Problem number one states, the drawing shows three flying arrows and nine balloons that do not move. When an arrow hits a balloon, the balloon pops and the arrow flies farther in the same direction. How many balloons will not be hit by arrows? Let's start with the top arrow. The top arrow is going to hit the green balloon, and as the problem states, the balloon is going to pop and the arrow is going to fly farther in the same direction. Same thing is going to happen when it meets the red balloon. The balloon is going to pop and the arrow is going to fly farther in the same direction. The second arrow is going to first hit the yellow balloon, fly farther, and then hit the blue balloon and fly off. Final arrow is going to hit the big yellow balloon, pop it, and then hit the small red balloon and pop it and finally fly off. Three balloons were not hit by arrows. The question asked, how many balloons will not be hit by arrows? The answer is three, letter A. Problem number two states, there are three objects on the table. What does Peter see if he looks at the table from above? The only real way to solve this problem is to use your imagination and to view the shapes from the top. This is how viewing the shapes from the top would look like. The problem asked us, what does Peter see if he looks at the table from above? The answer is shape C. Problem number three states, Diana got 14 points with two arrows on the target. The second time she got 16 points. How many points did she get the third time? Let's look at what information we can gather from the first target. We know that Diana got 14 points and that she hit the white region twice. From this information, we can extract how many points Diana would get if she hit the white region once. Let's let A represent the points earned from hitting the white region once. We get an equation 14 equals 2A, where 14 equals the points that Diana got, 2 represents the fact that she hit the white region twice, and A represents the points earned from hitting the white region once. In order to solve this equation for A, we just need to divide both sides by 2. Afterwards, we get that A equals 7. This means that when Diana hits the white region, she earns 7 points. Let's move on to the second target. In this target, we know that Diana got 16 points and she hit the white region once and the blue region once as well. Again, we'll let A equal the points earned from hitting the white region, and this time we'll let B equal the points earned from hitting the blue region. We get the equation 16 equals 1A plus 1B, where 16 represents the total number of points Diana got, the 1 in front of the A represents the fact that she hit the white region once, and the 1 in front of the B represents the fact that she hit the blue region once. Again, the variable A represents the points earned from hitting the white region, and the variable B represents the points earned from hitting the blue region. So let's solve. First, let's simplify the expression. And now it seems that we're stuck, but in reality, we're not, because we know that A equals 7, because A represented the number of points that you get from hitting the white region once, and we solved for that previously, so we just plug in 7 for A. Now in order to solve for B, we just subtract 7 from both sides, and we get that B equals 9. Again, B represents the number of points that Diana gets from hitting the blue region. And finally, let's figure out how many points Diana gets from the third target. So let's let P represent the total number of points that she earns, and B represent the points earned from hitting the blue region. So we get the equation P equals 2B. We can of course plug 9 in for B, because we just solved for the points earned from hitting the blue region. After simplifying, we get that P equals 18. The problem asked, how many points did she get the third time? The answer is 18 points, letter B. Problem number 4 states, a garden is divided into identical 1 meter by 1 meter squares. A fast snail, which we will represent in red, and a slow snail, which we will represent in yellow, move along the perimeter of the garden starting from the corner S, but in different directions. The perimeter of the garden is the line that's surrounding the garden, which we can mark here in cyan, and we'll mark the corner S in green. The slow snail moves at a speed of 1 meter per hour, and the fast one at 2 meters per hour. At what point will the two snails meet? So it is important to understand what it means for a snail to be traveling at 1 meter per hour and at 2 meters per hour. So the slow snail travels at 1 meter per hour, which means that in 1 hour, the slow snail is going to move 1 meter, and the red snail is traveling at 2 meters per hour, which means that in 1 hour, the fast snail is going to move 2 meters. In order to solve this problem, we are going to simulate the movement of these snails. So let's put the snails at the starting line and see how they move. At t equals 0, both snails are at point S, which was stated in the problem as the starting point. As 1 hour passes, the yellow snail moves up by 1 meter, since it is traveling at 1 meter per hour, and the red snail moves right 2 meters, since it is traveling at 2 meters per hour. As the simulation continues, it is clear that the snails end up at point B. This problem can also be solved analytically. Let's look at how this is done. In order to solve this problem, we are going to need to use an equation. The equation is x equals tr, where x represents the distance traveled, t represents the time traveled, and r represents the speed traveled. The question asked us, at what point will the two snails meet? So we are going to need to solve for the distance traveled. In order to solve for the distance traveled, we need to know the other two variables. We know what speed the snails are traveling at, but we don't know how long it takes them to travel, so let's solve for that. First let's solve for the perimeter of our garden. The equation for the perimeter of a rectangle is 2 times L plus W. L is going to be 10 meters, and W is going to be 2 meters. 