This is the Math Kangaroo Solutions Video Library, presenting solution suggestions for Levels 7 and 8 from the year 2017. 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 to 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. What is the time 17 hours after 5pm? At 5pm, there are 7 hours till midnight. So out of the 17, we go ahead 7 hours and we get to midnight. That leaves us with 10 more hours we have to move ahead. If we take midnight and go ahead 10 hours, we get the answer of B, 10am. Question 2. A group of girls stands in a circle. Xena is the 4th to the left from Yana and the 7th to the right from Yana. How many girls are in the group? Start off by drawing out the circle and placing Xena and Yana on points in the circle. Since we know that Xena is 4th to the left from Yana, we can go ahead and move. We have 1, 2, 3 girls in between them and Xena is the 4th. We can do this from the other side. Since she is the 7th, we have 1, 2, 3, 4, 5, and 6 girls in between Xena being the 7th. If we count up all the girls, we get the correct answer C, 11. Question 3. What number must be subtracted from negative 17 to obtain negative 33? If we write this out as an algebraic equation, we get negative 17 minus X equals negative 33. We can add 17 to both sides to give us negative X equals negative 16, which can be written as X equals 16. So 16 has to be subtracted from negative 17 to get negative 33. So the correct answer is C, 16. Question 4. The diagram shows a striped isosceles triangle and its altitude. Each stripe has the same height. What fraction of the area of the triangle is white? Take a closer look at the isosceles triangle and start dividing it into sections, like so. We notice that the bottom section is half white and half red. We do this for the next section, it is half white and half red. This one is also half white and half red. The next section is half green but also still half white. And this is true as we move towards it. We notice that every time we looked at a certain section, half of it was white and the other half was either red or green. But since we are asked what area is white, then we just have to look at the white. So the correct answer will be A, one half. Question 5. Which of the following equalities is correct? To do this we must solve each equality. So A will be 4 divided by 1, B will be 5 divided by 2, C is going to be 6 divided by 3, and 7 divided by 4, and 8 divided by 5. Solving these we get 4 divided by 1 gives us 4, 5 divided by 2 gives us 2.5, 6 divided by 3 gives us 2, 7 divided by 4 gives us 1 3 fourths, and 8 divided by 5 gives us 1.6. Look at all the answers, only one of them matches what we got, and that is B, 5 divided by 2 equals 2.5. Question 6. The diagram shows two rectangles whose sides are parallel. What is the difference in the lengths of the perimeters of the two rectangles? If you closer look at the rectangles, one of them is inside the other, you can start labelling the length and the width, which we can label as W and L. To get the perimeter you have to add together all the sides of a rectangle, so the blue inside rectangle's perimeter will be 2L plus 2W. The rectangle on the outside will be 2 times the sum of L plus 2 plus 3 plus 2 times the sum of W plus 3 plus 4. This can be simplified as such, even further so, as 2L plus 10 plus 2W plus 14. So the difference between the two perimeters will be 10 plus 14, which gives us the answer E, 24 meters. Question 7. Bob folded a piece of paper twice, and then cut one hole in the folded piece of paper. When he unfolded the paper, he saw the arrangement shown in the diagram. How had Bob folded his piece of paper? Looking at the placement of the two holes, we know that the way the paper must be folded must be in a diagonal fashion, like so. That excludes answers B, C, and E. So the answer has to be A or D. Since there are only two holes in the bottom left corner of the paper, then A cannot be the answer, because it folds over that corner, and there would be four holes. So the correct answer is D. Question 8. The sum of three different positive integers is 7. What is the product of these three integers? To get the sum of seven with three different integers, there are several different ways. Do 1 plus 1 plus 5. Do 1, 2, and 4. Do 1, 3, and 3. Do 2, 2, and 3. Only one of these is the sum of three different integers, 1, 2, and 4. So to get the product of the three integers, we just have to multiply them. We do 1 times 2 times 4. We get the answer D, which is 8. Question 9. The diagram shows four overlapping hearts. The areas of the hearts are 1 cm2, 4 cm2, 9 cm2, and 16 cm2. What is the shaded area? If we look at the largest heart, which has an area of 16 cm2, you notice it is shaded. The next largest heart, 9 cm2, is not shaded. 4 cm2, shaded. 