This is the Math Kangaroo Solutions Video Library, presenting solution suggestions for Levels 7 and 8 from the year 2014. 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. Each year the date of the kangaroo competition is the 3rd Thursday of March. What is the latest possible date of the competition in any year? We make ourselves a calendar with Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, and Saturday. We want to make Thursday the last day of the week. To do so, we have to make Friday the first day of the week. So that way Friday is the 1st, Saturday the 2nd, Sunday the 3rd, Monday the 4th, Tuesday the 5th, Wednesday the 6th, and Thursday is the 7th. Therefore, Thursdays will all be multiples of 7, of 14 and 21. Therefore, to make the date of the kangaroo competition the 3rd Thursday of March and as late as possible, then the correct answer will be D, March 21st. Question 2. How many quadrilaterals of any size are shown in the figure to the right? We can redraw the figure. All we have to do is count how many 4-sided shapes we can see. The first is marked by the yellow, that's one shape. Then the green, which is a second shape. Now since these are two rectangles overlapping, they will make more than just these two shapes, such as this blue one, which makes a third shape. And the fourth quadrilateral will be represented by this red. So in total, we can make D, four shapes. Question 3. What is the result of 2014 times 2014 divided by 2014 minus 2014? We can rewrite this using our order of operations. We can take out the 2014 divided by 2014, which is 10, and have this be 2014 times 1 minus 2014. Then we can multiply the 1 and the 2000 by 14, since that comes first before subtraction, at 2014 minus 2014, which leaves us with the correct answer, A, 0. Question 4. The area of rectangle ABCD is 10. Points N and N are the midpoints of sides AD and BC. What is the area of quadrilateral MBND? Take a closer look at our shape, and start off by realizing that AD times DC will give us 10, since that's how you find the area of a rectangle. You want to figure out DM times DC, since if we split this rectangle in half by the midpoints M and N, we notice that we get two smaller rectangles, which are half shaded and half not shaded. This means that the quadrilateral MBND is half the area of the whole rectangle. So with this, we know that DM will be half of AD. We multiply this out, and we get DM times DC, giving us 5. This means that our correct answer will be B, 5. Question 5. The product of two numbers is 36, and their sum is 37. What is their difference? To start this problem off, we must find each pair which has a product of 36. 1 and 36, 2 and 18, 3 and 12, 4 and 9, and 6 and 6. Now, we must find out their sum, so that we can match it up with a number that has the sum 37. 1 and 36 gives us 37, 2 and 18 gives us 20, 3 plus 12, 15, 4 plus 9, 13, and 6 plus 6 gives us 12. Only one of these had the sum of 37, so we know that our pair of numbers is 1 and 36. To find their difference, we must just simply subtract 36 minus 1, which leads us to our answer, which is E, 35. Question 6. Wanda has several square pieces of paper with an area of 4. She cuts them into squares and write triangles in the pattern shown in the diagram. She takes some of the pieces and makes the bird shown in the second diagram. What is the area of the bird? If we look at the square that Wanda has, we know that it has an area of 2, and we can label the small sections she cuts out. The largest triangle will have an area of 2, since it is half of the square. The smallest square is 1 fourth of the larger square, so it has an area of 1. The remaining two smaller triangles each have an area of half. Now if we look at the second diagram, the bird that Wanda made, we can just label each of the shapes to its corresponding area, which we predetermined. We get the following. Now if we simply add all these together, 0.5 plus 1 plus 0.5 plus 2 plus 0.5 plus 1 plus 0.5, we get our total, which is E, 6. Question 7. A bucket was half full. A cleaner added two liters to the bucket. The bucket was then three quarters full. What is the volume of the bucket? We can call x equal to 50 percent. x plus 2 will be 75 percent. Since when the cleaner was added, which was two liters, we got 75 percent, or three quarters. We do 75 percent is equal to 50 percent plus two liters. So we know that 25 percent of the bucket is two liters. Now we want to know the full volume of the bucket, which is 100 percent. To get from 25 to 100, we must multiply by 4. So 2 liters times 4 gives us our answer, which is B, 8 liters. Question 8. George built the shape shown using seven unit cubes. How many such cubes does he have to add to make the cube with edges with a length of 3? If we look at each plane, we can make it so. The white space represents where the cube is, and the black space represents where the cubes aren't and need to be filled in. The three levels will look like so, with seven cubes missing. Now if we count up all the black spaces, we get 8 in the first area, 4, and 8. If we total these together, we get our answer, which is E, 20. Question 9. Which of the following calculations gives the largest result? Let's start off by writing down each of the possible solutions. 