question 26. Several different positive integers are written on a blackboard. The product of the smallest two of them is 16. The product of the largest two is 225. What is the sum of all the integers? You can write out the four integers like so. A is less than B is less than C is less than D. Since we know that none of them are the same, they're all positive, they will be ascending. We know that the product of the smallest two is 16, so A times B will give us 16. We know that the product of the largest two is 225, so C times D is 225. We can get to 16 in three different ways. 1 times 16, 2 times A, and 4 times 4, which we can't do 4 times 4 since A has to be less than B. Now, 225 can be achieved in several ways, one of which is 1 times 225. However, this will not work since 1 will be either smaller or equal to A, and 1 can't be C if C is equal to or smaller than A, so that is off the table. We do 3 times 75. While 75, D will be the largest number, 3 might possibly be smaller than 16 or 8, which would be B, so this cannot be the answer either. 5 times 45, then still the same problem arises. C, which would be 5, would be smaller than B, which would be 16 or 8, so that can't be the answer. If we did 9 times 25, then we can get a set of integers that are ascending. We know it can't be 1 times 16, since 16B would be larger than 9 or C, so that can't be it. But if we have A as 2, B as 8, C as 9, and D as 25, this gives us our set. Since they are all ascending. 2, 8, 9, 25. Last part of the problem is to find the sum of all four integers. So A plus C plus D will give us our answer, which is 2 plus 8 plus 9 plus 25, giving us the sum of C, 44. Problem 27. The diagram shows a pentagon. Cipodae draws 5 circles with centers A, B, C, D, and A, such that the two circles on each side of the pentagon touch. The lengths of the sides of the pentagon are given. Which point is the center of the largest circle that she draws? If we look at our pentagon, we notice all the lengths of the sides. If we add them all up together, we get a total of 74. Now, since Cipodae is drawing circles, we can write out a formula. 74 equals 2 times the radius of circle A plus 2 times the radius of circle B plus 2 times the radius of circle C plus 2 times the radius of circle D plus 2 times the radius of circle E. With this, we can start writing out our formulas. 16 will be the radius of A plus the radius of B. 14, radius of B plus radius of C. 17, radius of C plus radius D. 13, radius D plus radius E. 14, radius E plus radius A, since it loops back around. Now, we know that 37 will be equal to the radius of A plus radius of B plus radius of C plus radius D plus radius E, since that is the same as our previous formula of 74, just divided by 2. This we can start setting up equations. 37 will equal the radius of A plus 14 plus 13, since we know that the radius of B plus the radius of C is 14 and the radius of D plus radius of E is 13. So with that, we know that the radius of A will be 30 minus 27 or 10. Radius A equals 10. After this, we can set it into the other formulas, such as 16 equals 10 plus radius of B or radius of B equals 6. 14 equals 6 plus radius of C. Radius of C equals 8. 17 equals 8 plus radius of D. Radius of D equals 9. And finally, 14 equals radius of E plus 10 or radius E equals 4. So we wanted to know which point has the largest circle drawn, that is the one with the largest area, out of which it is radius A with a radius of 10. So the correct answer will be circle A or just A. Question 28. Katie writes a different positive integer on each of the 14 cubes in the pyramid. The sum of the nine integers written on the bottom cubes is equal to 50. The integer written on each of the other cubes is equal to the sum of the integers written on the four cubes underneath it. What is the greatest possible integer that can be written on the top cube? We draw the bottom nine cubes, looking down on them. The numbers here represent the number of cubes that will be affected by it. So the cube in the middle will affect four different cubes above it. The cubes in the corners will only affect one cube each, and then the remaining cubes will will affect two cubes each. So the cube in the center is the most significant and wants to be the largest number. The cubes in the corners are the least significant. We can write out a formula for A plus 2B plus C, with A being the value of the cube in the middle, B being the value of the four cubes that will affect two cubes each above them, and C being the total value of the sum of