Problem number 11 states, in which of the four squares is the ratio of black area to the white area the largest? It is clear that squares A and B are not divided the same way that squares C and D are. So if we're going to be comparing the relative areas, we first need to make sure that the squares are cut up in the same ways. So let's cut up squares A and B like squares C and D are. Square A has 4 black triangles and 12 white triangles, making its area of black triangles to white triangles 4 to 12, which can be simplified to 1 to 3. Square B also has 4 black triangles and 12 white triangles, making its black to white ratio 4 to 12, or 1 to 3. Square C also has the same ratio as squares A and B, 4 to 12, which is simplified to 1 to 3. Lastly, square D also has a ratio of black area to white area of 4 to 12, or 1 to 3. The question asked us, in which of the four squares is the ratio of the black area to the white area the largest? The answer is the ratios are all the same, letter E. Problem number 12 states, 9 cars arrive at an intersection and drive off as indicated by the arrows. Which figure shows these cars after they leave the intersection? The easiest way to solve this problem is to simulate how the cars move, and then look at where they end up. Let's look at how a simulation might look like. Starting from the rightmost car as we will move counterclockwise around the intersection. We have one car on top, one car on the left, three cars on the bottom, and four cars on the right. The question asked us, which figure shows these cars after they leave the intersection? The answer is figure B. Problem number 13 states, four of the five numbers, 1, 2, 3, 4, and 5, were used so that both calculations following the arrows are correct. The numbers are not shown. Which number is covered by the spot with the star? So first, let's write two equations that model this scenario. Let's call the number in the green shape A, the number in the pink shape B, the number in the blue shape C, and the number in the red shape D. So our first equation is A plus B minus C equals 8. And our second equation is A times D divided by C equals 8. Let's start off with the first equation. In order for this equation to be true, A plus B have to be greater than 8. This is because if A plus B were less than 8, and C greater than 1, it is impossible for us to make 8. Let me show you some examples of what I mean. If A plus B equals 3, then that would mean that 3 minus C equals 8, and this is impossible, since C could only equal 1, 2, 3, 4, or 5. If A plus B equaled 8, this would also be impossible, because 8 minus 1, 2, 3, 4, or 5 doesn't make 8. The only way that we can make sure that A plus B is greater than 8 is by saying that A equals 5 and B equals 4, or that A equals 4 and B equals 5. Regardless of what method we choose, A plus B has to equal 9. If we plug in 9 for A plus B into our original equation, we get 9 minus C equals 8. Let's solve for C. In order to solve for C, let's subtract 9 from both sides, and let's multiply by negative 1. We get that C equals 1. Regardless of whether we said that A equals 5 and B equals 4, or that A equals 4 and B equals 5, numbers 1, 4, and 5 are used. Let's plug in all the information we know back into our graph. We know that A either equals 4 or 5, B either equals 5 or 4, and we know that C equals 1. So now let's look at the second equation. A times D divided by C equals 8. Since we know what C is, let's plug in 1 for C. Any number divided by 1 is that number, so we can just remove the divided by 1. The only numbers we can plug in for A is 4 or 5, and the only numbers we can plug in for D are the remaining numbers that we haven't used yet, so 2 and 3. The only way to make 8 out of those numbers is 4 times 2, so let's plug in 4 for A and 2 for D. Now we know that the number in the red shape is 2, the number in the green shape is 4, and the number in the pink shape is 5. In order to make sure our answers are correct, let's plug in all the numbers back into the original equations. 4 plus 5 is 9, minus 1 is 8, check. 4 times 2 is 8, divided by 1 is also 8, check. The question asked us, which number is covered by the spot with the star? The number covered with the star is the number in the pink shape, so 5. So the answer is 5, letter E. Problem number 14 states, a lion is behind one of three doors. A sentence is written on each door, but only one of the sentences is true. Behind which door is the lion? The sentences on the doors read, the lion is not behind this door, the lion is behind this door, and the sum of 2 and 3 is 5. Well, the sum of 2 and 3 is 5, so that means that door number 3 is telling the truth. The question told us that only one of the three sentences is true, which means that the remaining two sentences on doors 1 and 2 are lies. Door number 2 says the lion is behind this door. We've already established that this is a lie, which means that the lion is not behind door number 2. Door number 1 says the lion is not behind this door, and that is also a lie, which means that the lion is in fact behind door number 1. The question asked us, behind which door is the lion? The answer is door 1, letter A. Problem number 15 states, two girls, Ava and Olga, and three boys, Adam, Isaac, and Urban, are playing with a ball. When a girl has the ball, she throws it to the other girl, or to a boy. When a boy has the ball, he throws it to another boy, but never to the boy from whom he has just received it. Ava starts by throwing the ball to Adam, who will do the fifth throw. This is how the situation looks after the first throw. Ava had just thrown the ball to Adam. Since the problem stated that a boy can only throw the ball to another boy, it is impossible for any of the girls to get the ball ever again. So we can disregard them in this problem. At this point, Adam can choose who he wants to throw the ball to, either Isaac or Urban. We'll show what happens in both situations. If Adam initially throws the ball to Isaac, then Isaac cannot throw it