44 times 777, 55 times 666, 77 times 444, 88 times 333, and 99 times 222. Now we can simplify each of these, as they are all multiples of 11 as well as 111. So 44 times 777 is equal to 4 times 11 times 7 times 111. And we can do this for the following answers as well. Now once we get here, we can cross out 44 times 777 as well as 77 times 444, as those are both equal, so they cannot be the largest result. Next we simplify further by rewriting 11 times 111 times 9 times 2, 11 times 111 times 8 times 3, or 11 times 111 times 5 times 6. Now ignoring the 11 times 111 and focusing on the last two parts, 5 times 6, 8 times 3, and 9 times 2, we get the following, 30, 24, and 18. These values will also have to be multiplied by 11 as well as 111. But with this, we find out that 30 is the largest of these, since it is greater than 24, and it is also greater than 18. So our correct answer will be B, 55 times 666. Question 10. The necklace in the picture contains dark gray beads and white beads. Arnold takes one bead after another from the necklace. He always takes a bead from one of the ends. He stops as soon as he has taken the fifth dark gray bead. What is the largest number of white beads that Arnold can take? Look at the necklace. It doesn't matter which side we start off on. Either way, we'll have to take off one dark gray bead. Take off a bead from the left, which lets us take off one white bead. Take off another bead from the left. It allows us to take off three more white beads. Through this, we can go to the right side, take off one dark gray bead, so that we can take two more white beads. Taking off one dark gray bead, we lead to one white bead, and our final dark gray bead will be from the right side, allowing us to take off that white bead. So in total, the largest amount of white beads that Arnold can take is seven. So answer D. Question 11. Jack has a piano lesson twice a week, and Hannah has a piano lesson every other week. In a given quarter, Jack has 15 more lessons than Hannah. How many weeks long is their quarter? We make a chart like so, and label the number of weeks going up by twos. We label Hannah's lessons as well as Jack's lessons. Since Hannah has her lessons once every other week, and Jack has two every week, every two weeks, Hannah will have one lesson, and Jack will have four, like so. For four weeks, Hannah will have two lessons, and Jack will have had eight. Now, at this point, Jack has had six more lessons than Hannah. We want to get to 15 more lessons than Hannah, so we can continue along this path. Three and 12, four and 16, and then after 10 weeks, Hannah will have had five lessons, and Jack will have had 20 lessons. This means that Jack has had 15 more lessons than Hannah. So a quarter takes 10 weeks, so the correct answer is E, 10. Question 12. In the diagram, the area of each circle is one centimeter squared. The area common of two overlapping circles is one-eighth centimeter squared. What is the area of the region covered by the five circles shown? Take a closer look at the circles, and start labeling their areas. This first circle will have an area of one. The next circle, which has two parts overlapped with it, will have an area of six-eighths, since we subtract one-eighth twice. Next circle will have seven-eighths, since we subtract one-eighth. Same is true of the next circle, and the final circle has an area of one. We can add all these together to get a total of plus 20-eighths. Simplifies to 16-eighths plus 20-eighths, which is 36 divided by 8. Which, when simplified, gives us our answer, which is B, nine-half centimeters squared. Question 13. This year, a grandmother, her daughter, and her granddaughter notice that the sum of their ages is 100 years. Each of their ages is a power of two. How old is the granddaughter? Since each of their ages is two, let's start off with two squared, plus two to the power of five, plus two to the power of six, which simplifies to four plus 32 plus 64, which gives us 100. The youngest person, two to the power of two, is the granddaughter, and two to the power of two gives us four. So, the granddaughter is four years old. Answer C. Question 14. Five equal rectangles are placed inside a square with a side length of 24 centimeters. It's shown in the diagram. What is the area of one rectangle? We look at the diagram closer. We label the sides 24 and 24. Since we know that all the rectangles are exactly the same, we can divide this into three parts, the blue lines, and we know that the three blue lines together are 24 centimeters. Divided by three gives us eight, so each of the blue lines has a length of eight centimeters. We look at the other side and do two more eight centimeter length segments, which leaves us with eight more centimeters. Since the other two sides have equal length, we can do eight divided by two, so each of the red sides has a length of four centimeters. Now we know that the rectangle is red times blue, so that means we will do four times eight to get the area of one rectangle. Four times eight gives us our solution, which is E, 32 centimeters squared. Question 15. The