the four corner cubes, labeled 1 here. We also know that 50 is equal to A plus B plus C, since we know that the sum of the nine integers written on the bottom cubes is equal to 50. Now there are nine different integers. Each one is different, and we want the smallest numbers to be on the least significant cubes. So the value of C will be 1 plus 2 plus 3 plus 4. So each of the cubes in the corner of the 3x3 grid will be 1, 2, 3, and 4, regardless of which corner they are placed in. Then the four cubes that are in the other parts around the perimeter of the 3x3 grid will be the next smallest numbers, 5, 6, 7, and 8. We grab the total of those, then we get what we want left in the center cube, the one that will affect the most. So we can do 50 minus C minus B, or 50 minus 10 minus 26, which gives us A equal to 14. With this, we can plug that into our formula, 4 times 14 plus 2 times 26 plus 1 times 10, which gives us 56 plus 52 plus 10. We add all those together, we get our total of E, 118. Question 29. A certain train has five passenger cars, each containing at least one passenger. Two passengers are said to be neighbors if they are in the same car or they are in two adjacent cars. Each passenger has either exactly five or exactly 10 neighbors. How many passengers are on the train? If we draw our five different cars and we label each one A, B, C, D, and E, and we know that each car will have at least one passenger. So the first car will have one passenger plus whatever additional number of passengers inside of it. We can call this number A. So car A will have A plus 1 passengers, car B will have B plus 1, C will have C plus 1, D will have D plus 1, and E will have E plus 1. Now the amount of neighbors that a passenger in car A will have is A plus B plus 1. The amount of neighbors that a person in car B will have will be A plus 1 plus B plus C plus 1. And this can be said for C, D, and E will be similar to A having D plus 1 plus E. With this being said, we know that A will be less than B, E will be less than D. So that way, since each car will have exactly five or ten neighbors, and A is smaller than B, we know that A has to have five and B has to have ten, since five is less than ten, and there are no other options. We don't know what C is. However, with E being less than D, we can have the same be true as when we figured out A and B, having D equal 10 and E equal 5, since E will have less than D, and the only options are 5 and 10. And we know that 2A equals B, since 5 and 10. Now if we do this and write it out, we can have 2 times the sum of A, B, and 1 equal to A plus 1 plus B plus C plus 1. Simplified, we distribute the two, get 2A plus 2B plus 2 equals A plus B plus C plus 2. Or simply put, A plus B equals C. The number of passengers in C will be 1 plus B plus 1 plus A plus 1, or 1 plus the number of passengers in A plus 1, which means that C is greater than A, which means that C must be 10. We have 5 equals A plus B plus 1. We can write out 4 equals A plus B, C equals 4, or A plus B, and C equals 4, D plus E. And then once we add up all the total number of passengers in the train, we get A plus B plus C plus D plus E plus 5, or 4 plus 4 plus 4 plus 5, which leads us to our answer, which is C, 17. Question 30. A 3 by 3 by 3 cube is built from 15 black cubes and 12 white cubes. Five faces of the larger cube are shown. Which of the following is the sixth face of the large cube? We know that a cube, when laid flat, is composed of a net in this shape. We have our cubes and we lay out the faces. Then we can surmise that if we put the cross-shaped face as one of the bottom faces, then the face with the five different black cubes will have to be in the opposite side, since they cannot connect, since they do not share any sides. So therefore we know that it cannot be answer D. The other cubes can be placed in such a fashion and they will connect. And this leaves us with finding out the final cube. If we imagine folding our five cubes and leaving one spot open, then we know that the remaining cube will have to have two white spots on the same side, one white spot on the opposite side, and then it'll have to have one side that is completely black, and then the other side two black spots with one white spot. There's only one face that follows this criteria, so we know that the sixth face of the large cube will be answer A.

Math Kangaroo 2016

levels 7 and 8

Lucas Naleskowski

math competition

problem-solving

geometric figures

logical reasoning

arithmetic puzzles

spatial visualization