Hello and welcome to the Math Kangaroo Media Library. You are about to view interactive solutions to levels 1 and 2 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 this 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'm a past Math Kangaroo participant, and I hope you will find this presentation useful in preparing for your next Math Kangaroo competition. Question number 1. Look closely at these four pictures. Which figure is missing from one of the pictures? To find out which figure is missing from one of the pictures, I'll compare all the pictures together by marking off the figures that I can see in all pictures all at once. So let's begin with the larger red circle here in the picture on the left. Let's check it off that we see it in the first picture. In the second picture, we also see it. In the third picture, there is a large red circle, and also in the last picture on the right. So let's continue in this fashion, moving on to the smaller red circle. We see one in the picture on the left. There are two in the next picture. The next picture after that contains one small red circle and the last picture has one small red circle. Let's continue on to the next figure. Let's look at the green rectangle here, and mark it in purple. We see it in the picture on the left. The next picture over does not have a green rectangle, and the last two pictures have a green rectangle. So we see that where the question mark is in the picture second from the left, a green rectangle is missing. That is the figure that is missing from one of our pictures. The choice we should mark as the answer here is D, the green rectangle is missing. Question number 2, find the piece missing from the house on the right. Here we have several pieces, five of them, and here is a picture of a house where a piece of the roof is missing. So our job is to find the correct fitting puzzle piece that will fill this missing area of the roof. To do that, what I will do is I will draw in here the missing piece of the roof in the same color in black, like so, and then shade that area in, also in black. And that is the shape of the piece that we are missing. So let's try to draw it again here and see how it would look like. We have, if we were to enlarge this piece, we would have a straight edge like this, and a piece like this, and then slightly down to the right, slightly down again, and probably something like this, where I can erase this little bit here. So that is roughly how our missing piece looks like, and now let's compare it to the pieces that we have. Maybe rotating it will help also to make our job easier. So I will rotate this piece, and if I do that, it kind of starts looking like the shape represented in B. So I have copied the piece that we are missing, and it looks very much like the piece represented by the letter B. So we will make that our choice here, answer the question with B. Question number three, there are five ladybugs shown to the left, and we see that each ladybug has some spots, and it looks like each ladybug has a different arrangement of spots. How many spots are there on all the ladybugs together? So our job here is to count all the spots on all the ladybugs all together. So let's begin by looking at one ladybug at a time and counting the number of spots on that ladybug. So the first one here has one spot on each side for a total of two spots. So let's write that down next to this ladybug, two spots. The next ladybug has one spot on the left and two on the right, so that's a total of three. The next one has two spots on the left, one on the right for a total of three also. The ladybug on the bottom left here has three spots on the left, two on the right for a total of five. And finally, we see three spots on each side of the last ladybug for a total of six. So the total spots would be the sum of those five numbers. And let's write that down, two plus three plus three plus five spots, and the last ladybug had six. So maybe let's add these two at a time, two plus three is five, three plus five is eight, and then we have plus six. So that's five plus eight plus six. And again, two at a time, five plus eight gives us 13, then we have plus six. Finally, 13 plus six is 19. So we have a total of 19 spots on all the ladybugs all together, and that is answer C, there are 19 spots. Question number four, which of the following pictures can be rotated so that it will be the same as the picture shown on the right? And the picture on the right here represents a kangaroo. He's looking to our right, and our job is to rotate this picture either clockwise or counterclockwise. So imagine spinning it here either like this to the left, that would be counterclockwise, or maybe like this to the right, that would be clockwise. And we have to decide if one of those moves gives us back a picture that looks like one of the choices from A to B. So let's do that. Let's make a copy of our kangaroo right here, and then I will rotate this first in the blue direction, 90 degrees clockwise. And I have a little display on the top that tells me when I get to 90 degrees. So