Problem number one states, the largest of the following numbers is... So let's take a look at each of these numbers individually and evaluate their value. So first, let's start out with the first one, 2011 to the first power. 2011 times 2011 once is just going to be 2011. So the answer is 2011. 1 to the 2011 power is going to be 1 times 1 times 1 times 1, 2011 times. Of course, if you constantly multiply something by 1, you won't get any different number other than 1. So the answer here is 1. For 1 times 2011, it's pretty easy. 1 times any number is just going to be that number, so it's 2011. Next, 1 plus 2011 is just simple addition. Of course, we get 2012. And then 1 divided by 2011 is going to be 1 divided into 2011 equal parts, which is going to be a very small number, which is less than 1. The exact value doesn't matter since the problem asks for the largest number, and it's clear this is not going to be the largest one. So in this case, the largest number was 1 plus 2011, which evaluated to 2012. So the question asked us, which of these was the largest number? The answer was 1 plus 2011, letter D. Problem number 2 states, Ellie has 5 cubes and 3 tetrahedrons. How many sides do these solids have all together? So first, let's figure out how many sides there are on a cube, and then how many sides there are on a tetrahedron. So here we have a cube. We can see that on this side we have 3 faces. And then if we flip it over, on the other side we have another 3 faces. So in total, we know that a cube has 6 faces. Now for the tetrahedron. Here we can see that the tetrahedron has 3 sides. And if we flip it over, we see that it has one side on the back, so in total it has 4 faces. So the problem told us that the girl had 5 cubes and 3 tetrahedrons. That means that the total number of faces that she had was 5 times 6 plus 3 times 4. She had 5 cubes with 6 faces, and 3 tetrahedrons with 4 faces. Evaluating this, we get 30 plus 12, which then simplifies down to 42. So the question asked us, how many sides do these solids have all together? The answer is 42, letter A. Problem number 3 states, a pedestrian crosswalk consists of alternating white and black stripes. The crosswalk begins and ends with a white stripe. How wide is the crosswalk if every stripe is 50 centimeters wide, and there are 8 stripes in all? So the question told us that we have 8 stripes in all. And there are 8 stripes in all. So the question told us that we have 8 white stripes in total, and each white stripe is surrounded by a black stripe. This means that in total we have 15 stripes. We know that each stripe is 50 centimeters long, or 0.5 meters. So if we multiply the length of each stripe by the number of stripes, we get the total length of the crosswalk. So 0.5 meters times 15 stripes is 7.5 meters. So the question asked us, how wide is the crosswalk if every stripe is 50 centimeters wide, and there are 8 stripes in all? The answer is 7.5 meters, letter B. Problem number 4 states, a certain broken calculator divides instead of multiplying, and subtracts instead of adding. I enter the expression 12 times 3 plus 4 times 2. What answer will the broken calculator display? So we know that the person entered 12 times 3 plus 4 times 2. We know that multiplication switches to division, and addition switches to subtraction. So instead we'd have 12 divided by 3 minus 4 divided by 2. First let's do what's in the parentheses, so 12 divided by 3 would be 4, and 4 divided by 2 would be 2. Then we have 4 minus 2, which is 2. So the question asked us, what answer will the broken calculator display? The answer is 2, letter A. Problem number 5 states, at exactly 2011, Peter Lowe was the first person to use a calculator. At 2011, Peter looked at his digital watch, which was set to display time in 24-hour mode, military time. Note that this is 8-11 p.m. in 12-hour mode. What is the shortest amount of time, in minutes, after which the displayed time will again contain the digits 0, 1, 1, and 2? So in order to solve this problem, let's just take all the answers, since we know one of them has to be true, and let's add them to the original time, and then we can figure out which one of them contains the digits 2, 1, 1, and 0. So first let's start out with 40. If we add 40 minutes to the current time, we get 2051, which does not contain the correct digits, so 40 cannot be the answer. Next let's do 45. 45 plus 2011 would be 2056, which doesn't contain the right digits, so let's continue on to 50. If we add 50, we get 2101, which does contain a 0, two 1's, and a 2, so 50 seems to work. Let's continue on with 55. 