Hello, and welcome to the Math Kangaroo 2015 Level 7 and 8 Solutions video series. First of all, I would like to congratulate you for participating in the Math Kangaroo. I used to take the kangaroo tests myself throughout all of the years of my elementary school and high school, so I know how big of a challenge it is. It's definitely a great mathematical task, and you should feel proud for taking it on. The purpose of this video series is to show you a way to logically solve every single one of the questions on the Level 7 and 8 test. Now granted, the methods that you see in the video series here are not necessarily the only methods that can be used to solve the problems, but they are the ones that I think are the most practical. However, if you solved the questions using your own methods and still got the correct answer, then perhaps your method is equally good, or possibly even better, given some circumstance. Before moving on to the video series, I would like to offer two pieces of advice. First, it would probably be best for you to try working out a problem on your own before going on to the video solution. This way, you can go ahead and try comparing your own thought process to the thought process and the logic that is displayed in the video. And through this method, you can potentially learn some new patterns, new tricks, new techniques that you may not have been aware of before, or perhaps even notice some sort of mistake in your own thought process. The other piece of advice I have to offer is to please, please, pretty please show your work when you actually try going through all of these questions. All too often, we have smart mathematicians, the kind of mathematicians that like to work on math kangaroo tests, try too hard to do math in their head because they think it will save time. And I myself am certainly guilty of this. But when you try doing math in your head, very often you catch yourself making silly mistakes. So it is easiest to just write out all of your work, as this will maximize your benefit and really help you see where you made a mistake and help prevent some of the sillier mistakes that can be done just by keeping it all in your head. If you have any questions about the solutions in this video series or any questions about this particular test in general, please feel free to email me at thomas at mathkangaroo.org, the email that is listed below. On a final note, I hope you enjoy the video series and that you find it greatly beneficial to your learning. Have fun! Problem number one, 20 over 15 equals what? First, notice that 20 over 15 simplifies to 4 over 3, which means that our goal is to find out which other fraction, given here, also simplifies to 4 over 3. When we check answer choice A, we see that the numerator adds up to 8 and the denominator adds up to 6, and 8 over 6 also simplifies to 4 over 3. This means that the correct answer is A. Problem number two, a journey from Koitse to Poprad through Preov lasts 2 hours and 10 minutes. The part of the journey from Koitse to Preov lasts 35 minutes. How long does the part of the journey from Preov to Poprad last? First, we're going to convert 2 hours and 10 minutes into 130 minutes, which is the total duration of the trip. Then we subtract 35 minutes from this, and we see that the remainder of the trip from Preov to Poprad is 95 minutes. We then convert this back into hours and minutes, and we see that the correct answer is A, 1 hour and 35 minutes. Problem number three, four identical small rectangles are put together to form a large rectangle as shown. The length of the shorter side of the large rectangle is 10 centimeters. What is the length of the longer side of the large rectangle? First, notice that we're working with this long side right here. We could also work with the top one, but it doesn't really make a difference, so we'll look at the bottom one instead. We know that this middle segment right here is 10 centimeters long, because it's identical to this segment right here. So we can go ahead and label that. Next, we notice that this segment right here and this segment are both equal to each other and equal in length to this segment and this segment, because all of the rectangles are identical. So we can label all four of these segments X. Next, notice that these two segments, which add up to 2X, are equal in length to the segment that is given to us, which we know is 10 centimeters long. This means that 2X is equal to 10 centimeters. Now if we want to find the length of this long segment here, all we have to do is perform the equation 10 plus 2X, which is equal to 10 plus 10, which is equal to 20 centimeters. So our final answer is B. Problem number four. Which of the following numbers is closest to 2.015 times 510.2? It is easiest to round this multiplication problem to 2 times 510, which is equal to 1,020. The only answer choice that comes close to 1,020 is E, which is 1,000. So that is our answer. Problem number five. The net of a cube with numbered faces is shown in the diagram. Sasha correctly adds the numbers on