1 cm2, not shaded. To find the area that is shaded, we just have to take the not shaded and subtract it from the shaded. So 16 cm2 subtract 9 cm2 gives us 7 cm2, which is shaded. The same is done for the 4 and 3, which gives us 3 cm2. We add them together to get the correct answer of B, 10 cm2. Question 10. Yvonne has 20 euros. Each of her four sisters has 10 euros. How many euros does Yvonne have to give to each of her sisters so that each of the five girls has the same amount of money? To do this, we just have to take the sum of all the euros, so the 20 from Yvonne and the 10 from each of the four sisters. That gives us a total of 60. Then we can divide this by 5 to give us how many euros each sister should have, which is 12. And since Yvonne had 20 euros and each of the four other sisters had 10 euros, Yvonne will have to give every sister 2 euros so that they all have the same amount. Question 11. Annie the ant started at the left end of the pole and crawled two thirds of its length. Bob the beetle started at the right end of the same pole and crawled three fourths of its length. What fraction of the length of the pole are Annie and Bob now apart? If we look at the pole, we've got Bob and Annie. Bob has traveled three fourths of the length of the pole. But if we want to look at it from the same distance that Annie crawled, we can subtract one from three fourths and say that Bob is one fourth of the pole's length away from the beginning of the pole. And Annie is two thirds. Next, to find the distance between the two of them, we have to find a common denominator. So two thirds is also known as eight twelfths, and one fourth is three twelfths. To find the distance that they are apart, we just have to do eight twelfths minus three twelfths, and we get the answer E, five twelfths. Question 12. One sixth of the audience in a children's theater were adults. Two fifths of the children were boys. What fraction of the audience were girls? First, to start off with, it's easy to assume a number. I'm going to assume that there are 60 total people in the audience. The random number chosen, because in the question we are mentioned one sixth and two fifths, so an easy number to choose will be one that can be divided by six and five, such as 30, 60, 90, or so on. For simplicity, I chose 60. We know that one sixth of the audience are adults. So if we have one sixth adults, we can multiply our total audience of 60 by one sixth, and that would mean there are 10 adults in this supposed audience. That means there are 10 adults and 50 children. We know that two fifths of the children are boys, and since there are 50 children in the audience and two fifths of them are boys, we can multiply both values. We find out that there are 20 boys in the audience. Out of the 50 children, there are 20 boys, so that stands to reason there are 30 girls in the audience. We want to find out what the fraction of the whole audience is girls. So we take 30 girls and divide it by 60, the total number of people in the audience, and we get the answer A, one half. Question 13. In the diagram, the dashed line and the black path form seven equilateral triangles. The length of the dashed line is 20. What is the length of the black path? If we look at the path and the dashed line, as well as the triangles that they make up, we know that there are seven equilateral triangles. One of the properties of equilateral triangles is that every side is the same. The dashed line is in such a place that it is one side of every unique equilateral triangle. So the length of the dashed line is equal to one side of every triangle added together. If we take the length of the dashed line and multiply it by two, we will get the other two sides of the triangles. So two times 20 will give us the length of the black path, which is the answer D, 40. Question 14. Four cousins, Emma, Eva, Rita, and Zena are 3, 8, 12, and 14 years old, although not necessarily in that order. Emma is younger than Rita. The sum of the ages of Zena and Emma are divisible by five, and the sum of the ages of Zena and Rita are also divisible by five. How old is Eva? First, we know that Emma is younger than Rita, so the value of Emma's age is less than Rita's. We also know that the sum of the ages of Zena and Emma are divisible by five. To get a number that is divisible by five with the different ages of 3, 8, 12, and 14, there are only two possibilities, adding 8 to 12 or adding 3 to 12. Know that the sum of the ages of Zena and Rita is also divisible by five. The same applies, either 12 plus 8 or 12 plus 3. With this, we know that Zena will be 12, since the 12 appears in all possibilities. Next, we take into account that Emma is younger than Rita. So that means the sum of Zena and Emma will be less than the sum of Zena and Rita. So Emma will be 3, and Rita will be 8. That only leaves Eva being 14 years old, so the answer is A, 14. Question 15. This year there were more than 800 runners participating in the kangaroo hop. Exactly 35% of the runners were women, and there were 252 more men than women. How many runners were there in total? We take the total of 100%. We can take away 35% of women