2011 plus 55 is going to be 2106, which doesn't contain the correct digits. Let's move on to 60. If we add 60, we get 2111, which doesn't contain the right digits either, so the only one that works is 50. So the question asked us, what is the shortest amount of time in minutes, after which the displayed time will again contain the digits 0, 1, 1, and 2? The answer is 50 minutes. Problem number six states, the picture on the right shows three squares, large, medium, and small. The vertices of the medium square lie on the midpoints of the sides of the large square, and the vertices of the small square lie on the midpoints of the sides of the medium square. The vertices of the medium square lie on the midpoints of the sides of the large square, and the vertices of the small square lie on the midpoints of the sides of the medium square. The area of the small square is 6 centimeters squared. What is the difference between the area of the large square and the area of the medium square? The area of the small square in the middle has an area of 6 centimeters squared. And because the vertices of the small square are the midpoints of the sides of the medium square, the area of the small square is half that of the medium square, which means that the medium square has an area of 12 centimeters squared. Applying this same logic, but this time to the medium square and to the large square, we can see that the large square has an area of 24 centimeters squared. So if the large square has an area of 24 centimeters squared, and the medium 12 centimeters squared, the difference in the areas is 12 centimeters squared. So the question asked us, what is the difference between the area of the large square and the area of the medium square? The answer is 12 centimeters squared. Letter C. Problem number 7 states, There are 17 houses on the street on which I live. The houses on the right side of the street are numbered with consecutive even numbers starting with 2, and the houses on the left side of the street are numbered with consecutive odd numbers starting with 1. I live in the last house on the right side of the street, and my house number is 12. What is the number of the last house on the left side of my street? So we know that the first houses have a number 1 and a number 2. And as we go along the street, both numbers increment by 2 each time we move one house down. So it would be 3 and 4, and then 5 and 6, and on and on and on, all the way until we reach to the house where the person in the problem lives. House number 12. It may seem that we're done here, but we're not. Because we only have 12 houses here, and the problem specifically stated that there are 17 houses on the street, so we have to keep on going on the odd side, since we know that 12 is the last number on the even side. 17 for 15 numbers, 19 for 16, and lastly, 21 for 17. So the last number on the other side of the street is 21. So the question asked us, what is the number of the last house on the left side of my street? The answer is 21. Letter E. Problem number 8 states, All 4-digit numbers, the sum of whose digits is 4, were arranged in the order from least to greatest. In which position is the number 2011? Note, a 4-digit number cannot begin with a 0. So first, let's try to figure out all the possible ways of which we can create the sum of 4. So for example, we could have a 1 and a 3. We could have two 1's and a 2. We could have two 2's or four 1's. We could also have just one 4, and we would only have the number 4000, but that's greater than 2011, so we just don't have to include that. So for the 1 and the 3, 1003, 1030, and 1300 are the only possible combinations that are less than 2011. For 1, 1, and 2, we have these combinations. For 2 and 2, the only possible combinations that are less than 2011 is 2002. And for 1, 1, 1, 1, there's really only one combination, 1111. So in total, we have 11 numbers that are less than 2011, whose digits add up to 4. So that means that 2011 is in place 12. So the question asked us, in which position is the number 2011? The answer is 12, letter C. Problem number 9 states, numbers A, B, and C have the following properties. The arithmetic mean of A and B is equal to 17, and the arithmetic mean of A, B, and C is equal to 15. The number C is equal to... So here are the formulas of the arithmetic mean for A and B. We just have A plus B over 2. We gotta add up the numbers A, B, and then divide by the number of numbers that there are, which is 2. And we know that's equal to 17. And for the arithmetic mean of A, B, and C, we add up A plus B plus C, and then divide it by the number of values there are, which are 3. So A plus B plus C over 3 has to equal 15. So now on the left, let's multiply both sides by 2, and on the right, let's multiply both sides by 3. After this, we know what A plus B is, so we can substitute that in on the right side of the equation to get 34 plus C equals 45. Then we subtract 34 from both sides to get that C equals 11. So the question asked us, what's the value of C? The answer is 11. Letter C. Problem number 10 states, the sum of the smallest three-digit numbers, whose digits add up to 8, and the largest three-digit numbers, whose digits add up to 8, equal to... So first, let's try to figure out the smallest number, whose digits add up to 8. So of course, the smallest three-digit number that's possible is 100. And as we keep on going up, 101, and so on, we eventually reach 107. The sum of the digits of 107, 1 plus 0 plus 7 is going to be 8, so 107 is the lowest number that we can have. Now for the largest number, let's start with the largest three-digit number, 999. Of course, as we go down, we'll have digits 9 plus A plus B, and this of course cannot be equal to 8, since 9 already is above 8, and if we add two more, it's going to be even greater than 8, so we can just skip all the 900s and go to the 800s. For the 800s, the only way that 8 plus A plus B is going to be equal to 8 is if A and B are both 0, which means that the largest number that follows this pattern is going to be 800. 