Hello and welcome to the Math Kangaroo Media Library. You are about to view interactive solutions to levels 5 and 6 of the 2015 competition. You will likely notice that some of the solutions are slightly different from the suggested solutions you may have already reviewed. So as you follow along, compare your own solutions and the suggested solutions with the current presentation and make sure you understand the differences. If you have any questions or comments, feel free to contact me at the address provided. My name is Luke and I am a past Math Kangaroo participant, and I hope you will find this presentation useful in preparing for your next Math Kangaroo competition. Question number one. Which figure has one half of its area shaded? Looking at the five figures, the triangle in B immediately attracts our attention. Let's underline it and discuss why the other possibilities are incorrect. In C and D, we can clearly see that three quarters or one quarter of the area are shaded. And in figures E and A, we have an odd number of fields. So those figures cannot be shaded exactly in half if they're divided already into an odd number of fields. So indeed, our intuition was correct and B is the answer. Question number two. My umbrella has kangaroo written on top. It is shown in the picture to the right. Which of the pictures below does not show my umbrella? So here is the top view of the umbrella with the word kangaroo here spelled on top. And we have to decide which view of these five is not possible when looking at the umbrella from the side. And the way we're going to do this is to match the orientation of the letters with the correct spelling. So here in picture A, A and G, as viewed from the top, would have to appear in a clockwise orientation. So that looks like this piece of the umbrella. Here we have A followed by N followed by G. Now we have to pay attention to how these letters are positioned next to one another. We see that the downslope here of A and the line that starts N, they are in the correct position here in our diagram. And likewise, the curve of G is facing the correct side of the letter N. So A is possible. We make a note of that here saying that there is a checkmark by A that is possible. Let's move on to B. Let's use a different color here. Double O has to be followed by K, so it looks like in B we have this piece. And is K correctly positioned? We have an upright here segment of K, and that is directly next to O, and that looks like it's correct. So B, we don't see any problems with that. Let's move on to C, which represents GAR. And that looks like this segment of the umbrella. G followed by A followed by R, like so. And now let's look at the orientation of the letters. We have G here. The open side of G is facing A, so that's okay. And then in the next segment, we have A, the slant side of A facing the upright side of R. And here, where I'm going to put the blue dot, that is backwards. The down slope of A is not facing the straight edge of R, so the R is reversed here. And so this picture is the one we're looking for. This is not a correct representation of the umbrella viewed from the side. And so we can answer the question right away. The correct answer here is C. In other words, C is not a correct representation. It does not show the umbrella. Question number three. Sam painted the nine squares as shown in the figure to the right using the colors black, white, and gray. So here is Sam's picture. He has a three-by-three grid with squares either in black, white, or gray. At least how many squares does he need to repaint so that no two squares with the common side are the same color? And we observe that the three colors do border each other, and sometimes we have a run-on. And the black squares here make an L shape because we cannot really distinguish the boundary between them. The bottom square here does border two other black squares on an edge. So that square at least would have to be repainted. So let's put a cross through it to indicate that it would have to be repainted. And the color it would have to become is not white, otherwise we would have a conflict here with the white square. It could become gray because the only gray square here would border this repainted one on the corner, and that is okay. They would not have the same side color. And the second such situation appears over here, where we have two gray squares in the upper right-hand corner that share an edge. One of them would have to be repainted, and we choose to repaint the corner square over here. It would have to be repainted black because the white color here would border on an edge or on the side with the white square I'm pointing to. So essentially, this repainting could be accomplished by switching the two colors, gray and black, in the squares that are marked with a cross. And at least these two squares have to be repainted, so the answer here is A. Sam has to repaint two squares to accomplish this task. Question number four. There are ten ducks. Five of these ducks lay an egg every day. The other five lay an egg every other day. How many eggs do the ten ducks lay in a period of ten days? There are two rates in this problem, and we need to consider each one