I'm almost there. There we go, that's 90 degrees. Now the kangaroo is looking down. His tail is pointing up and his feet are pointing to the left. And let's move this picture about and see if it looks like any of the ones that are our choices. And the only two with the tail pointing up are choices D and E, and clearly our picture looks most like choice E. So with one rotation here, let me erase the red arrow. With one rotation, 90 degrees in the clockwise direction, we obtain picture E. So let that be the answer here to this problem, E. Question number five. What does the tower shown to the right look like from above? Here's our tower, which looks like maybe the top of a pencil or a crayon. And our job is to imagine looking at this structure from above and decide which of these shapes would appear when the tower is viewed from that point of view. So let me copy this picture here. So it's a little easier to work with. Let's scale it up a little bit. And now let's imagine how such an object would be constructed. So in order to build such a tower, first of all, we would need a circular base. So let's draw in here the portion of the picture that we cannot see. That would be a circle on the top also as well as on the bottom. And so that would be a cylinder, maybe like the shape of a can. And then here we have a circular cone sitting on top of it. So therefore viewed from above, the base of this cone would be a circle. It could not have any corners. So it could not be the triangle in A, could not be the square in C or the triangle in D or the rectangle in E because the front of it is rounded by symmetry. The back of it, which we cannot see here also has to have the same shape. So the best choice here is the circle in B. Question number six. The diagram to the left shows six numbers. What is the sum of the numbers outside the square? And what we have in the diagram is a square in red. We have a circle in blue and then two rows of numbers, or if you prefer, we have three columns of numbers for a total of six numbers. We have to sum up the numbers that are outside of the square. So let's look at the square. Let's look at the numbers on the inside. And to avoid confusing ourselves since we're not allowed to use these numbers, let's just color them in black. Let's shape them so we don't see them. We're not allowed to use the nine or the four or the three or the seven because these are positioned inside of the square so what we have left over are two numbers that are now inside of the circle we have the we have the 2 and we have the 8 and what we need to do is compute their sum so the sum is 2 plus 8 and that will give us 10 and the answer here to question number 6 is e the sum of the numbers outside of the square is the sum of 2 and 8 and that gives us 10 question number 7 half of a movie lasts half an hour how long does the whole movie last we are trying to watch the whole movie and the whole is made up of two halves so let's write that down one whole will be equal to one half plus another one half and the problem we are told that a half of a movie will last a half of an hour so we can substitute that information into our calculation and have here a replacement of 30 minutes another name for half an hour so we have now one whole movie is equal to 30 minutes plus another 30 minutes and adding these together we obtain 60 minutes and another name for 60 minutes is just one hour so the whole movie lasts the whole hour and the answer is c one hour question number 8 eric has 10 identical metal strips like the ones shown here below he used screws to connect pairs of them together into five long strips for example in a the two screws connect two strips together with an overlap of one hole in between them to make a longer strip the question is which connected strip is the shortest we see that the configuration of the strips here differs by the number of holes between the screws or the length of the overlap between the two strips that's how these are distinguished so let's count that number first in a we have exactly a length of one there is one hole here for that overlap in b the screws are separated by one two three four five holes so let's put a five over here for the length of that overlap in c we count three holes in d we count two and finally in e we count four and we can see by looking at the metal strips here that the length of the connected strip is determined by the length of the overlap let me illustrate this i copied two of the strips and then i will move them end to end if they're positioned like this then there is no overlap we would just need one screw here to make a longer strip and if i create an overlap then two screws are needed to connect this into a solid longer piece and the overall length gets shorter the more overlap i shorter the more overlap i have so the more overlap the shorter the piece let's write that down and then let's answer the question the larger the overlap the shorter the piece and so in our case the largest overlap here is in figure b with five holes separating the two screws we have the largest overlap and