8 plus 0 plus 0 equals 8. So the question asked for the sum of these two numbers, so 107 plus 800 is 907. So the question asked us, the sum of the smallest three-digit numbers whose digits add up to 8, and the largest three-digit numbers whose digits add up to 8, are equal to 907, letter B. Problem number 11 states, what is the value of 2011 times 2.011 divided by 201.1 times 20.11? In order to solve this problem, let's break it up into two division problems, and then let's multiply the results at the end. So first, 2011 divided by 201.1. If we multiply the denominator by 10, we get the numerator. This means that the numerator is 10 times greater than the denominator, so the value would be 10. The opposite is true on the right side. If we multiply the numerator by 10, we get the denominator, which means the numerator is one-tenth the size of the denominator, which means the value of the multiplication is one-tenth. 10 times one-tenth is 1. So the question asked us, what's the value of 2011 times 2.011 over 201.1 times 20.11? The answer is 1. Letter C. Problem number 12 states, during three consecutive games, the soccer team FC Kangaroo made a total of three goals, and had one goal scored against them. This soccer team won one of the three games, lost one game, and tied one game. What was the final score of the game that FC Kangaroo won? So here we have a table. On the left are all the matches they played. They have the match that they won, that they tied, and that they lost. And for the vertical columns, it's going to be the score. So on the left, it's the goals scored by the FC, and on the right, it's going to be the goals scored by the opponent. In the very bottom match, we know that they lost, which means that the goals by the opponents has to be greater than the goals by the FC. So the goals by the opponents cannot be a value of zero, so it must be a value of one. In order to lose, FC must have scored zero goals. The total number of goals scored by the opponent in all three games is one, which means that the amount scored when they won a match and when they tied a match must have been zero. In the middle match, since the match was tied, the goals scored by the FC must have also been zero. The goals scored by the FC in all three rounds is three. So far we have zero, so when they won a match, they must have scored three goals. So the question asked us, what was the final score of the game that FC Kangaroo won? The answer is 3-0. Letter E. Problem number 13 states, a jeweler had nine pearls, which weighed 1 gram, 2 gram, 3 gram, 4 gram, 5 gram, 6 gram, 7 gram, 8 gram, and 9 grams respectively. He made four necklaces, and in each necklace sat two of these pearls. The pearls in each necklace weighed 17 grams, 13 grams, 7 grams, and 5 grams respectively. What is the weight of the pearl that was not set in any of the necklaces? So this problem is deceptively easy. Let's figure out how much in total all the pearls weigh individually. So 1 plus 2 plus 3 plus 4 plus 5 plus 6 plus 7 plus 8 plus 9 is 45. So all the pearls together weigh 45 grams. Then let's figure out how much all the necklaces weigh together. 5 plus 7 plus 13 plus 17 is 42 grams. Which means there is a difference of 3 grams between the weight of the necklaces and the weight of the pearls. Which means that the pearl that wasn't used must have weighed 3 grams. So the question asked us, what is the weight of the pearl that was not set in any of the necklaces? The answer is 3 grams, letter C. Problem number 14 states, the circles in the map of the labyrinth on the right represent gold coins. An adventurer is going through a labyrinth in search of gold coins, but he may not visit any portion of the labyrinth more than once. What is the largest possible number of gold coins the adventurer can collect? So first let's take a look at the gold coin in the bottom left. This is not reachable. Because once you go into it, you have to go back the same way that you came, and therefore the adventurer would revisit a cell twice. So that one cannot be reached. Now let's take a look at these three coins. And let's say that the adventurer goes like this. This means that it is impossible for him to reach the last coin in the row, with reaching the exit as well, because he would have to come back the same way that he already went. So in these three, only two out of the three can be reached. Next let's take a look at these ten. Let's say the adventurer goes like this. Only nine out of the ten coins can ever be collected in the bottom ten. Which means that we skip the one in the top row, the one in the middle row, and the one in the bottom left corner. Which means we have to skip three coins. But the remaining thirteen coins the adventurer can get. So the question asked us, what is the largest possible number of gold coins the adventurer can collect? The answer is thirteen. Letter