heart and the arrow are in the position shown in the figure. The heart and the arrow start moving at the same time. The arrow moves three places clockwise, and the heart moves four places counterclockwise, and then they both stop. They continue the same moves over and over again. After how many such moves will the heart and the arrow land in the same triangular region for the first time? We look at our shape and we put in our heart and arrow. We know that we have to move the heart four places counterclockwise and the arrow three places clockwise, which puts them like so. This is done again. They will be sitting like so. Now the way the shape is designed, moving four places counterclockwise is the same as moving three places clockwise, and moving four three places clockwise is the same as moving four places counterclockwise, which means that the heart and the arrow will keep skipping each other, and they will never move to touch each other on the same spot. So our answer will unfortunately be E. It will never happen. Question 16. The diagram shows the triangle ABC, in which BH is a perpendicular height, and AD is the angle bisector at A. The measure of the obtuse angle between BH and AD is four times the measure of angle DAB. See the diagram. What is the measure of angle CAB? We take a closer look at the given diagram of the triangle. We can see that HB is a straight line and it is divided by AD, which gives an obtuse angle of 4A. So we know since it is a line, it will have 140 minus 4A. Since a line has an angle of 180, the side next to it will have the angle 180 minus whatever it is next to it. So we have 180 minus 4A. Now we look at this triangle. The right triangle has an angle of 90 as well. So 180, which is the total angle measurement of the three angles in a triangle, is equal to A plus 90 plus 180 minus 4A, which is 180 equals 270 minus 3A. Or negative 90 equals negative 3A. We divide both sides by negative 3. We get A equals 30. Now we were looking for angle CAB. CAB is equal to 2A. 2 times 30 gives us our answer, which is C, 60 degrees. Question 17. Six boys share an apartment with two bathrooms in which they use every morning beginning at 7. There is never more than one person in either bathroom at one time. They spend 8, 10, 12, 17, 21, and 22 minutes at a stretch in the bathroom respectively. What is the earliest time that they can finish using the bathrooms? Now we start off at 7. To do this, we want to get rid of the largest time wasters first, so the 21 and the 22 minutes. So after so, it'll be 7.22 and 7.21. Then we want to get rid of the next largest time users, 17 and 12. We want to give the larger, 17, to the earlier time, 7.21, like so. 12 will go to 722. It will be 738 and 734. We are left with 8 and 10 minutes. We will give the 10 minutes to the earlier 734 and the 8 minutes to the later 738, like so. After such, it will be 744 after the first bathroom is done of use and 746 after the second bathroom is done being used. Therefore, the earliest time they can finish using the bathrooms will be B, 746. Question 18. A rectangle has sides of length 6 centimeters and 11 centimeters. One long side is selected. The bisectors of the angles at either end of the side are drawn. These bisectors divide the other long side into three parts. What are the lengths of these parts? Draw a rectangle 6 centimeters by 11 centimeters, and we draw the angled bisectors, like so. Since they are angled bisectors, we know they make a 45 degree angle. Now, this tells us an important detail. This creates an isosceles triangle, meaning the other leg of this triangle will have a length of 6 centimeters as well, represented by the red segment. The opposite triangle that is made by the other bisector is the same, 6 centimeters, represented with the red line. Now, these red lines overlap. Since we know that the length of this one side of the rectangle is 11 centimeters, we know that this overlap, represented by the blue segment, is equal to 1 centimeter. So, 6 minus 1 will give us 5, meaning the lengths of the three parts created will be answer E, 5 centimeters, 1 centimeter, and 5 centimeters. Question 19. The pirate Captain Sparrow and his pirate crew dug up some gold coins. They divided the coins among themselves so that each person got the same number of coins. There had been 4 less pirates, then each person would have received 10 more coins. However, if there had been 50 coins less, then each person would have received 5 coins less. How many coins did they dig up? Let G represent the amount of gold coins, and P the amount of pirates. Now, there had been 50 coins less, then each person would have received 5 coins less. That means we can divide 50 by 10. This means there are 10 pirates. Now, if there had been 4 less pirates, each person would have received 10 more coins. That means 10 times 6 will give us 60, which means that 4 pirates had an equal worth of 40 gold. It means 4 pirates equals 60 gold. If we multiply this out, we get 10 pirates equals 60 gold times 2.5, which gives us our answer, D, 150. Question 20. The average of two positive numbers is 30% less than one of them. By what percentage is the average greater than the