separately for a separate calculation. Let's begin with the five ducks that lay an egg every day. And so the calculation would look like this. We have five times, the rate is one egg per day. And then we're looking at a period of ten days. So multiplying these together, we have exactly a total of fifty eggs. Now, the other half of our ducks lay an egg every other day. And we can think of this, mathematically at least, as the rate of one half of an egg per day. And that's how we will calculate the amount of eggs here. We have five of these ducks. The rate at which they're laying eggs is one half of an egg per day. And then we multiply that by the same number of days, or ten. And the result is exactly half of the previous amount of eggs, or twenty-five. And then we add fifty with twenty-five, for a total of seventy-five eggs in total after ten days. So that's answer A, seventy-five eggs. Question number five. The figure to the right shows a board where each small square has an area of four centimeters squared. What is the length of the thick black line? So here in our diagram, we see some squares. We know their area. We see a thick black line. And I'll just make a quick copy of this diagram and enlarge it here so we can see better what's going on. We have two pieces of information. Each square has an area of four centimeters squared. So that allows us to calculate the length of one of these segments. That would have to be exactly two centimeters. So that the other length is equal to it. That's also two centimeters. And so the area is then four centimeters squared. And each square is exactly the same as every other square. And then we need to count the number of edges that the black thick line occupies. So let's do that. We have one, two, three, four, five, six, seven, eight, and nine edges. And then we can calculate the length. That is nine edges times two centimeters per edge, as we have discovered earlier. And that gives us a total of 18 centimeters for the length of the thick black line. And that, in our choices, is answer B, 18 centimeters. Question number six. Which of the following improper fractions is smaller than two? Let's recall first that an improper fraction is a fraction where the numerator, the number on top, is greater than the denominator, or the number on the bottom. In choice A, we have 19 over 8. 8 times 2 is 16. So the numerator is more than twice as large as the denominator. Consequently, the fraction in A is actually larger, strictly larger than 2. Likewise in B. 9 times 2 is 18, which is less than 20. The numerator is more than twice as large as the denominator. And so here the fraction represents a number greater than 2. In C, the gap is smaller. We have 10 times 2 is 20, which is just one off. But still, that difference is more than twice the denominator. So this is greater than 2. Here we have a number that's exactly equal to 2. 2 times 11 is 22. Or in other words, 22 is divisible by 11 exactly twice with no remainder. And finally, we have an E, 23 over 12, and that number is strictly less than 2. Since 2 times 12 is 24, and that number is now greater than the listed numerator here, 23. So our answer is E. The fraction 23 over 12 is smaller than 2. Question number 7. A pumpkin and a watermelon together weigh 8 kilograms. The watermelon is 2 kilograms lighter than the pumpkin. How much does the pumpkin weigh? Let's begin by assigning variables to our quantities. We have two weights, the weight of a pumpkin and the weight of a watermelon. So I will use P for the weight of the pumpkin and I will use W for the weight of the watermelon. And then we know that together they weigh a total of 8 kilograms. And this I can express as P plus W is equal to 8. The total weight is 8 kilograms. Next we know that the watermelon is 2 kilograms lighter than the pumpkin. So the weight of the watermelon, W, is, that is the equal sign, and then 2 kilograms lighter than the pumpkin would be the weight of the pumpkin minus 2. And that is our second equation. So we have W is equal to P minus 2. And we can solve this system of equations. I will proceed by substituting in my first equation here. I see P plus W is equal to 8. And then we have another name for W, that is P minus 2. That will be substituted in here for W. And next I write P plus P minus 2 is equal to 8. On the left-hand side we combine like terms. We have 2P minus 2 is equal to 8. Add 2 to both sides to isolate the variable. 2P is equal to 10. We finally divide by 2 on both sides to isolate P. We obtain P is equal to 5. And with the units we can conclude that the pumpkin weighs exactly 5. And the units are kilograms. And that comes out to be answer D in our problem. 