and 35% more, since that is the men minus the 250. So that means the 252 is 30% of the runners, since we took away the men and the value of the men, not counting the extra 252. Since we know that 252 is 30% of runners, or 3 out of 10, to find out the total number of runners, we just do 252 times 10 over 3. That gives us our answer of E, 840. Question 16. Ria wants to write a number in each box of the diagram shown. She has already written two of the numbers. She wants the sum of all the numbers to equal 35, the sum of the numbers in the first three boxes to equal 22, and the sum of the numbers in the last three boxes to equal 25. What is the product of the numbers she writes in the gray boxes? To go a closer look at the set of five boxes, we can set different variables for the three empty boxes, as in x, y, and z, for simplification of finding out the answer. We know that the sum of all the numbers will be 35. To simplify this, we can write out x plus y plus z will be 28. We also know that the sum of the first three boxes, or 3 plus x plus y, will be equal to 22. If we take our first formula of x plus y plus z equals 28, we can write it as x plus y equals 28 minus z. And we can put this formula in our second formula, and get 3 plus 28 minus z equals 22, since we're replacing the x plus y with a 28 minus z. That gives us the answer of negative z equals negative 9, or z equals 9. Going on further, since we know that the sum of the last three boxes is 25, we can write out y plus z plus 4 equals 25. We do the same thing here, knowing that z is 9, we get the answer of y plus 9 plus 4 equals 25. That is simplified to y plus 13 is 25. That'll tell us that y equals 12. All that's left to find out is the x. So we can go back to our formula of 3 plus x plus y equals 22. We now know the y value, and we get x plus 15 equals 22, or x equals 7. All that's left to do is to multiply x and z, 7 times 9, and that gives us our answer, which will be 63. And so the correct answer is a 63. Question 17. Simon wants to cut a piece of thread into nine pieces of the same length, and marks his cutting points. Barbara wants to cut the same piece of thread into only eight pieces of the same length, and also marks her cutting points. Carl then cuts the thread at all the cutting points that are marked. How many pieces of thread does Carl obtain? Imagine our rope, then we can see the cutting points where Carl will be making cuts. These are where Barbara as well as Simon have made their marks. All that's left to do is to count out the different segments. So each segment will be counted out separately, based on where the different cutting marks from Barbara or Simon were. So we see that we have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, and 16 segments. So the correct answer is going to be b, 16. Question 18. Two segments, each one centimeter long, are marked on opposite sides of a square with a side length of 8 centimeters. The ends of the segments are joined as shown in the diagram. What is the shaded area in centimeters squared? Look at the square with the triangles more closely. We can start solving the problem. Now we know that the height of the first triangle plus the height of the second triangle is equal to 8 centimeters, since that's the height of the square, and both triangles are inside the square without overlapping. Now to find the area of a triangle, we must just do the simple formula of 1 half base times height. Now we can do 0.5 times 1, since we know that the base of the triangles is 1 centimeter, times h1, and for the second triangle times h2. We can add these together since we are trying to find the total shaded area. We can simplify this into 0.5 times 1 times the sum of h1 plus h2, and as we previously established, h1 plus h2 is equal to 8. So all we have to do to find the shaded area in centimeters squared is multiply 0.5 by 1 by 8. That gives us 0.5 times 1, which gives us the answer of b, 4. Question 19. Tycho wants to prepare a schedule for his jogging. He wants to jog exactly twice a week and on the same days every week. He never wants to jog on two consecutive days. How many different schedules can he prepare? We look at all the days of the week. We notice that there are seven, so on the first day of the week he has seven different options to choose for his jogging. Now next, we have to choose another day of the week that isn't the same day and isn't on a day directly before or after, since he doesn't want to jog on consecutive days. So every time we do this, we notice that if he chooses Thursday, for example, we can't choose Thursday, Wednesday, or Friday for the next second jogging of the week. So that always takes away three of our options. So that means for the second day he only has four options. After this, we just have to do the simple formula of 7 times 4 divided by 2, and that will lead us to the answer b, 14. Question 20. Emily wants to write a number into each cell of a 3x3 table so that the sum