This is the Math Kangaroo Solutions video library, presenting solution suggestions for levels 5 and 6 from the year 2019. These solutions are presented by Lukasz Nowoszkowski. 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. To reading and listening to the question, pause the video and try to solve it on your own. Question 1. Carrie has started to draw a cat. She added more to her drawing using her black pen. Which of the figures below can be her finished drawing? To do this, we must look at Carrie's original cat and start eliminating the different possibilities. Looking at cat A and its nose, we notice that it is larger than Carrie's original cat, so this cannot be the answer. Cat C has ears with extra white spots, which could not have been filled in afterwards, so it can't be C. Cat D has an upside down nose, so it is also eliminated. And cat E has a circle for a nose, so this cannot also be the cat. The only one we are left with is B, so our answer is cat B. Question 2. The Mayan people wrote numbers using dots and bars. A dot is written for 1, and a bar for 5. How did they write 17? Now if we take the number 17, we can divide it into 2 plus 15. Now with the 2, we know we will have to draw 2 dots, since each one represents 1. With this, we can reject A, B, and D, since they do not have 2 dots. Next we have to draw a number of bars equal to 15. Since each bar is worth 5, we will need 3 bars, totaling 15. So our finished drawing will look like this, which gives us the answer C. Question 3. A digital clock in 24 hour mode shows the time 2019. What will the clock show the next time it uses the same digits? We know that a 24 hour clock will only go up to 24 hours, and the minute side will only go up to 59. So while A goes up to 2091, the minutes 91 are not possible, so we can reject A. The next largest number is 2109. This is a valid number for a clock to be at, and it is the next soonest time we use the digits 2, 0, 1, and 9 in a different order, 2, 1, 0, and 9. So the answer is C. Question 4. There are 14 girls and 12 boys in a kindergarten class. If half of the children go for a walk, at least how many of them are girls? To find out how many half of the children are, we first have to take the total. 14 plus 12 gives 26. So there are 26 students in the kindergarten class. Half of that is 13. Now, we want to see at least how many of them are girls. That means we want the lowest possible number of girls. So if we assume that all the boys went, all 12 of them, we do 13 people going on a walk, minus 12 boys, leaving us with 1 girl. So that means at least 1 girl went for the walk. E. Question 5. The sum of the dots on opposite faces of an ordinary die is equal to 7. Which of the following could be an ordinary die? Since we know that the faces that are opposite of each other total 7, no other two faces can be 7. So if we look at die A, and we see the faces 5, 2, and 4, this cannot be an ordinary die, since faces 5 and 2 give you a value of 7. The same thing can be said about die B, which has 3, 4, and 1 as the visible faces. And 3 plus 4 gives you a value of 7. And these are not opposite sides. Die C has the same problem. 4, 3, and 5. 4 plus 3 gives us 7. The faces are not opposite of each other, so this cannot be an ordinary die. Finally, die D has 6, 1, and 5. 6 plus 1 gives us 7, so it cannot be die D. The only one remaining is die E. So our correct answer for the ordinary die is die E. Question 6. Which of the following geometric figures is not in the design to the right? If we take a closer look to the design, we can start eliminating what shapes we do see to find the one that we don't see. First, we notice a triangle, so we know that the answer won't be A, since it is on the design. We also see a square, so we can get rid of B. Next, we find a regular hexagon, so we can get rid of C. Then, we look at the larger shape, which is the dodecagon, so we can get rid of answer E, leaving us with the regular octagon. So that means that the regular octagon is not on the design, so we get answer D. Question 7. Laura wants to color one 2x2 square of this figure. How many possibilities are there? We look at the figure that Laura drew, we can start making our 2x2 squares, which will be represented by a red 2x2 grid. This is the first possibility that Laura can do. Next, we know that Laura can put it here, also she can put it in here, and finally she can put the color in the other corner. So we have found four ways so far. However, we need to find all possible ways, which there are more. This is the fifth, sixth, seventh, and this is the final way