back to Adam, because the question says that the boy can never throw the ball back to whom he had received it from. So Isaac has to throw the ball to Urban. And if Adam initially threw the ball to Urban, then Urban has to throw it to Isaac, because he received the ball from Adam. So after the third throw, this is how the situation looks like. If Adam initially threw the ball to Isaac, after the third throw, Urban has to throw it to Adam, because he just received it from Isaac. And similarly, in the other situation, Isaac would also have to throw the ball to Adam, because he just received it from Urban. This is how the simulation looks like after four throws. After the fourth throw comes the fifth throw, and Adam has the ball in both situations, which means that Adam has to do the fifth throw. The question asked, who will do the fifth throw? The answer is Adam, letter A. Problem number 16 states, Emily wants to enter a number into each cell of the triangular table. The sum of the numbers in any two cells with a common edge must be the same. She has already entered two numbers. What is the sum of all the numbers in the table? Before we can tackle on this problem, let's see a simpler abstract example. Let's say the center value in these four cells is x, and the value to the left is y. The only way to make x plus y, if we are already given x, is with another y. So the other two cells have to also be y. This might be a little confusing, so let's see a concrete example. Let's say the number in the middle is 4, and the number to the left is 3. The sum on this edge is 7. The only way we can make 7 with 4 is a 3, so the remaining two cells have to be 3's. Let's check our work. This edge adds up to 7, 4 plus 3 here is also 7, and 4 plus 3 here is also 7. Now let's try to solve the original problem. Let's start off by looking at this 3. This 3 has to be surrounded by the same number on all three sides. We'll call this number x. Now let's look at the 2 in the bottom left corner. All of the cells that surround it also have to have a common value. We'll call this value y. Now let's look at this y. All of this cell's neighbors have to have the same value. This cell already has two neighbors, x and 2. This means that the remaining cell has to have a value of x. This also means that x has a value of 2. Let's replace all occurrences of x with 2. Now let's look at this cell. All of this cell's neighbors also have to have the same value, which means that y is equal to 3. Again, let's replace y with 3. Now we just need to fill in the remaining cells. Let's start off by looking at the 2 at the bottom. One of its neighboring cells has a value of 3, so the other must also have a value of 3. Now let's look at the cell that we just filled in. Both of its neighbors are 2, so its third neighbor also has to be 2. In the end, we have 6 2's and 3 3's, which means that the total value of our triangle is 21. The question asked us, what is the sum of all the numbers in the table? The answer is 21, letter C. Problem number 17 states, On Monday, Alexandra emails a picture to five friends. For several days, everybody who receives the picture, emails it the next day to two friends who haven't received the picture yet. On which day does the number of people who have received the picture become greater than 100? Here we have a model of the situation. On Monday, Alexandra sent the photo to five people. Then, on Tuesday, the five people who got the photo initially from Alexandra send the photo to two people each. So now, ten new people received the photo on Tuesday, which means that fifteen people received the photo in total. Fifteen is still not over a hundred, so we need to continue. Today, the ten new people that received the photo yesterday sent the photo to two new people each, meaning that twenty new people received the photo today. In total, thirty-five people received the photo on Wednesday. Thirty-five is still not greater than a hundred, so we still need to continue. On Thursday, the twenty people that received the photo yesterday sent it to two new people each on Thursday. Forty people received the photo on Thursday, which means in total seventy-five people received the photo. Seventy-five is still not greater than a hundred, though, so we need to move on to the next day. Eighty new people received the photo today, meaning that by Friday, a hundred and fifty-five people received the photo. The question asked us, on which day does the number of people who have received the picture become greater than one hundred? The answer is Friday, letter C. Problem number eighteen states, the faces of a cube are painted black, white, or gray so that opposite faces are of different color. Which of the following is not a possible net of this cube? First, let's determine what faces on the cube net are on different sides of the cube. So when the cube net is turned into a cube, these two edges are going to combine, which means that from the top, it's going to look like a square. This means that these two faces are opposite, and that these two faces are opposite. The remaining faces are the caps of the cube, and they are also opposite. It does not matter where the caps are on the net, they will still be on opposite sides. Let's look at the cube net for letter A. Again, we are trying to find a cube net whose opposite faces are the same color. Again, we are trying to find a cube net whose opposite faces are the same color. So let's identify opposite faces here. As we can see, both A's are different colors, both B's are different colors, and both C's are different colors. So therefore, A cannot be the answer. If we move on to B, and we identify the faces, we again see that A has different colors, B has different colors, and C also has different colors. So it cannot be B either. If we look at cube net C, we can again see that opposite faces have different colors. The A's are different colors, the B's are different colors, and the C's are different colors. In cube net D, the same thing can be observed. A's are different colors, B's are different colors, and C's are different colors. If the answer is not A, B, C, or D, that means that the answer has to be E. But just to make sure, let's check. And it looks like opposite faces C are the same color, which means that this is not a possible cube net. So the question asked us, which of the following is not a possible net of this cube? The answer is E. Problem number 19 states, John does a calculation using four digits represented in the addition by A, B, C, and D. Which digit is represented by B? In order to solve this problem, let's first figure out what D is equal to. Let's first figure out the maximum value that D can be. So let's maximize A, B, and C to all be 9. 