of the numbers in any two cells that share an edge are the same. She has already written two numbers as shown in the diagram. What is the sum of all the numbers in the table? Look at the table. We see that Emily has already written a 2 and a 3 in two cells. We can start off by labeling each cell with a different variable from a, b, c, d, e, f, and g to help us determine what the values will be. Now, since they will have to share sides and be the same values, we know that 2 plus a will equal a plus d, which will equal D plus 3. Or 2 plus A equals A plus D. With that we can take away the A and say that D equals 2. Now if we look at the other side A plus D equals D plus 3. We can take away the D from both sides and have A equals 3. Now if we continue on A plus D equals D plus F and A equals F so F will be 3. Now we know that C is 3, E is 2, B is 2, G is 2. We know this since the sides are all sharing a pattern and they will always be equal to 2 or 3. So to find the solution we must do 2 plus A plus B plus 3 plus D plus 3 plus 2 plus F plus G pattern is either a 2 or a 3 sharing sides with the opposite number. That is simplified into 5 plus 5 plus 5 plus 5 since there are four sets of 2 plus 3 and since it is an odd number of cells 9 there will be one remaining 2. And then we can get the answer by adding them together which will be D 22. Question 21 The numbers of degrees of the angles in a triangle are three different integers. What is the minimum possible sum of the smallest and largest angles? With this we can immediately throw out an equilateral triangle which would have 60 degrees for each angle making it 120 degrees. Next up we know that A has to be less than B which is less than C since we will have three different angles. Now A plus B plus C will be 180 degrees as is true of any triangle and we know that A plus B plus C will be greater than A plus B plus B since C is greater than B and with that we also know that A plus B plus B will be greater than 2B. Now with this we can reason that B has to be less than 90 degrees since if it is more than that then C will have to be even greater and that will not be a triangle since C would have to be at least 91 degrees and that would be 181 degrees without A. So B is somewhere between 2 degrees and 89 degrees. Now that A plus C is equal to 180 degrees minus B since this is just a different way of writing A plus B plus C equals 180. With that we can amplify our formula and write out 91 degrees is less than or equal to the difference of 180 degrees and B which is less than 178 since it will have to be a triangle. That is possible. With that we know that A will be 1 degree, B will be 89 degrees, and C will be 90 degrees. This fits all our criteria of A being less than B being less than C, A plus B plus C equaling 180 degrees, as well as B being less than 90 degrees and allows for the minimum possible sum between the smallest and largest angles. So our answer will be A plus C or 90 plus 1 which is the answer C 91. Question 22. 10 kangaroos stood in a line as shown in the diagram. At some point two kangaroos standing side-by-side and facing each other exchanged places by jumping past each other. This was repeated until no further jumps were possible. How many exchanges were made? To solve this problem we have to have all the kangaroos facing the same way as the kangaroos that are facing the same way as they are. They cannot be facing each other. We can start off by writing down each kangaroo. You can also draw arrows instead if this helps you visualize better or even draw the kangaroos over. But the R's represent the kangaroos looking to the right so we have three R's then two L's, kangaroos that are looking to the left, three more R's and then two more L's. In each one exchange we'll switch an L and an R as they move like so. They continue to switch like this until the L makes it all the way to the left side and cannot stare at any other R's. The other L's must follow suit. This can also be seen as the R's making their way towards the right since they will also not be looking at the L's. But I chose to do it like this. Once we have all the L's get to the end then we know that all kangaroos are staring either one way or another without staring directly into each other. This took 18 different moves so that tells us that the answer was C, 18. Question 23. Diana has nine numbers 1, 2, 3, 4, 5, 6, 7, 8, and 9. She adds 2 to some of them and 5 to all the others. What is the smallest number of different results she can obtain? To do this we want to add 5 to the smallest numbers so that we get numbers that we will get once we add 2 to the larger numbers meaning 1 plus 5 will give us 6 which is a value we will be able to get later on. 2 plus 5 as well, 3 and 5 to give us 8. And once we reach 4 we want to add 2 so that we can get 6 which we have gotten previously by doing 1 plus 5. 5 plus 2 will give us 7 which we've already had. 6 and 2 will give us 8 which has already occurred. 