eric will therefore make the shortest strip by arranging his two metal pieces like he did in b and that is the answer to the question here b question number nine there are 11 flags on a straight race track the first flag is at the start and the last flag is at the finish distance between each flag is four meters the question is how long is the track that's a lot of information to keep track of over here so let us draw a picture i have prepared a picture of 11 flags arranged in the line let me copy it over here there we go we can count the flags if we if we wish so we would have one two three four five six seven eight nine ten and one more for 11 flags we also know that the distance between each flag is four meters so i have marked that over here there are four meters between each flag and if the start is here on one end let's label this as start and at the other end we have the finish then that picture represents the racetrack and our job is to count its length so since there are four meters between the consecutive pair of flags we can simply count how many spaces there are between the flags so let's do that this would be one two three four five six seven eight nine and finally 10 spaces so we have 10 spaces times four meters space which would give us 10 times 4 meters and 10 times 4 is exactly 40 meters and so there we go we can also just add up 4 plus 4 plus 4 plus 4 plus 4 10 times and obtain the same answer of 10 times 4 meters or in other words 40 meters which is that as the length of our racetrack so the answer here is the 40 meters question number 10 marco has 9 pieces of candy and tomo has 17 pieces of candy how many pieces of candy does tomo need to give to marco so that each boy has the same number of pieces of candy our two friends marco and tomo have some pieces of candy between them and they need to split them evenly between themselves so let us first discover how many pieces of candy they have how many pieces of candy they have together so we see that marco has 9 and tomo has 17 pieces since 9 plus 17 is equal to 26 they have all together exactly 26 pieces then we see that 26 divided by 2 if they were to split them evenly between each other is equal to 13 so each boy should have exactly 13 pieces and so with this information we see that marco needs 13 minus 9 or 4 pieces tomo can give 17 minus 13 or 4 pieces to marco and then if that happens each boy will have the equal number of 13 pieces of candy so tomo needs to give marco exactly 4 pieces take away 4 from 17 you obtain 13 add 4 to 9 you also 9 you also have 13 and so the answer here is c 4 pieces of candy question number 11 martha built six towers using gray cubes and white cubes as shown in the picture to the right she made each tower using five cubes cubes of the same color do not touch how many white cubes did she use looking at martha's construction over here and the picture to the right we see clearly the construction of four of the towers and the two towers in the back are constructed from you a key piece of information that we're going to use here is that cubes of the same color do not touch so therefore the two white cubes that we see cannot be next to one another they have to be separated by a gray cube like so and gray cubes cannot also touch they must be separated by a white cube in that case if there are five cubes in each tower we can conclude now with certainty that each tower looks exactly like every other so in other words the tower here that i have selected is identical to the other towers well i will write that all towers look like this one and now we can see clearly that there are one and two white cubes per tower so we have all together six towers that's given the problem and each tower has two white cubes you We have 6 times 2, or that would be 12, of the white cubes. And so that is the answer here. Martha used exactly 12 white cubes, as we have counted, by following her diagram. And the answer is C, 12. Question number 12. When written as 5-5-2015, the date May 5th, 2015, has 3 fives. The next earliest date that will have 3 fives is which of the following? We have 5 choices here, 5 dates. Our first task would be to rewrite each one of them in the numerical format given in the example. So we have here, for A, May 10th is 5-10, the year 2015. For B, that's April, the 4th month, 25th, 2015. In C we have, again, May, the 5th month, the 25th day, 2015. In D, January, the 1st month, the 5th day, and the year is 2055 here. And finally in E, we're back to May 15th, and 2015 is the year. So looking at these numerical formats of the given dates, we see that choices A and B contain only 2 fives. So those are not possible, we need 3 fives. Choices C, D, and E do contain 3 fives, so we will now distinguish between them. The next piece of information that we need is that the next earliest date that will have 3 fives is in reference to today. So that's from today being the day of the competition for March 19th in the year 2015. That is 3-19-2015. So we can discount choice D, the year 2055 is 40 years from now, and we have 2 choices in the current year, both