B. Problem number fifteen states, each region in the picture must be colored in such a way that neighboring regions are of different colors. The available colors are red R, green G, blue B, and orange O. Three regions have already been colored as indicated by the letters in the picture. What will be the color of region X? So first let's take a look at this region. This region neighbors an R, a G, and a B. This means that it must be an O. This means that it must be an O. Next, let's take a look at this region. This region neighbors an O, a G, and a B. Therefore, it must be an R. Let's take a look at this region. This region neighbors an O, a B, and an R. So it must be a G. Next, this region neighbors an O, a G, and an R. So it must be a B. This region neighbors a B, an R, and a G. So it must be an O. And then this region neighbors a B, G, and O. So it must be an R. So the question asked us, what will be the color of region X? The answer is red, letter A. The answer is red, letter A. Problem number 16 states, Peter, who is an avid fisherman, caught 12 fish over three consecutive days. On each day, except the first day, he caught more fish than on the previous day. On the third day, he caught fewer fish than the total number from the previous two days. How many fish did Peter catch on the third day? So let's say the number of fish Peter caught on the first day is A, on the second day is B, and on the third day is C. We know that the total number of fish that he caught is 12. The problem told us that on each day, except the first day, he caught more fish than on the previous day. This means that C is greater than B, which is greater than A. Then the problem told us that on the third day, he caught fewer fish than the total number from the previous two days. So we know that A plus B is greater than C. We know that on the last day, he caught more than the average of the other days. So the average of all the three days is that he caught four fish. But on the last day, he must have caught more than four fish. Now in our very bottom equation, let's add C to both sides. To get A plus B plus C is greater than 2C. We won't worry about the greater than 4 right now. So of course we can substitute A plus B plus C for 12. To get 12 is greater than 2C, and then to get 6 is greater than C. So now we know that C is sandwiched between 4 and 6 exclusive. And since Peter couldn't have caught a fractional number of fish, he must have caught five fish on the third day. So the question asked us, how many fish did Peter catch on the third day? The answer is 5. Letter A. Problem number 17 states, which two numbers can be removed from the numbers 17, 13, 5, 10, 14, 9, 12, and 16 so that the arithmetic mean remains unchanged? So in order to solve this problem, let me show you a trick with means. Here we have the numbers 1, 2, 3, 4, 5. If we add them all up and divide by 5, we get that the average is 3. Now if we remove two numbers which are equidistant from the average, so for example 1 and 5, the average will not change. If we do it again, if we remove 2 and 4, the average will not change again. So if the average of the numbers that we're removing is the average of all the numbers, the average won't change. So now let's apply this to the problem. Here are the numbers we had. Obviously let's find the average, so let's add them all up and divide it by 8, and we get an average of 12. So now from our possibilities, there's only one whose values average to 12, and that's 14 and 10. Obviously 14 plus 10 over 2 is going to be 12. So if we remove those two numbers, it is in fact true that the average doesn't change, and it is still 12. So the question asked us, which two numbers can be removed from these numbers so that the arithmetic mean remains unchanged? The answer is 14 and 10. Letter E. Problem number 18 states, consider a plane containing the line segment DE of length 2. How many points F in this plane have the property that the triangle DEF is a right triangle with an area equal to 1? So here we have our line segment DE, and we know that it has a length of 2. Now if we put our point F right here, and we say that FD is equal to 1, we get an area of 1 half base times height, so we have an area of 1 of FDE. The triangle is right because FDE is a right angle, so it works. That's one combination. To get another combination, we can just flip it horizontally over the vertical axis, and for another one we can flip it vertically over the horizontal axis, and then flip it back over for a fourth combination. So now we have four combinations, but there's one combination left, if we do it like this. The angle DFE is 90 degrees, and we know from using 45-45-90 triangle proportions that FD is root 2, and FE is root 2. The area is 1 half base times height, the base and the height are both root 2, so we have an area of 1. Of course we can flip it over for a second combination, and those are our six possible combinations. So the question asked us, how many points F in this plane have the property that the triangle DEF is a right triangle with an area equal to 1? The answer is 6, letter C. Problem number 19 states, the number A is positive and less than 1, and the number B is greater than 1. Which of the following numbers is the greatest? So in order to do this, let's just say that A is going to be 0.5, and B is going to be 2. Let's bring up our possible expressions, and let's substitute in A and B, and then let's evaluate all of them. We can quite clearly see that 2.5 is the greatest. 