opposite faces of this cube. What three totals does Sasha get? In order to find the solution, you need to be able to visualize the cube being folded up. For some people, this is a little difficult to do mentally, and at home, it would probably be easiest to make some note cards with the numbers on them and tape them together as they are in the net, and actually try folding them up physically. This should help practice the visualization. And once you do this, either in your head or with actual note cards, you'll notice that the pairs of numbers that are across from each other are 1 and 3, 2 and 4, and 5 and 6. And when we add up these sums, we get 4, 6, and 11. So our final answer is A. Problem number six. Which of the following numbers is not an integer? What the problem is really asking is which of the fractions does not simplify to a whole number. So we can check these one by one using some well-known rules of division. For example, we know that answer choice A is clearly going to be an integer since any number divided by 1 is just itself. So we can eliminate answer choice A. Answer choice B, 2012 over 2. Since we're dividing an even number by 2, that's also going to be a whole number. So we can eliminate B. Answer choice C. When we're dividing a number by 3, if we want to see if it's perfectly divisible by 3, we add up the digits. And if the sum of its digits is divisible by 3, then the original number is also divisible by 3. This always works. So when we add up the digits, we see 2 plus 0 plus 1 plus 3 equals 6. And 6 is clearly divisible by 3. So we would get an integer there as well. So we can eliminate answer choice C. When we check answer choice E, 2015 divided by 5, any number that is divided by 5, it is perfectly divisible if in the ones digit you have a 5 or a 0. And since that is the case for 2015, since there's a 5 in the ones digit, we know that that would also be an integer. This leaves us with answer choice D. Problem number 7. A triangle has side lengths of 6, 10, and 11. An equilateral triangle has the same perimeter. What is the side length of the equilateral triangle? First we will find the perimeter, which is equal to 6 plus 10 plus 11, which equals 27. Since this is also the perimeter of the equilateral triangle, we can divide it by 3 to get the side length. And 27 divided by 3 is equal to 9. So our final answer is D. Problem number 8. The diagram shows the net of a triangular prism. Which edge coincides with edge UV when the net is folded to make the prism? This problem is rather similar to the earlier problem that involved folding up the net of a cube. And again, here, if it's difficult for you to visualize how it should be folding up, go ahead and try making a real life 3D model of this net using paper, tape, or whatever other resources you have at your disposal. But here, if we try explaining how this folds, the triangle with edge UV first folds up and then sort of rotates as you fold it in this direction. So VW here will rotate over and it will coincide with XW. And then the triangle will continue to rotate and then UV will rotate over and it will coincide with XY. Again, if this is difficult to visualize, it would probably be best to try making a real life model and convince yourself that this is how it works. So again, UV, once it folds up, will rotate over and coincide with XY. So our final answer is C. Problem number 9. When Simon the squirrel comes down to the ground, he never goes further than 5 meters from the trunk of his tree. However, he also stays at least 5 meters away from the doghouse. Which of the following pictures most accurately shows the shape of the region on the ground where Simon might go? First we will mark the region where Simon would go if he could walk anywhere, if it weren't for the doghouse. We would mark this down with this blue circle right here. Now there's also a circular region around the doghouse which Simon should never cross. And we know that this circle will overlap to some degree with the circle around the tree, like so. So Simon is willing to walk around in any part of the blue circle that does not overlap with the red circle. If we shade that region in, this is the area that Simon would be willing to go to. And the answer choice that resembles this is answer choice C. So that is our answer. Problem number 10. A cyclist rides at 5 meters per second. The wheels of his bicycle have a circumference of 125 centimeters. How many complete rotations does each wheel make in 5 seconds? First note that 5 meters per second is the same as 500 centimeters per second. If we want to find out how many times a wheel of the bicycle rotates in 1 second, we should divide 500 centimeters by 125, the circumference of the wheel. And this gives us a total of 4 rotations per second. Since the question is asking for how many rotations each wheel makes in 5 seconds, we perform the operation 4 times 5, and that gives us 20 rotations in 5 seconds. So our final answer is D.

Math Kangaroo 2015

problem-solving

math test solutions

simplifying fractions

logical approach