in May, the 5th month, and the earliest of these 2 will happen on the 15th day rather than on the 25th. So here E is our correct answer. The next earliest date that will contain 3 fives of the choices given is May 15th, 2015, and that's choice E. Question number 13. Emil placed the numbers 1, 2, 3, 4, and 5 correctly in the boxes in the diagram on the right. What number did he place in the box with the question mark? Here in our diagram we have spaces for 5 numbers. The middle number here is represented by the question mark, and the arrows indicate operations. So here there is an unknown number to which we add another number and obtain this number we're looking for. The second operation is subtraction, and again we have 2 choices here. So let's do an example to see what Emil could have done. We have, let's say, in the first box the number 1, in the second box the number 3, so their sum is 4. And then of the numbers we are given, there remain the numbers 2 and 5. So if we place 2 in this box, the answer in the last box is 2. That is not possible. We cannot use 2 twice, so let's erase those. If we instead use 5, the other remaining number, then here we have a negative number, minus 1, and that is not a possibility. So this is not a possible arrangement. 4 cannot be the number in the middle. So let's just discard 4 right away. OK. Thinking about the possible choices for the number hidden behind the question mark, we can look at the arrows a little bit closer. Here the arrow in the direction from left to right represents addition, and the arrow from left to right in the second step represents subtraction. So if we were to undo this arrow, or reverse its direction, here we would have plus. That is, we would have to add these 2 numbers in these 2 boxes to obtain the question mark, and likewise add 2 numbers here again to obtain the same number hidden behind the question mark. In other words, let me just write this down quickly. This number, this unknown number, is obtained as a sum of 2 pairs of the given numbers. And these are 1, 2, 3, 4, and 5. Now thinking about choosing 2 numbers at a time from the 5 given, and making a number in 2 different ways by means of addition, we can pretty quickly see that the number 5 is obtainable in 2 different ways from the remaining 4 numbers. So if we were to place the number 5 here where the question mark is, as a sum of 2 numbers, we can add 1 plus 4, or 2 plus 3 in these boxes on the left. Likewise, on the right, we could have a 2 plus 3, or a 1 plus 4. The details are outlined in the written suggested solutions, but let's complete this example. If I were to place a 1 and a 4, or a 4 and a 1 on the left, then I would obtain a 5 as their sum, and then 5 minus 2 would be equal to 3, or 5 minus 3 would be equal to 2 on the right hand side. So through various permutations, we have lots and lots of possible correct arrangements, but in either case, the number 5 here has to be replaced by the question mark each time. So that is the answer, E, the number 5. Question number 14. Vera invited 13 guests to her birthday party. She had 2 pizzas, and each of them was cut into 8 slices. Each person at the party ate 1 slice of pizza. How many slices of pizza were left over? Since we're counting slices of pizza, let's find out how many we had to begin with. So, we know that there were 2 pizzas, each of them cut into 8 slices. From here, we can calculate that there are a total of 2 times 8, or 16 slices of pizza. So, Vera has 2 times 8, which is 16 slices of pizza. Now, how many of them were eaten? Each person at the party ate 1 slice, and we have 13 guests. So, the 13 guests plus Vera make 14 persons, and each person eats 1 slice. There are 14 people at the party. So, combining this information together, each of the 14 people eats a slice, and of the 16 that we have, we subtract 14, and we cover 2 slices of pizza. So, 2 slices remain. And that is the answer to the question, which is choice D. 2 slices of pizza are left over. Question number 15. Don has 2 identical bricks, as shown in the picture here to the right. Which figure can he not build using these 2 bricks? And here we have 5 figures. Our task is to see which of these figures may be built using the 2 bricks, and which one is not possible. So, let's copy the 2 bricks over and over, and try to follow the instruction here. So, for A, I have my 2 bricks. What I'll do is I'll simply rotate one of them, and move it down, like so. And we see that through this rotation and translation, we obtain a figure that's very much like figure A. So, figure A is possible to construct. So, we cross that off. That's not a choice for this question. Let's move on to figure B. Again, we take the 2 bricks. Now, it should be an obvious move. We simply move these a little closer together, and there we go. That's how to construct figure B. So, figure B is not an answer, again. We cross it off. C, after a moment's thought, it's not quite obvious what to do with figure C, so let's move on. Perhaps this is our answer. Let's see if the other 2 are possible. In D, let's take our 2 