other numbers? This means A plus B, all divided by over 2, is 70% of A. That means A plus B equals 140% of A. It's going to be simplified to A is equal to 5B halves. If A plus B equals 5B halves plus B, then we can write out the sum of A and B divided by 2 is equal to 5B divided by 2 plus 2, all divided by 2, which then leads to this. Now, A plus B all over 2 will be 7B divided by 4, which gives us 7B divided by 4. If we divide 7 by 4, that gives us 1.75, or 175%. We subtract 100%, we are given 75%, which gives us our answer. The percentage that the number is less than the average is A, 75%. Question 21. Andy enters all the digits from 1 to 9 in the cells of a 3 by 3 table so that each cell contains one digit. He has already entered 1, 2, 3, and 4 as shown. Two numbers are considered to be neighbors if their cells share an edge. After entering all the numbers, he notices that the sum of the neighbors of 9 is 15. What is the sum of the neighbors of 8? Draw out the table that Andy drew. We can start theorizing where the 9 is placed. If we put the 9 in this corner, then we would have to put 8 in the middle, since we know that the sum of the neighbors of 9 is 15, we have to put the largest possible number. However, in this spot, 8 plus 1 plus 3 does not give us 15, it gives us 12, so 9 cannot be placed in the top middle. The same is true if 9 is next to 1 and 2, as well as if it is next to 2 and 4. However, when 9 is placed next to 3 and 4, and 8 is put down in the middle, we get 8 plus 4 plus 3. So the sum of the neighbors of 9 is 15. To find the sum of the neighbors of 8, we simply have to add together the 4 neighboring tiles, which we know will have values of 5, 6, 7, as well as the 9 we already know, since those are all the other remaining numbers. When added together, we get our answer, which is E, 27. Question 22. An antique scale is not working properly. If something is lighter than 1,000 grams, the scale shows the correct weight. However, if something is heavier than or equal to 1,000 grams, the scale can show any number above 1,000 grams. We have five weights, each under 1,000 grams, weighing A grams, B grams, C grams, D grams, and E grams, respectively. When they are weighed in pairs, the scale shows the following. B plus D equals 1,200. C plus E equals 2,100. B plus E equals 800. B plus C equals 900. A plus E equals 700. Which of the weights is the heaviest? Now, since we are given these values, we can start to figure out which scale or which weight weighs more. Since we know that B plus E is 800 and C plus E is 2,100, we can tell that C is greater than B. Since the E's are the same, the C would have to be larger. Same with B plus D is 1,200, and B plus C is 900. So this means D is greater than C. Using the same method, B will be greater than A, and C will be greater than E. With that, we can write out this. A is less than B, which is less than C, which is less than D. This will stand to reason that D is the largest. However, we are left with E. But since we proved that C is greater than E, we don't have to worry about that. So we will find out that the heaviest weight is D. Question 23. Quadrilateral ABCD has right angles only at vertices A and D. The numbers shown are the areas of two of the triangles. What is the area of ABCD? Now, we know that the area is base times height. A triangle ABD will be equal to a triangle ABC, since they share base and height. So ABD is equal to 15, which means ABC is also equal to 15, as we can tell here. Label this E. Now, BCE is equal to 10. And one of the properties we know is they will be congruent. So AED divided by ECD is equal to ABE divided by BCE, which can be simplified to 10, which is AED divided by ECD, the unknown, equal to 5 divided by 10, which can, if we multiply 5 divided by 10 by 2, we get 10 divided by 20. So ECD must equal 20. To find the full area of ABCD, we just add up all the total areas of the triangles, which gives us 10 plus 10 plus 5 plus 20, which gives the answer B, 45. Question 24. Liz and Mary compete in solving problems. Each of them is given the same list of 100 problems. For any problem, the first of them to solve it gets four points, and the second to solve it gets one point. Liz solved 60 problems, and Mary also solved 60 problems. Together, they got 312 points. How many problems were solved by both of them? Let's say x is equal to one point problems, and y equal to four point problems. With this, we know that 1 times x plus 4 times y is 312. And we also know that x plus y will be 120. Now, x plus 4y gives us 312. We know that x equals 120 minus y. So if we substitute that in for x, we get 120 minus y plus 4y equals 312, or 3y equals 192, which gives us y equals 64. Since there were 120 problems solved in total, we do 120 minus 64, we get 56. That means there were 56 one point problems. That means 56 were both solved by Mary and Liz. Question 25. David rides his bicycle from his home to his friend's house. He was going to arrive at 3pm, but he spent two thirds of the planned time covering three fourths of the distance. After that, he rode more slowly and arrived exactly on time. What is the ratio of the speed for the first part of the trip to the second