7 and 2 will give us 9 which unfortunately we have no way of duplicating unless we added 5 to 4 which would mean the 6 is not duplicated so it would be the same answer. 8 and 2 will give us 10 since there is no other number. If we added 5 it would also be a different number and 9 and 2 will also give us 11 since 9 and 5 would also be a unique number that cannot be duplicated. So if we add up all the different results we had 6, 7, 8, 9, 10, and 11 we get the answer B, 6. Question 24. Buses leave the airport every 3 minutes to drive to the city center. A car leaves the airport at the same time as one of the buses and drives to the city center by the same route. Takes each bus 60 minutes and the car 35 minutes to drive from the airport to the city center. How many buses does the car pass on its way to the center not including the bus it left with? Know that the bus and the cars will take the same routes. We imagine that each bus leaves between 3 minutes then we can say that the first bus on this line left 3 minutes ago and the bus before that left 6 minutes ago, 9 minutes ago, and so on. Now the car will take 35 minutes to get to the destination. A bus which left 3 minutes ago will have a total distance of 60 minutes as all buses do but since it left 3 minutes ago it will have 57 minutes left of traveling. That means the car will get there before the bus does since the car will be there in 35 minutes and the bus will be there in 57 so the car will pass that bus. So if we take 60 minutes minus the time traveled we will get the time left and any time that the time left is greater than the car's time left which is always 35 minutes then we know that the car will pass that bus. So if we have a bus that has been traveling for 30 minutes it will have 30 minutes left of travel when the car leaves. This means that after 30 minutes the bus reaches its destination however after 30 minutes the car still has 5 more minutes of travel so it cannot pass that bus. Same thing is true for a bus that has left 27 minutes ago since that bus only has 33 more minutes left of traveling so it will reach the destination 2 minutes before the car. However, a bus that left 24 minutes ago will reach the destination in 36 minutes so the car will pass that bus with 1 minute to spare. We count out all the buses that were before 36 minutes then we can look at our plot line of 3 minute, 6 minute, 9 minute, 12 minutes, 15 minutes, 18 minutes, 21 minutes, and 24 minutes and with that we get a total of the answer A, 8. Question 25. Felicia's tablecloth has a regular pattern as shown in the diagram. What percentage of the tablecloth is black? We look closer at the tablecloth we notice we can divide it into squares. We make a grid such as this we can divide the tablecloth into a 5 by 5 piece. Now only the outside perimeter boxes have black in them so 16 out of 25 of the 5 by 5 box have black in them. However, each of the 16 is identical and they are not fully black. They are only half black so we will divide the 16 by 2. With this we will have 8 so 8 out of the 25 is the percent is the fraction for our black tablecloth. If we do this we get an answer of D, 32 percent. Question 26. Each digit in the sequence starting with 2, 3, 6, 8, 8 is obtained in the following way. The first two digits are 2 and 3. Afterwards each digit is the last digit of the product of the two preceding digits in the sequence. What is the 2017th digit in the sequence? So for this all we do is multiply the first two digits and then we take the last digit of the product and that is the next number in our sequence. So we start off with 2 and 3 multiply them get 6. Then we multiply 3 and 6 and get 18. Now since this number has more than one digit we look at the last digit of the product and erase everything else and we are left with 8. So that means we do 6 times 8 which gives us 48 which we get rid of the 40 and we are left with 8. Then we have 8 and 8 which gives us a 4 and so on and so on. Now we see when we repeat this pattern we have 6, 8, 8, 4, 2, 8, 6, 8. This is a pattern that's going to repeat 6, 8, 8, 4, 2, 8. The sequence repeats on a loop. We are looking at the 2017th digit in the sequence. We can take 2017 and subtract the first initial two digits the 2 and the 3 of the sequence since they are the starting values and do not repeat. So we are looking at 2015. We divide 2000 by 15 the number of digits in our sequence then we get 335 with a remainder of 5. This means that the sequence of 6, 8, 8, 4, 2, 8 will repeat 335 times and the remainder of 5 means that we just have to move over five more spots. So the 6 will be the first, 8 will be the second, 8 will be the third, 4 will be the fourth, and 2 will be the fifth. And that is our remainder 5. So that means we are left with the answer A, 2. Question 27. Mike had 125 small cubes. He glued some of them together to form a big cube with nine tunnels leading through the whole cube as shown in the diagram. How many small cubes did he not