999 plus 999 equals 1,998. 1,998 is less than 2,222, which means that D has to be less than 2. This means that D can either be 0 or 1. If we add two positive numbers, it is impossible for us to make 0, so D cannot be equal to 0. This means that D is equal to 1. Now let's start solving for the other digits. Let's first look at this digit. It is a 1, which means that A plus C has to either be equal to 11 or 1. Let's first say that A plus C is going to be equal to 1, so let's let A equal 1 and C equal 0. It looks like we won't be able to make a number in the thousands, but just to make sure, let's maximize digit B. Let's say that digit B equals 9. If we add 190 and 91, we get a result of 281, which is less than our goal of 1,111. This means that it is impossible for us to make 1,111 if A plus C equals 1. Therefore, A plus C has to be equal to 11. For this example, we will say that A equals 4 and C equals 7. 7 plus 4 is 11, and we carry the 1. Again, our goal is to make 1,111, so we want all 1's. Let's just say that B equals 0, and let's just carry the 1 down. Then finally, 4 plus 7 is again 11, and we get our target number, 1,111. The question asked us, which digit is represented by B? The answer is 0, letter A. Problem number 20 states, four ladybugs sit on different cells of a 4x4 grid. One of them is sleeping and does not move. Each time you whistle, the other three ladybugs move to a free neighboring cell. They can move up, down, right, or left, but they are not allowed to go back to the cell they just came from. Which of the following images might show the result after the fourth whistle? In order to solve this problem, let's first figure out which ladybug is the sleeping ladybug. This is going to be the ladybug that doesn't move. As you can see, this ladybug stays in the same position throughout the duration of the simulation, which means that this ladybug is the sleeping ladybug. We will mark it in green. Now let's start tracking individual ladybugs. Let's start off with this one. This ladybug can either move up, right, or down. If we project the possible movements onto the next cell, we can see that one of its possible positions overlaps with an unidentified ladybug. This means, of course, that that is the same ladybug. We will color this ladybug in blue. Now let's look at how the blue ladybug will continue to move. The blue ladybug can only move to the right. It cannot move down because that is the cell where it came from, and the problem told us that a ladybug cannot go to a cell from which it came from. If we use the same technique we used earlier, we can see that the blue ladybug moves to the right. Now the blue ladybug can either move right or down. Again, it can't move to the left because that's the cell where it just came from. And if we project its possible movements, it is clear that the blue ladybug moves down. The question asked us about how the grid would look like after the fourth whistle. So let's also predict the movement after the fourth whistle. The blue ladybug can only move to the left or to the right. It cannot move down because that cell is already occupied by the sleeping ladybug, and it cannot move up because that is the cell from which it came from. Now let's check another ladybug. We will color this ladybug in orange. The orange ladybug can either move up or down. It cannot move to the right because that is already the position of the sleeping ladybug. By overlapping the possible positions, it is clear that the ladybug moves down. From this point, the ladybug can only move to the right. Again, it cannot go up because that is the cell from which it came from. Again, this ladybug can only move to the right. It cannot move up because that is where the sleeping ladybug is, and it cannot move to the left because that is where it came from. Since we will be looking at the simulation after the fourth whistle, let's also look at where the ladybug can move. The ladybug can either move up or to the right. Now we have identified all of our ladybugs. But to make our job easier, let's predict the final motion of the red ladybug. The red ladybug can either move up or to the left. The blue ladybug and the red ladybug share a possible cell, so we will color that cell purple. The red ladybug cannot move down because it came from that cell. Finally, we can solve the problem. First, let's check to see if the sleeping ladybug is still in its position in all of the possibilities. In image E, it is not present, so image E cannot be the answer. Next, let's check to see if some of the ladybugs are in impossible positions, positions to where they could not have moved to. In image B, it is impossible for a ladybug to be at this position, so the answer cannot be B. In image C, it is impossible for a ladybug to be at this position, so it also cannot be C. In image D, it looks like all the ladybugs are in the same position. But if we look at the orange ladybug, it could have either moved up or to the right. But in image D, it didn't move either up or to the right. It teleported somewhere. So it cannot be letter D. This means that it has to be letter A. The question asked us, which of the following images might show the result after the fourth whistle? The answer is letter A.

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