Laura could color a 2x2 square. So when we add up all the ways that Laura can do this, we find that there are 8 possibilities. Question 8. The six smallest odd natural numbers are written on the faces of a die. Tony throws it three times and adds the results. Which of the following numbers cannot be the sum? The six smallest odd natural numbers that we can produce are 1, 3, 5, 7, 9, and 11. However, we can disregard this. We know that the properties of odd and even numbers are such as odd plus odd will give you an even number, and an odd plus an even number will give you an odd number. Now, since all the numbers on the die are odd, we will have to do an odd plus an odd plus an odd number, or an even plus an odd number after we turn the two odds into one even. There, all the answers are odd except for 20, which is an even number, and an even plus an odd value will give us an odd number. So the sum of the numbers cannot be even. So our answer will be C, 20. Question 9. The sum of the ages of a group of kangaroos is 36 years. In two years, the sum of their ages will be 60 years. How many kangaroos are in this group? So in two years, we have to have the kangaroos age 24, since 60 years minus 36 is 24, meaning the group total age will increase by 24 in two years. Now, if we take 24 divided by 2, we will get our answers, which is how many kangaroos are in the group, since our answer will be 24 divided by 2, which gives us B, 12. So 12 kangaroos will have to age two years each for them to increase by 24 years. Question 10. Michael paints the following solids made out of identical cubes. Their bases are made of eight cubes. Which solid needs the most paint? So if we look at each of the cubes from the top, left, bottom, and right, they will all look like this, sharing the same amount of faces. However, if we look at them from the front and the back, they will all have four cubes, except for B, which will look like so. So it will have one less, two less faces to paint, so it cannot be B. And all the others will look like this. No matter if we look at them from the front or the back, we will see four faces. If we look at them from the top, right, bottom, or left, they will have two by four, or eight faces. However, answer A has two faces pointing inwards, which we will not see. We would see all the other faces except those. And on all the other cubes, we will see their faces. So A will require two more paints than all the others. So our answer is A. Question 11. A three-digit number is written on each of the three pieces of paper. Two of the digits are covered. The sum of the three numbers is 826. What is the sum of the two covered digits? Make ourselves a table with the sum being 826. We can fill in the other three numbers, which are 243, 100 something 7, and something 26. Now we must simply add these together. So 3 plus 7 plus 6 gives us 16. So we leave the 6 on the bottom, and we move 1 over to the tens place. Then we do 1 plus 4 plus 2 plus a number, which will give us a number ending with a 2. The only possible number we can have here that is a one-digit number is 5. So 1 plus 4 plus 5 plus 2 gives us 12. So then we can move over to the hundreds place, adding in another 1. Now 1 plus 2 plus 1 plus something has to give us 8. 1 plus 2 plus 1 gives us 4. Then 4 plus 4 will give us 8. So this number will be a 4. Now we found the two covered digits, and we just have to find their sum. So 4 plus 5 gives us our answer, which is C, 9. Question 12. Riri the frog usually eats five spiders a day. When Riri is very hungry, she eats 10 spiders a day. She ate 60 spiders in 9 days. On how many days was she very hungry? Now if we take 60 spiders and divide it by 5, we would get 5 spiders a day for 12 days. Or we can show this by doing 5 plus 5 plus 5 plus 5 plus 5 plus 5 plus 5 plus 5 plus 5 plus 5 plus 5 plus 5. 5 added to itself 12 times. Now to make this 9 days, we have to cut down 3 of these days. We know that when Riri is very hungry, she eats 10 spiders, which is 5 plus 5, which we can write down like so. Meaning we will have 5 plus 5 plus 5 plus 5 plus 5 plus 5. Six times she will eat five spiders, and then three times she will eat 10 spiders. So we know that Riri was hungry on three days, C. Question 13. Pia is playing with a folding yardstick made of 10 parts. Which of the following figures cannot be formed with this folding yardstick? We look at all the parts. We can start to decide which one will not be able to be formed. Since it is made up of one folding yardstick, the pieces have to be in a line without going over, since it cannot be broken up. The