bricks. One is upright, the other one we will rotate and arrange in front of the upright piece. So, let's move it and rotate it slightly, and then move it in front. And that's how figure D would be constructed. And the construction of figure E should be obvious. We take the 2 pieces and simply stack them one on top of the other. That is how figure E would be made, like so. So, we managed to eliminate figures A, B, D, and E, because we showed exactly how these are constructed. Figure C is the remaining one, and we have no way of making it, given the 2 identical bricks that Don has. So, that is the answer to this question. It is C. Question number 16. Which piece is missing from the puzzle to the right? Our puzzle to the right consists of triangular puzzle pieces, and the pattern is of 3 types of stars based on how many times they're outlined. There are stars outlined just once, stars with a double border, and stars with a triple border. And in the missing triangular piece, we have a type of each star that needs to be matched. So, let's label these. A single border star is a star of type 1, 1 for 1 trace. A double border star, let's call that the star of type 2. And with 3 outlines, let's call it a star of type 3. Now, piece A over here has a star of each type, so that's good, that is a possibility. Piece B also has a star of each type. Piece C is missing a star of type 1, so C is not a choice, it will not fit here. Piece D does not have a star of type 2, so we also cannot use D. And piece E has no star of type 2. Again, we cannot use piece E. So, how do we decide between pieces A and B? Let's look at the missing piece here in the puzzle, and let's decide on an orientation. Going clockwise like this in the direction of the blue arrow, what we have is an arrangement of stars that looks as follows. We have 1, 2, and then, excuse me, 1 followed by 3, followed by 2, like so. And our piece has to have the same arrangement. So, let's look at that. In piece B, we have 1 followed by 2, followed by 3. And in piece A, we have 3 followed by 2, followed by 1, which, after a rotation, does match this arrangement that we're looking for. So, 1 followed by 3, followed by 2. Or, if we continue our pattern, we would have 2 followed by 1, followed by 3. Or, said in another way, 3 followed by 2, followed by 1. So, in that order, we're looking for the arrangement of the stars, and that happens to be only in puzzle piece A. In puzzle piece B, we have the counterclockwise arrangement. So, our answer here is A. Question number 17. In one jump, Jake the kangaroo jumps from one circle to a neighboring circle along a line as shown in the picture to the right. He cannot jump into any circle more than once. He starts at circle S and needs to make exactly four jumps to get into circle F. In how many different ways can Jake do this? So S here on the left represents his starting position and F on the right his goal. He has to make exactly four jumps moving from one circle to another along these solid lines here, but he may not move across like so. So to illustrate all the possible paths, I'll just make a copy of the diagram here and draw arrows to represent Jake's movements. So suppose that starting from S, he makes one jump and goes up, makes a second jump and continues to the far side until one more jump down takes him to the finish line at F. So that is four arrows, so four moves and one possible path for Jake. Now he can also choose to travel along a similar route but in the opposite direction. So instead of moving up, he would go down and then jump across to the end in two moves for a total of three and then one more move would take him to the finish line. And so that is a total of two paths like this. Can he move any other way? Well here is another copy of our diagram and starting at S, Jake can simply head for the finish line right away. That's one move and a second move would take him to the finish two moves ahead of time. So he cannot go that way, let's undo that, let's send him up instead. So if he goes up, that's two moves, across three and then down for four. Now again by following a symmetrical pattern, he can move down on the second move, across and then up and that's two more paths. And finally we can have one more pair of possibilities. Starting at S, initially Jake can go up and then across, that's two moves, and then down and finally reach the finish line. So that's four moves here. By symmetry, he would move down and then once he's over there, move across to the center and finally to the finish line again in four moves. So that's finally two paths. And it looks like we have a total of two plus two plus two or six paths and we have a seventh path. Is E a possibility? And the answer is no. We have exhausted our choices from the start. We moved up and to the right as many times as we could have. If we introduce any more moves, we would possibly reach the finish line but not in four jumps. We would use less or more jumps than four. So it is not possible to have one more path. If indeed we did find another way for Jake to move to the finish line by this