5 kilograms. Question number 8. Each plant in John's garden has either 5 leaves only or 2 leaves and 1 flower. In total the plants have 6 flowers and 32 leaves. How many plants are there? And here to the right we have a picture representing the two types of plants in John's garden. We notice that here is a plant with 2 leaves and exactly 1 flower. And we have a total of 6 flowers. So there must be exactly that many plants containing a flower. So we have 6 flowers and that would be attributed to 6 of these plants with a flower. So let me make a copy here of this picture. And we have 6 of these. And we can say that this sort of a plant times 6 contains 6 flowers which is the total number of flowers present. And 6 times 2 or 12 leaves. And now how many leaves remain? And we can say that there remain 32 minus 12 or 20 leaves on plants that look like. And now I'll make a copy of the first type of a plant in John's garden and put it over here. In other words 20 leaves divided by 5 leaves per plant means there are exactly 4 of the flowerless plants. Okay and now we can add up the total number of plants. We have here first of all 6 of these plants and now 4 of the flowerless plant. So together there are 10 plants and that is answer A. How many plants are there in John's garden? John has 10 plants. Question number 9. Alva has 4 paper strips of the same length. She glues 2 of them together with a 10 centimeter overlap and gets a strip 50 centimeters long. So here's a diagram of her construction. There are 2 strips identical in length but we don't know what that length is. We place them not back-to-back but with an overlap of 10 centimeters on top of one another and the resulting strip here has a length of 50 centimeters. Now the question is with the remaining 2 paper strips Alva wants to make a strip 56 centimeters long. How long should the overlap be? And we will need to calculate the length of each of her strips before solving the problem. So let's let that unknown quantity be X. Here we have the length of one of these strips that is unknown. Let's call it X. To make a 50 centimeter strip we take 2 of these strips of length X. We first arrange them back-to-back and then slide them 10 centimeters on top of one another. So we have to subtract the overlap 10 centimeters from the total length. That's our equation here. We can solve this for X by adding 10 centimeters to both sides. We have 60 centimeters is twice the unknown length. So each of Alva's strips measures exactly 30 centimeters in length. And now we'll follow the same calculation but our overlap is not known. So I'll say let now Y be the length of the overlap. And we're trying to construct a 56 centimeter strip. So following the same formula we have the length that we want 56 centimeters. That is going to be twice the length of each of these strips so twice 30 centimeters. And now we're subtracting the overlap which in this case we don't know. And so we can solve this equation 56 centimeters is 60 centimeters minus the overlap. And so from 60 we must subtract 4 in order to obtain 56. So 4 centimeters is Y. That is the length of the overlap necessary for this construction. That is choice A. Alva needs to create an overlap of 4 centimeters to make a 56 centimeter strip. Question number 10. Tom used six squares with the side length of one to form the shape shown in the picture. What is the perimeter of the shape? Here we have this shape. It has three layers of three squares and then followed by two and one on the very top. And we know there's no symmetry here in this diagram so symmetry being what we usually use to calculate perimeter makes our calculation here a little bit more difficult. I'll make a copy of this shape and enlarge it so that we can see the construction. The scale is about twice the size here. And what I'll do is I'll pick off the layers one at a time beginning with the with the top piece. I'll take that off here like so and move it over. Okay so that has a perimeter of 4, one on each side. Here the perimeter is 4. The middle piece, I'll try to copy that out of this picture. So the middle piece over here has different dimensions. The perimeter will be slightly larger with one on the sides, two on top, also two on the bottom. Two on the top, two on the bottom. So the perimeter here, that comes out to be six. And finally the largest piece on the bottom, let me copy it, move it over. So clean up this area. The largest piece has dimensions of one on the side, three on the top and bottom. So its perimeter is eight. And the way we're going to calculate the figure's perimeter is to add up all these numbers and account for the over counting that we're performing. So how are we over counting? We have a surface that is counted twice or over counted. The bottom of the middle piece, where I emphasize that in green, is sitting exactly on top of the bottom piece. So these green surfaces touch and they do not contribute to the total perimeter. So that length of two has to be subtracted twice from the total perimeter. So that is minus four here that we're going to include in the calculation. And likewise for the middle and the top blocks touch, I'll emphasize that in blue here like that, those lengths do not contribute to the perimeter. We have to subtract them twice and that is going to be minus two over here from the calculation. So now with that, the total perimeter here is, and we have our individual pieces, eight plus six plus four. And we're subtracting the overlaps, minus four and minus two. And that gives us an answer of twelve. So the perimeter of Tom's figure, once he stacks up his pieces, is twelve. And D is the answer to number ten.

Math Kangaroo

interactive solutions

geometry

algebra

solution strategies

student preparation