to the speed for the second part? We have t equal to time and d equal to distance. Set up an equation like so. Three fourths d divided by two thirds t. So that's equal to 3d divided by four times three divided by 2t, which is 9d over 8t. The same can be done by the remaining distance, which is one fourth d and one third t, which gives us 3d divided by 4t. We divide these 9d divided by 8t, all divided by 3d divided by 4t, or the first part of David's biking divided by the second part of David's biking, we get the ratio of three to two. So our correct answer is c, three to two. Question 26. We have four identical cubes. They are arranged so that a big black circle appears on one face, shown in the last picture. What can be seen on the opposite face? If we draw a net, which is a cube unfolded, and we draw a quarter of the large circle, we take the four cubes, as shown in the question, we can start seeing what pattern is on what side. The line across will be here, the square will be here, the smaller circle will be in this place. That only leaves the smaller square and the smaller diagonal, which means the smaller diagonal is left right here. When this net is folded up, that will be the opposite from the larger circle. So that means our answer is a. Question 27. A group of 25 people consists of knights, serfs, and damsels. Each knight always tells the truth. Each serf always lies. And each damsel alternates between telling the truth and lying. First, 17 of them said yes, when asked, Are you a knight? Then 12 of them said yes, when asked, Are you a damsel? Finally, eight of them said yes, when each was asked, Are you a serf? How many knights are in the group? The group has 25 people. Since 12 of them said yes, when asked, Are you a damsel? We can subtract 12, since we know those 12 will not be knights. Since eight of them said yes, when asked, If you are a serf, you can subtract eight more, knowing those will also not be knights, since knights cannot lie. This gives us 25 minus 12 minus 8, which leads to our answer, which is b, 5. Question 28. Several different positive integers are written on the board. Exactly two of them are divisible by two, and exactly 13 of them are divisible by 13. Let m be the greatest of these numbers. What is the smallest possible value of m? Since we want as few numbers as possible, we will have to compound those numbers which are divisible by two and 13, meaning we will have to have two numbers that are both divisible by 13 and two. So the first smallest number, which is divisible by 13, is 13. Next will be 26. It is divisible by 13 and two. Next will be 39, then 52, which is also divisible by two and 13. This fulfills the two numbers which are divisible by two. So we just need to keep going with nine more numbers which are divisible by 13, in ascending order, to get the smallest value of m. Here's 65, 91, 117, 143, 169, 195, 221, 247, and 273. So that means the smallest possible value of m is c, 273. Question 29. On a pond, there are 16 water lily leaves and a four by four pattern as shown. A frog sits on a leaf in one of the corners. It then jumps from one leaf to another, either horizontally or vertically. The frog always jumps over at least one leaf and never lands on the same leaf twice. What is the greatest number of leaves, including the one it is sitting on, the frog can reach? Draw the four by four pattern of lilies with the frog. We can start marking each lily that the frog has been with a red circle. And the frog leaps to the next corner, which will also be marked by a circle. Then it leaps to the next corner, and the next corner. Then it leaps to the closest lily pad, which it can. Then it leaps, skipping one lily pad. It turns, leaps, then it turns 180 degrees to leap here. It turns 180 degrees more to leap back. Then it turns and leaps down to turn back and leap all the way to the top. So it can turn down, leap here, turn to the left and leap, turn to the right to leap to the end, turn back to leap to the left. And finally, it will leap to the top, meaning it covered all lilies in the four by four pattern 16 lilies. So the correct answer is a 16. Question 30. A five by five square is made from one by one tiles, all with the same pattern as shown. Any two adjacent tiles have the same color along the shared edge. Perimeter of the large square consists of gray and white segments of width length of one. What is the smallest possible number of such gray unit segments? Start off with the five by five grid and start filling it out with the one by one tiles. Since we want the largest amount of white faces showing, we will organize them to do as such. Show as many white faces as possible. Once we do three sides, it's slightly more complicated, as we need to have them turning inwards, like so. And to maximize the amount of white faces, we will have to organize the tiles like so. Then once we count up the gray faces, we have one, two, three, four, and five gray faces on the outer edge. So the smallest possible number of gray unit segments is answer B, five.

Math Kangaroo 2014

Levels 7 and 8

Lucas Naleskowski

math problem-solving

geometry and algebra

competition strategies

educational video