symmetry where we had arrows in red and in blue, instead of seven paths, we would always create one more. So there must be an even number of paths. We have here six possibilities and it is not possible to make one more path of four jumps. So six is the final answer here. That's choice D. Question number 18. A ship was attacked by pirates. One by one, the pirates climbed a rope to get to the ship. The pirate captain was the eighth pirate to climb and there were as many pirates in front of him as behind him. How many pirates climbed the rope? Since there is a lot of information in this problem, let's focus on it piece by piece and our attention is drawn to the pirate captain. So as we go along, we'll identify the important pieces of information and draw a picture to keep track. So the pirate captain was the eighth pirate to climb the rope. So let's have a picture of him. I'll draw him in red and the best I can do unfortunately is stick figures. So that's captain and he's number eight in line. So that means there are seven pirates in front of him. I've done this before. Here are the seven pirates and let's move them here to the left of the captain and we have that group that has already climbed. So we have eight pirates in total and again using what we know from the problem, we can continue reading and we see that there were as many pirates in front of the captain as behind the captain. So we need another group of seven pirates here to move in front of the captain and that's our total party that climbed the rope. So now we can count them up. Together we will have a total of seven pirates in front plus the pirate captain which is one pirate plus the seven pirates in the back of the captain and that is a total of 15 pirates that climbed the rope. And so we have the answer to our question and that would be the 15. Question number 19. For three days Joy the cat was catching mice. Each day Joy caught two mice more than the previous day. On the third day Joy caught twice as many mice as on the first day. In total how many mice did Joy catch during the three days? This problem contains a lot of information. We will focus on each piece one at a time until we have enough information to solve the problem. We begin by noticing that each day Joy caught some mice and each day he caught two more than before. So for us that means the numbers we're looking at are two apart and so either they're all even or they are all odd. The number of mice caught is always odd. Let's begin with odd. So having caught one mouse on day one, three on day two, five on day three, seven and then nine and so on. Or the other possibility is that the numbers are always even. So since he catches mice each day perhaps he caught two initially and then four the next day, six on the third day, eight and ten and so on. Now we don't know which three days we're talking about so let's use more information here. In the next sentence it says that on the third day Joy caught twice as many mice as on the first day. So let's see if we can make that work. Let's take days that are a day apart. So day one let's say and day three and let's try to multiply by two. So here we see that one times two is not five. So that sequence of days is not correct but maybe sometime in the future we would find two days one apart where the cat caught or doubled his catch and can we actually find a sequence of days like that. So let's just move this arrow about here. One times two is not five. Three times two is not seven. Five times two is not nine. So I don't think this sequence is going to work. We must move to the possibility that there are even numbers and so here we have two times two is not six but four times two is eight and again six times two is not ten. So this is the only possibility. This would be day number one. This would be day number two and this would be the catch for day number three. So that four times two is eight and Joy indeed doubles his catch on these days. So that's our sequence and the question is how many mice did Joy catch during these days. So now we can add up these numbers and we have that Joy caught four plus six plus eight which gives us a total of 18 mice. Now we're ready to answer the question and the answer here will be C. Catch of 18 mice. Question number 20. The numbers 3, 5, 7, 8, and 9 were written in the squares of the cross pictured here to the right so that the sum of the numbers in the row is equal to the sum of the numbers in the column. Which number was written in the central square? So of the So of the five numbers that we are given we have to decide upon the number that is missing here from the central square. So let me label that square with a question mark so that the sum across the row and the sum across the column are equal. And what we notice is that the number that we have represented with the question mark appears in both sums. In that respect this question is very much like number 13 where we have to decide on two different ways of adding two pairs of numbers to obtain the same sum. So in other words this number represented by the question mark we may