first shape can be created by folding the yardstick in a way following the arrows. The next star-shaped one can be done this way as well. This one can be formed like so, and then this one can be formed like so. Answer C. However, if we start going in this direction, we notice that we get to a part where we don't know which way to go, and the yardstick cannot fold into two ways at the same time, such as the red arrows. One of the ways will have to be left, so we know that answer C cannot be formed. Question 14. Five equal squares are divided into smaller squares. Which square has the largest black area? We look at each of the squares, make them slightly larger. We can count up the black and the white squares. So here we will do black squares over the total number of squares. So the first answer A will be 2 divided by 4, since there are two black squares and four squares in total. The next shape has five black squares and nine total squares. Next we have 8 divided by 16, 13 divided by 25, and 18 divided by 36. Now we should simplify these fractions so it is easier to gauge which has the largest area of black squares. The first one is equal to 1 half, the next one is 1 half plus 1 18th, then 1 half, then 1 half plus 1 50th, and then 1 half. Three of the answers A, C, and E are all 1 half. B and D are greater than those. Since we know that 1 over 18 is greater than 1 50th, we get our answer, which is B. Question 15. A big triangle is divided into equilateral triangles, as shown in the figure. The side of a small gray triangle is 1 meter. What is the perimeter of the big triangle? Take a closer look. We can see our gray triangle, which has a side length of 1 meter. Each of the triangles inside the triangle is equilateral, so all sides will be equal. So we can represent 1 meter by a red line. Now the next triangle above this is the same size as the gray triangle, so it is also 1 meter, which when we add together gives us 2 meters, which is represented by our orange lines. So all these orange lines are 2 meters. Now the bottom side of the large equilateral triangle will be 1 plus 2 plus 2, since it is a red line and two orange lines, or 1 plus 2 plus 2, which gives us 5. Since it is an equilateral triangle, all the other sides will also be 5 meters. So we simply do 5 plus 5 plus 5 to get our answer, 15 meters, A. Question 16. In the garden of the witch, there are 30 animals, dogs, cats, and mice. The witch turns six dogs into cats. Then she turns five cats into mice. Now the number of dogs, the number of cats, and the number of mice in her garden are all equal. How many cats were there in the beginning? Since we know that there are 30 animals and the number of mice, cats, and dogs is equal, we know there will be 10 of each animal at the end. Now we have to just go backwards. Since we know that previously she turned five cats into mice, we will have to take away five mice and add five cats, meaning we'll have 10 dogs, 15 cats, and five mice. Then we know that the witch turned six dogs into cats, so she would have originally six more dogs and six less cats. So we end up with 16 dogs, 9 cats, and 5 mice. This answers our question. How many cats were there at the beginning? C, 9. Question 17. With blocks of dimensions 1 cm by 1 cm by 2 cm, you can build towers as shown in the picture. How high is a tower that is built in the same way using 28 blocks? First, we look at the fourth tower in the picture. Then we add another level to it, like so. And then to make it use 28 blocks, we need to add one more level, like so. Now when we count up all the blocks, there are 28 in total. Since we know that each wooden block is 1 by 1 by 2 cm, we can calculate our height. The first set of blocks will be at 2. Then these ones will be at 1, represented by the blue and the red line. So then we just have to total these off, like so. And we get 2 plus 1 plus 2 plus 1 plus 2 plus 1 plus 2. Simplified, 3 plus 3 plus 3 plus 2, or 9 plus 2. With this, we get our answer. The highest tower that we can build in this way using 28 blocks is answer B, 11 cm. Question 18. Bridget folded a square sheet of paper twice, then cut it twice as shown in the figure. How many pieces of paper did she get? We take a little square after the two folds, and we cut it like Bridget did. Then we will have one, two, three, four pieces. Then we must unfold it, like so. This will result in 5 and 6 pieces, so two more pieces when we unfold it. Then we have to unfold it one more time and it will look like so. This will give us seven, eight, and nine pieces when the sheet is unfolded. So with this we know