2.5 is A plus B, so that one would be the greatest. Obviously this is just a specific example, Obviously this is just a specific example, not a concrete proof. So now let's do the proof. So here we have all of our expressions, and we know that A is going to be sandwiched between 0 and 1, and B is going to be sandwiched between 1 and A plus B. This means that it is impossible for B to be greater than A plus B. Then we know that B is going to be sandwiched between B minus A and B plus A. So B minus A has no chance of being greater than B plus A ever. Next, for multiplication, we know that A is going to be less than A times B, because B is going to be greater than 1, so it's going to magnify A, and we know that A is less than 1, so it's going to make B smaller, so A times B is going to be less than B. B plus A is obviously greater than B, so A times B can never be greater than A plus B either. Next, for A divided by B, we kind of have the opposite case, where 0 is less than A divided by B, since B is a large number, it splits up A into many small parts, so it would have to be less than A. Obviously we know that 1 is going to be greater than A, and B is going to be greater than 1, and then B plus A is going to be greater than B. So A divided by B has no chance of being greater than B plus A. So the answer must be A plus B. We proved it both with an example and with a concrete proof. So the question asked us, which of the following numbers is the greatest? The answer is A plus B, letter B. Problem number 20 states. Each of the four squares in the picture is identical. We draw one more such square in such a way that the resulting figure has an axis of symmetry. And how many ways can this be done? So obviously the square that we're going to add is going to be touching the other squares, so these are the only possible places where the square can be. So first let's start out at the top and work clockwise. Here it is not symmetrical. Let's move on to the next one. Here it does have an axis of symmetry, this horizontal one right here. So that one works. Then for the next one, we don't have an axis of symmetry. Let's continue on here. This one actually does have an axis of symmetry. It's this 45 degree angle right here, so that one actually works. This one doesn't have an axis of symmetry, so let's move on. This one doesn't have an axis of symmetry either, so let's continue. This one does have an axis of symmetry, this vertical line right here, so that one works. This one doesn't have an axis of symmetry, so let's continue. And finally, this one doesn't have an axis of symmetry either, so that one doesn't work. So we only have three spots where we can put the square for it to have an axis of symmetry. So the question asked us, we draw one more such square in such a way that the resulting figure has an axis of symmetry. In how many ways can this be done? The answer is three ways. Letter C. Problem number 21 states. A square piece of paper was cut into six rectangles as shown. The sum of the perimeters of these six rectangles is equal to 120 centimeters. What is the area of the sheet of paper? So first let's call the width and the height of the square piece of paper x. Next, we know that the perimeters of all the individual pieces altogether is 120 centimeters. This includes all of the vertical side lengths, there are three of them, because the middle one is counted twice because all of the perimeters are counted individually, and then all the horizontal ones as well. There are six because there are two on the outside and two in the middle which are counted twice, which is a total of six. So in total, the 120 centimeters is equal to 4 plus 6x or 10x. Therefore, one side length of the piece of paper is equal to 12, so the area x squared is 144. So the question asked us, what is the area of the sheet of paper? The answer is 144 centimeters squared, letter D. Problem number twenty-two states, the five-digit natural number 2, 4, x, 8, y is divisible by 4, 5, and 9. The sum of the digits x and y is equal to... So we have the number 2, 4, x, 8, y. We know that it's divisible by 4, 5, and 9. Since it's divisible by both 4 and 5, that means it's divisible by 20. All numbers that are divisible by 20 must end in a zero, so therefore y must be a zero. So we know our number is 2, 4, x, 8, 0. In order to get the last number, I'm just going to guess and check. So we know that the number has to be divisible by 120 since it's divisible by 4, 5, and 9, right? We multiply all of them together. And then let's divide 24,000 x 80 by 180. So let's just do division. Let's do it with 24,080. We'll use the lower bound of zero, so this will give us a lower bound estimate. So now let's just divide. 180 goes into 240 once. Multiply by 1, we get 180. Subtract, we get 60. Let's bring down the 8. 180 goes into 608 three times. 