ignore it because it appears in both sums and then in that case the numbers here that are paired represented by the blue question marks and those represented by the green question marks they have to have the same sum regardless of the presence of the number with the red question mark. So we have our number 3, 5, 7, 8, and 9 and then we think of pairing these up two at a time so that both pairs yield the same sum. And if we disregard the number 8 after a moment's thought we see that the number 8 after a moment's thought we see that 3 plus 9 gives us 12 and so does 5 plus 7. So those are the two pairs that we're looking for. In blue we'll have one pair of numbers let's say 5 and 7 will make a total of 12 and then the second pair that we need for our row for example would be 3 and 9 and that also gives us a sum of 12. Disregarding 8 marked here in red representing the number given the red question mark. So let's see how that would work. We can make ourselves a short example here. Let me just clear the cross and let's write 8 here in the center, 5 and a 7 would go in the column and a 3 and a 9 in the row. So the sum is 20 each time. 5 plus 8 plus 7 is 20 and then 3 plus 8 plus 9 is 20. And so we have our answer here. The number in red, the number 8 has to be written in the central square. And the answer is D. Question number 21. My grandmother has a dog named Atos as well as some ducks, hens and geese. She has 40 animals altogether. She has four times as many geese as ducks. Atos and the hens make up one half of all her animals. My grandmother therefore has how many and which type of animal? Let us break up this problem into several manageable pieces. We have information that grandma has 40 animals and these are broken up into some ducks, some hens and some geese. So let's write that down first. We have 40 animals. 40 animals. That's one dog plus some number of hens, plus some number of geese, plus some number of ducks. And then let's continue reading. We know that grandma has four times as many geese as ducks. So let's focus on the number of geese and ducks. Since whatever number ducks there are, there are four times as many geese, we know that there are an even number of geese. the number of geese is equal to four times the number of ducks so let's go through our choices and eliminate all of those where there are an odd number of geese and that's choice C and well so far we can only eliminate C so let's keep reading and move on to the next piece of information we have the next sentence telling us that the dog and the hands make up one half of all the animals so the dog plus the hands that's 20 animals half of 40 since there is just one dog we have 19 hands which is the case in choice A and choice D so the other choices may be eliminated we cannot have 20 hands which are the possibilities in B and E so we are down to two between A and D now let's count the animals carefully in choice A that gives us 19 hands one dog five ducks and four times the number of geese so 20 geese of 20 plus 5, 25 plus 20, that's 45 animals. And we cannot have that, because grandma has a total of 40 animals altogether. So choice A must be eliminated, and that leaves us with choice D. Grandma, in fact, has 19 hands, one dog, 16 bees, which is four times the number of ducks, so let's write in that there are four ducks to complete the problem. Question number 22. One of the six stickers shown below was placed on each of the six faces of a duck. The next picture shows the die in two positions, and we note that of the six stickers, five are visible, and only the kangaroo sticker is not yet present. Which picture is on the face opposite the face with the kangaroo sticker? To answer the question, we will have to pair the stickers on opposite sides of the faces and decide which picture is opposite that of the kangaroo. So let's do that by imagining some rotations of the die. The first thing I will do is take the die on the left and imagine that I'm rotating it about an axis to the center like this, and the rotation will be viewed from the top in the clockwise direction, like so, so that the green triangle moves to where the yellow starts. So let me draw that. Here I have a copy of the die, and after the rotation, what will happen is the green triangle should be on the front face, so let me draw that in. Here is the green triangle. The top face, after rotation, will still look like a square, so let's draw that in on top. That's still our square, roughly like this. And then what happened to the yellow star? Well, now that star is on the face. We cannot see, but we know it is on the left, so let me draw here the yellow star, roughly that. That's on this face. So what happened? Now, can we locate any of the other pictures? Yes, we can. Opposite the yellow star, we have the brown arrow, and the brown arrow will be pointing up in this case, so let's draw that in. A brown arrow that will be pointing up, so roughly here, something like that. And we can see that after a rotation here of the picture of the die on the right. About a third axis here, so let me draw that in. If