10 plus 2 is 12, and 2 times 12 is 24. The perimeter of our garden is 24 meters. We know that the total distance the snails need to travel is 24 meters, and we know that the rate at which they travel together is 3 meters per second. This value is gotten from adding their two speeds. In order to solve for t, we just need to divide both sides by 3, and we get t equals 8 hours, which means that it takes the snails 8 hours to meet up. Let's use our equation again, but this time to solve for the distance traveled by the snails individually. So we're solving for x in this case, so we need to know t and r. So I'm going to plug in 8 for t, and for r you can plug in either the speed of the slow snail or the speed of the fast snail. I'm going to show both ways. And now we solve for x, and we get that the slow snail travels 8 meters and the fast snail travels 16 meters. So let's count 8 meters and let's see where the snail ends up. The slow snail ends up at point B. The fast snail moved the other way, so if we count 16 meters in the other direction, we also get that the snail ended up at point B. This makes sense since the snails were supposed to meet up at the same location. The question asked us, at what point will the two snails meet? And regardless of what strategy we used to solve it, the question asked us, at what point will the two snails meet? Regardless of what method you used to solve it, you should have gotten point B, letter B. Problem number 5 reads, Alice subtracted two two-digit numbers, then she painted two cells. What is the sum of the two two-digits in the painted cells? Let's refer to the two digits in the painted cells as x and y. Let's solve for y first. The only digit that yields 5 when 3 is subtracted from it is 8. Here are some examples, 13-8 equals 5, 23-18 equals 5, and 33-28 equals 5. Therefore, y must be 8. Now let's solve for x. If we let a represent the whole first number, we get an equation of a-28 equals 25. In order to solve for a, all we need to do is add 28 to both sides, and we get that a equals 53. In order to get x, we just take the first digit, so x equals 5. The question asked us, what is the sum of the two digits in the painted cells? 8 plus 5 is 13, so the answer is 13. Letter D. Problem number 6 states, a star is made of four equilateral triangles and a square. The perimeter of the square is 36 centimeters. What is the perimeter of the star? So we are told in the problem that the perimeter of the square is 36 centimeters. The equation for the perimeter of a square is P equals 4a, where P is the perimeter of the square, and a is the length of one side of the square. If we plug in 36 for P, and we solve by dividing both sides by 4, we get that each side is 9 centimeters long. The star is made up of equilateral triangles. In an equilateral triangle, all sides have equal length, meaning that if one side has a length of 9 centimeters, all sides have a length of 9 centimeters. This means that all the line segments that make up our star are 9 centimeters long. We were asked to solve for the perimeter of the star. The perimeter is this line right here marked in yellow. The perimeter consists of 8 9 centimeter long sides. So therefore, our perimeter is 72 centimeters. The question asked us, what is the perimeter of the star? The answer is 72 centimeters. Letter E. Question number 7 reads, the picture shows the calendar of a certain month. Unfortunately, an ink spot covers most of the dates. What day is the 25th of that month? We can solve this problem by simply writing in the missing dates. This way, it is easy to see that the 25th lands on Saturday, but this method is time consuming and there is another way. If we know that dates in the same column are 7 days apart, and we know that the 3rd lands on a Friday, then we know that the 10th lands on a Friday, and we also know that the 17th lands on a Friday. We also know that the 24th lands on a Friday. If the 24th lands on a Friday, then we know that the 25th must be a Saturday. The question asked us, what day is the 25th of that month? The answer is Saturday. Letter D. Problem number 8 states, how many times do we have to roll a regular die to be sure that at least one result will be repeated? This question is quite complicated, so first let's solve a simpler question. How many times do we have to flip a coin to be sure that at least one result will be repeated? The first time that we flip a coin, it can either be heads or tails. The question wants to know how many times we have to flip a coin in order for one result to be repeated. In this case, there is only one result, so it is impossible for the result to be repeated. We have to flip the coin at least twice. When we flip the coin twice, some results repeat, but the combinations in the red boxes don't. So therefore, we have to move on to a third flip. After the third flip, we have at least one repeated result in all of the combinations. A coin can either be heads or tails, but if it already was heads and tails, then the next flip has to result in a repeat. It doesn't matter if it's heads or tails. Let's go back to the original problem. How many times do we have to roll a regular die to be sure that at least one result will be repeated? Let's assume that we rolled the dice 6 times, and each time we rolled the dice, we got a unique result. If we roll the dice a 7th time, it doesn't matter what we roll, because the result will have to repeat. The question asked us, how many times do we have to roll a regular die to be sure that at least one result will be repeated? The answer is 7 times, letter C. Problem number 9 states, there are 3 squares in the figure. The side length of the smallest square is 6 centimeters. What is the side length of the biggest square? In order to solve for the largest square's side length, we first need to solve for the medium square's side length. The medium square's side length is 2 centimeters greater than the side length of the smallest square's side length. So therefore, the medium square's side length is 8 centimeters. The largest square's side length, which we will mark in A, is equal to the medium square's side length, plus the smallest square's side length, minus the 2 centimeter overhang. 8 plus 6 is 14, minus 2 is 12, so the largest square's side length is 12 centimeters. The question asked us, what is the side length of the biggest square? The answer is 12 centimeters, letter C. Problem number 10 states, in the figure on the right, the circles are light bulbs connected to some other light bulbs. Initially, all light bulbs are off. When you touch a light bulb, this light bulb and all its neighbors are lit. At least how many light bulbs do you have to touch to light all the light bulbs? So for example, if we touch this light bulb, then it lights up, and all of its neighbors also light up. In order to minimize the number of light bulbs touched, we need to make sure we turn on as many light bulbs as possible per touch. In the light bulbs is written the number of light bulbs that will be turned on when that light bulb is touched. For example, with light bulbs with a 4 written inside of them, that light bulb will light up, and all of its 3 neighbors will also light up. With light bulbs with a 3 in them, that light bulb will light up, and all of its 2 neighbors will also light up. Since there are 8 bulbs, and the maximum number of bulbs we can light up per touch is 4, that means that it might be possible for us to only touch 2 light bulbs, and light up all 8 light bulbs. So let's try to do this. Let's touch the light bulb in the middle, so it lights up, and all of its 3 neighbors also light up. Four light bulbs remain that are not lit up. Luckily all of these light bulbs share a common neighbor. So we can just touch that neighbor, and all of the remaining light bulbs will light up. The question asked us, at least how many light bulbs do you have to touch to light up all the bulbs? The answer is 2 light bulbs, letter A.

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