that Bridget gets C, nine pieces of paper. Question 19. Alex, Bob, and Carl go for a walk every day. If Alex doesn't wear a hat then Bob wears a hat. If Bob doesn't wear a hat then Carl wears a hat. Today Bob is not wearing a hat. Who is wearing a hat? We know that Bob is not wearing a hat since the problem tells us. Next we look at our first piece of information. If Alex doesn't wear a hat then Bob wears a hat. But since Bob is not wearing a hat that means Alex must be wearing a hat. Also we know that if Bob doesn't wear a hat then Carl wears a hat. We were told that Bob is not wearing a hat therefore Carl is wearing a hat. So our answer is A, both Alex and Carl are wearing hats. Question 20. Each of the following pictures shows the net of a cube. Only one of the resulting cubes has a line with connected endpoints drawn on it. Which one? We take a closer look at each of the nets A, B, C, D, and E. Then for each of the nets below the sides of small squares that are not shared with another small square are labeled from A to G in such a way that the sides with the same label form one edge of the cube obtained by folding the net as demonstrated in front of you. The net D is the only one that shows midpoints of the two sides with the same label labeled B as the endpoints drawn for the line. Meaning net D is the only one when folded into a cube that will have the lines connect to each other. As the other nets for example net B has the endpoints of the line B, C, and E meaning they will not connect to each other. So our answer is D. Question 21. The cube shown in the figure has a positive integer written on each face. The products of the two numbers on opposite faces are the same. What is the smallest possible sum of the six numbers on the cube? The question tells us that one face multiplied by the face opposite to it will be equal to another face multiplied by the face opposite to it. So with the faces 5, 10, and 15 we must find a common number we can multiply to. The smallest such number for this is 30. 5 times 6 will give us 30. 10 times 3 will give us 30. 15 times 2 gives us 30. So that means the opposite side of 5 is 6. Opposite of 10 is 3 and opposite of 15 is 2 to get us the smallest number 30. Now we just have to add these together. 2 plus 3 plus 5 plus 6 plus 10 plus 15 gives us 41. So the smallest possible sum of the six numbers on the cube is C 41. Question 22. Six identical black beads and three identical white beads are arranged on weighing scales as shown in the picture above. What is the total weight of these nine beads? We represent black beads as B and white beads as W. We can write out the formula for the first scale. Two black beads are equal in weight to two white beads plus 6 grams and three black beads and a white bead are equal in weight to a black bead plus 30 grams. Now we simplify our first formula to two black beads minus 6 grams equals two white beads. Second formula one white bead is equal to a black bead minus 3 black beads plus 30 grams or a white bead is equal in weight to 30 grams minus 2 black beads. We put these two together we get 2 black beads minus 6 grams is equal to 2 times 30 grams minus 2 times 2 black beads or 2 black beads minus 6 grams is equal to 60 grams minus 4 black beads. Further simplified 6 black beads total 66 grams meaning each black bead is 11 grams in weight. If we put this back into our first formula then we can get this 2 times 11 grams is equal to 2 white beads plus 6 grams or 22 grams is equal to white beads plus 6 grams. Two white beads are 16 grams meaning each white bead is 8 grams. Since we want to find the total weight of 6 black beads and 3 white beads we just have to add them together now. So we do 6 black beads each 11 grams. We do 11 plus 11 plus 11 plus 11 plus 11 plus 11 plus 3 white beads 8 plus 8 plus 8 and we get 66 plus 24 which is the total weight of the 9 beads. Which one added together we get 90 grams. So our answer is E 90 grams. Question 23. Robert made five statements A to E. Exactly one of the statements is false. Which one? A. My son Basil has three sisters. B. My daughter Anne has two brothers. C. My daughter Anne has two sisters. D. My son Basil has two brothers. And E. I have five children. If we look at statements B and D my daughter Anne has two brothers and my son Basil has two brothers we can tell they are mutually exclusive meaning that one of them has to be false for the other to be true. So with this we know that A, C, and E will all be true statements. Now we know that Robert has five children and we look at statement B. Anne has two brothers. Then we look