use? Now the cube is made of a 5 by 5 by 5 pattern. Each of the tunnels will be 5 cubes deep. If we look at the first three tunnels on one side they each take away 5 cubes which means they take away 15 cubes in total. So negative 15 from the 125. Now the next set of tunnels will take away 4 since those tunnels will intersect with the previous tunnels we've already dug out. So we don't have to remove 5 we just have to remove 4 each time since one of every tunnel is already removed for us. So that gives us negative 12. The third set of tunnels we do the same thing since we already have two sets of tunnels made for us we can take we don't have to take away as many cubes so we only take away 3 cubes for each tunnel which gives us negative 9. We add all these together negative 15 or take away 15 take away 12 take away 9 cubes from the 125 then we get a total of 39 which is our answer D. Question 28. Two runners are training on a 720 meter circular track. They run in opposite directions each at a constant speed. The first runner takes four minutes to complete a full lap. The second runner takes five minutes. How many meters does the second runner run between two consecutive meetings of the two runners? If we look at the first runner and we notice that they have a speed of 720 meters for every five minutes. The second runner is a speed of 720 meters every four minutes since it is a 720 meter circular track and they are taking five and four minutes. Now we have a ratio of four to five or the second runner to the first runner. Now all we have to do is do four divided by the sum of four and five times 720 which is the length of the track. When we do this we get 4 divided by 9 times 720. This can be simplified by dividing the 720 by the 9 giving us 4 times 80 and that leads us to the solution which is E, 320 meters. Question 29. Sarah wants to write a positive integer in each box in the diagram so that each number above the bottom row is the sum of the two numbers in the boxes immediately underneath. What is the largest number of odd numbers that Sarah can write? If we look at the pyramid it has several levels and we know that the two boxes below one box when added together will be the box above it and we are only looking at this in terms of odd and even numbers. Know that a property of even numbers is an even number will only occur when it is the sum of two even numbers or two odd numbers and an odd number will only occur when it is the sum of one odd and one even number. So to maximize the amount of odd numbers that Sarah can write we have to make the top number even. This means the next two numbers will be odd and odd since two odds when added together will make an even number. Now to make an odd number we have to have an odd and an even. Since this level has three spots that both share odds right above them we can use 1E to facilitate odd numbers as results on the two above spots. We just have to do this continuously going down the level knowing that every odd number is the sum of an even and an odd number like so and every even number is the sum of two odd numbers or two even numbers. To maximize odd numbers we will make them odd instead of even. Once we do this we get a pyramid that looks like this and it is filled up with all these E's and O's or even and odd numbers. When we total up the numbers we find out that the largest number of odd numbers that Sarah can write is answer D, 10. Question 30. The diagram shows parallelogram ABCD with area S. The intersection point of the diagonals of the parallelogram is O. Point M is marked on DC. The intersection point of AM and BD is E and the intersection point of BM and AC is F. The sum of the areas of the triangles AED and BFC is 1 3rd S. What is the area of the quadrilateral EOFM in terms of S? We are given the fact that AED plus DME give us half of DA times DM since that is the triangle. We also know that triangle BFC plus triangle MCF is equal to 0.5 times DA times MC. Similarly enough it is also a triangle. Now with this we can have the formula triangle AED plus triangle DME plus triangle BFC plus triangle MCF is equal to half of DA times the sum of DM plus MC. And we know that 1 half DA times DC since DM plus MC is also just simplified DC. And that will give us half of S or half of the entire area of the parallelogram ABCD. Now triangle DME plus triangle MFC gives us an area of half of S minus 1 3rd of S. DME plus MFC will give us 0.166 repeating S. Furthermore we can get to DME plus MFC plus the quadrilateral EOFM will give us 1 4th of S. And now all we have to do is subtract DME and MFC from the quarter of S to be left with quadrilateral EOFM to get our answer. So that means that 0.25 S minus 1.6 repeating S will give us 0.083 repeating S. When we return that back into a fraction we get answer D, 1 12th S.

Math Kangaroo

Lucas Naleskowski

arithmetic operations

geometric properties

pattern recognition

algebraic equations

problem-solving

mathematical comprehension