we were to rotate about an axis to the center like that, in the following direction, so facing the arrow up, rotating like this, we would obtain the picture that I have begun to draw on the left. And so then, we know that the blue circle after this rotation is opposite the red square, so let's draw that in here. The blue circle would be on the bottom face, like so, perhaps. And we have now located the face opposite the green triangle, that is the face obstructed here on the right, on the bottom, and on the left, in the middle picture. That is where the kangaroo must be located, that is by the process of elimination. So the kangaroo is on the face opposite the green triangle. And that is our answer B. Question number 23. Sylvia, Tara, Una, and Vanda went out for dessert. They stood in line one after another. Each one of them ordered one of the following desserts, ice cream, waffle, bun, and cake, and each one ordered a different item. We know that the first girl did not buy the ice cream or the waffle. We know that Una was not last in line and she bought the cake. And we know that Sylvia, who was standing behind Tara and in front of Una, did not buy the waffle. So which of the following is true? We have to decide on the order of the girls and what they purchased. Let's take this very complicated problem and break it up into pieces. We already have three separate statements, and let's read them carefully again to decide on the order of the girls. It seems like the third statement here contains the best starting information. Sylvia was standing behind Tara and in front of Una. So from here, we can arrange at least these three girls into the following order. We have Sylvia standing behind Tara, so Tara is first, then Sylvia, and then Una. That is what we know. Now moving on to the middle statement, we know that Una was not last in line. And here we have Una last in line, which is obviously not true. What must happen then is Wanda is in fact last, following Una. So there we go. That means that Wanda is here last in line. From the second statement, we also know that Una did buy the cake. So let's write that down. Una has the cake. And that finishes all the information in statement number two. Okay, what is left in statement number three? We know that Sylvia, who was standing behind Tara and in front of Una, did not buy the waffle. So who bought the waffle? Well, it wasn't Sylvia, so let's make a note of that information. Not the waffle. And we know that the first girl, by the way, this finishes the information in statement number three, so let's underline that. Sylvia did not buy the waffle. And let's move on to statement number one, which tells us the first girl did not buy the ice cream or the waffle. Since the cake is accounted for, the waffle here is not bought by Sylvia, nor by the first girl, nor did she buy ice cream. The only thing that's left over is the bun. So statement number one tells us that Tara has in fact purchased the bun. So let's get rid of that piece of information. And we know that Tara has the bun. So if Tara has the bun and she's first, Una has the cake and she's third, Sylvia does not have the waffle, then Wanda must have the waffle. And there is only one piece of dessert left. Sylvia does not have the waffle, she must have the ice cream. So let's write that in next to Sylvia. And now we have the correct order. So which of the statements are indeed true? A. Wanda was first in line. No, that is false. B. Tara bought the ice cream. That is also false. C. Wanda bought the waffle. That is so far true. D. Una was second in line. That is false. And E. Sylvia bought the bun. Well, no, she did buy the ice cream. So statement B is false. And the only true statement is C. That Wanda bought the waffle. And we also figured out as a bonus that she was last in line. But we can answer our question here now. The answer is C. Question number 24. We left for a summer camp yesterday at 4.32 p.m. and got to our destination yesterday at 4.32 p.m. and got to our destination today at 6.11 a.m. How long did we travel? Let's draw up here a timeline of our travels. We will have at the very left the departure time of 4.32 p.m. and on the right the arrival time of 6.11 a.m. the following day, which means that somewhere in between we passed midnight. Let's mark that in red as 12 a.m. And we know that from 12 a.m. until 6.11 a.m. there is a block of time there that's exactly 6 hours and 11 minutes. 6 hours and 11 minutes. Now, how long is it from 4.32 p.m. until midnight? There is a landmark hour here in between. For easy bookkeeping, let's call that our 5 p.m. And so from 4.32 p.m. until the next hour we have 28 minutes. And then from 5 p.m. until midnight we have 7 hours. And we can add all of these up. We will be traveling for a total of 28 plus 11 minutes, which gives us 39 minutes. And we will be traveling for a total of 7 plus 6 hours for a total of 13 hours. And there we go. 13 hours 39 minutes. That is the total travel time. So the answer to question number 24 is A.

Math Kangaroo

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