at D. Basil has two brothers. And we look back to C saying Anne has two sisters and Basil has three sisters. That means Robert will have three daughters in total. And if he has five children as stated in E that means he will only have two sons. If D was true and his son Basil had two brothers Robert would have a total of three sons and three daughters making six children which cannot be the case. So that means that D is the false statement. Question 24. Benjamin writes an integer in the first circle and then fills the other five circles by following the instructions. How many of the six numbers in the circles are divisible by three? We look at the circles we see we'll have to add one, add one, multiply by three, add two, and multiply by two. We start off with our first integer being a number that is not divisible by three such as four. We add one, get five, add one, get six. Meaning one of these three is divisible by three. However if we try a different integer let's say five. We have five plus one, six, which is divisible by three. Seven which is not. Again we only got one number divisible by three. Try one more time. Six is divisible by three. Seven and eight are not. So we got one number divisible by three in the first three boxes. Then we will multiply any number by three. We know that this number will always be divisible by three. So so far we have two numbers that are divisible by three. Now if we add two to a number divisible by three, nothing will be accomplished. The number will not be divisible by three. Same thing if we multiply by two. The number will stay not divisible by three. So that means there are only two possible ways. So our answer is C, two numbers divisible by three. Question 25. The cardboard shown on the right is folded into a two by one by one box. One of the pictures below does not show this box. Which one? We take a look at the net. We can start folding the picture. Now this can only be folded into a two by one by one box in such a way that the two black boxes will form a two by one wall. Now with this information we know that the white box next to the two black boxes will be a side box, meaning it'll be on the one by one plane. The other white box will also be there. So that means that the two one by one sides of the two by one by one box have to be white. Out of the possible choices A, B, C, D, and E, only B has no white face on its side. So that means this cannot be a possible box, B. Question 26. Emily took selfies with her eight cousins. Each of the eight cousins is in two or three pictures. In each picture there are exactly five of Emily's cousins. How many selfies did Emily take? We label K as the number of cousins in two pictures, and 8 minus K as the cousins in three pictures, and we can set up a formula. 2 times K plus 3 times 8 minus K, or simplified, 2K plus 24 minus 3K. This we can simplify to 24 minus K. Since we know that there are five cousins in each of the photos, this number has to be divisible by 5. And since K has to be between 8 or 0 and an 8, that means there is only one possibility for K, and that is 4, since no other number, once subtracted from 24, is divisible by 5. So our answer is B, 4. Question 27. Jet and Willie are throwing balls at two identical pyramids of 15 cans. Jet knocks down six cans with a total of 25 points. Willie knocks down four cans. How many points does Willie score? We look at the pyramids after Jet's throw and after Willie's throw. We can see that they will overlap, and we can put together a more complete pyramid. However, we are still missing the top can. We know that Jet knocked over six cans with a total of 25 points. If we circle the cans that she knocked down on this pyramid, plus the one missing, we can find out what the missing can is equal to. We just do 25 points, which she scored, minus 3, 8, 2, 3, and 4, and we get that the missing can is worth 5 points. Now, with this, we can find out how many points Willie scored, since we know what four cans he knocked down. He knocked down the 8, the 4, the 9, and the 5 can. When added together, we get the answer. Willie scored 26 points. D. Every digit on my 24-hour digital clock is composed of at most seven digital display parts, as follows. But unfortunately, for every digit display, the same two digital parts are not working. At this moment, my clock shows. What will it show after 3 hours and 45 minutes? If we take a look at what the clock is currently showing, we can start to find out what the actual time is. Know that two digital parts are not working on every slot, and they are the same two parts every time. The first digit of any 24-hour clock has to be 0, 1, or 2. With the three slots we already have filled out, this can only be one number, a 2. Next, the number can only be a 0, 1, 2, 3, or a 4. Since the clock will only go up to 24 hours, and we are at the 20th something hour. Again, there are two slots missing. We do this. We find out these are the two slots. We find out that the next number is a 3. The next number will have to be a 4, and then a 7. So we know that the hour is 2347, and we want to find out what the clock will show after 3 hours and 45 minutes. So 3 hours and 45 minutes after 2347. We just have to add this, like so. And we get 2692. Since every hour has 60 minutes in it, we cannot be at minute 92. So we just carry over the extra 60 minutes and make it hour 27 and 32 minutes. But as we've established, it is a 24 hour clock, so it cannot go up to 27. So that means we take away 24, and we start the new day, and it'll be 332. Now we just have to find out what this will look like with the parts missing. We just draw out these digitally. We get the 0, 3, 3, and 2, and then take away any slot that is already circled. As we've established, those parts are not working. It'll look like so. With this, we know our answer is A. Question 29. Linus builds a 4x4x4 cube using 32 white and 32 black 1x1x1 cubes. He arranges the cubes so that as much of the surface of his large cube as possible is white. What fraction of the surface of his cube is white? We take a look at a 4x4x4 cube made up of 1x1x1 cubes. We can start to find out how much of the cube is white. Now, any given cube will have 8 corners, and each corner will have 3 faces, even though it is just made up of 1x1x1 cube. So, this cube will have 8 corners, and if we use 8 white cubes on them, they will net 3 white faces for each one cube used. Now, the same can be said about edges of a cube. Any cube will have 12 edges. So, if we use white cubes on the edges, which each have 2 faces for the cost of 1 cube, we will maximize the white area, like so. And since we will be using 2 cubes for every edge, that will be 24 cubes. 24 and 8 cubes will net us our total 32 white cubes, which we can use. So, to maximize the white area, we have to use 8 cubes for the corners, and 24 cubes for the edges of the cube. Now, if we take 24 edge cubes and multiply by 2 the number of faces each cube will provide, we get 48. If we take 8 and multiply it by 3, we get the number of faces each corner cube will provide, 24. When we total this, we get 72 white faces on the 4x4x4 cube. Now, to find the total number of faces that this cube has, we have to do 4 times 4, the number of faces on one side, times 6, the number of sides of a cube, which gives us 96 total faces. So now, to find out the fraction of the surface which is white, we must do 72 white faces divided by 96 total faces. When we do 72 divided by 96 and simplify it, we get d, 3 fourths. Question 30. Zef has 2 machines. One makes 1 white token into 4 red tokens, while the other makes 1 red token into 3 white tokens. Zef starts with 4 white tokens. After exactly 11 exchanges, he has 31 tokens. How many of those are red? We know that 1 white token can be turned into 4 red tokens, and 1 red token can be turned into 3 white tokens. So, if we're starting off with 4 tokens, we must keep track of the number of exchanges. The first exchange will get 4 red tokens for 1 white. If we do this for each of the white tokens, we will end up with 16 red tokens after 4 exchanges. If we turn 1 of the red tokens into white tokens, we have a total of 5 exchanges. Now, if we were to take every last one of our exchanges, which we have 6 more of to make 11, turning red tokens into white tokens, then we would end up with 9 red tokens and 21 white tokens, which gives us 30 tokens, and we need 31. So, that means all 6 of our remaining exchanges cannot be red into white tokens. However, if you notice, we are only missing 1 token, and with this, we know we can gain an additional token if 1 of those exchanges was instead a white to red token exchange. With that, we would have 10 red tokens and 18 white tokens, and then with our additional exchange, we can have 4 more red tokens at the cost of 1 white token, totaling 11 exchanges. With this, we have 17 plus 14, giving us 31 total tokens and 11 exchanges, and to find out how many of those are red, we just look up and see 14 red tokens. So, our answer is C, 14.

Math Kangaroo 2019

Lukasz Nowoszkowski

problem-solving skills

Mayan numeral system

mathematics competition

logical deduction

pattern recognition

step-by-step explanations