180 x 3 is going to be 540. Subtract that from 608, we get 68. Bring down the zero. That goes in about three times as well. 180 x 3 is 540. Subtract, and it doesn't go in perfectly. So our answer is 133 and then some spare. So this is a good starting point. We know that 133 x 180 is going to be about the value of 24 x 80. Of course, this is the lower estimate, so if we just increment 133 by 1, we can check again. By doing this multiplication, we get 24,120, which does not match our number. So let's continue. 135. It doesn't match our number again, so let's continue. When we get to 136, we get the number 24,480, which does match the pattern of our number. That means that x is equal to 4. So the sum of the two digits x and y is 4. So the question asked us, the sum of the digits x and y is equal to? The answer is 4. Letter E. It can be placed in the unshaded region of the grid in such a way that there will be no room left to place any of the remaining four shapes. Any shape may be rotated or turned over before it is placed in the grid. So the most interesting section to me is this bottom section. Because if we fail to cover it up, we can simply take the 1x5 piece and cover it up with that. So at least one of the squares in this bottom region must be covered up. So first, let's start out with the 1x5 piece. The 1x5 piece can be placed here and only here. It cannot be placed vertically because there is not enough room. And it is quite easy to see that another piece can be placed here. As a matter of fact, all the pieces can be placed somehow in there. So that one doesn't work. Next, let's try the cross. Again, we have to cover up the yellow line. So let's try it like this. This is the only possible combination as well. And we can use this piece like this so the cross doesn't work. Next, let's try this piece. Of course, there are a couple combinations that we can do. Like this, like that, and like this. But in all of these combinations, we can just place this piece like this, and it covers it up. No matter how we orientate that piece. So that one doesn't work. Next, let's try the C one. We have to cover up the yellow line. So we can do it like this. Again, the piece fits in here. We could do it like this, but the piece still covers up the unshaded region. Like this, still the same problem. Or like this, and we still have the same problem. So that one doesn't work. By elimination, it must be the last one, but let's just make sure it works. If we place it like this, we can see that we segment the unshaded region into sections of four blocks. Obviously, it's impossible to put a shape with five blocks into a section with four blocks without overlapping. So this one works. So the question asked us, Which of the following shapes, each also made of five squares, can be placed in the unshaded region of the grid in such a way that there will be no room left to place any of the remaining four shapes? The answer is this one, letter B. Problem number 24 states. Inside a square with a side length equal to 7 cm, there is a square with a side length equal to 3 cm. A third square with a side length equal to 5 cm intersects each of the other two squares. See illustration. The difference between the area of the black region and the sum of the areas of the shaded region is... First, let's name all of our regions. We'll call Ws the white region inside the small square, Shn the shaded region inside the large square, Shout the shaded region outside the large square, Wm the white region inside the medium square, and B the black region. So now we can see that the small square with a side length of 3 is composed of the white region inside the small square and the shaded region inside the large square. So Ws plus Shn is equal to 3 squared. Next, for the medium square, we can see it is composed of the white region inside the medium square, the shaded region inside the large square, and the shaded region outside of the large square. That is of course equal to 5 squared. And then lastly, 7 squared, the big square, is equal to the white region inside the small square, the white region inside the medium square, the shaded region inside the large square, and the black region. So now the question asks us to find the difference between the area of the black region and the sum of the areas of the shaded region. So, let's start out with the only equation that contains the area of the black region, and try to get that expression. So first, let's subtract out the middle equation. That's what we get. Let's combine like terms. And we can see that we have an extra Ws, and we need an extra Shn. In order to get this, let's subtract the first one, Ws plus Shn equals 3 squared. And then combine like terms. And we get the desired expression, B minus Sh out minus Sh in. So if we evaluate the right side, we get a value of 15. So the question asks us, the difference between the area of the black region and the sum of the areas of the shaded region is 15 cm squared. Letter D. Problem number 25 states, during target practice Bob can earn 5, 8, or 10 points for hitting the target. He hit the 10 point mark as many times as he hit the 8 point mark. Altogether, Bob managed to earn 99 points while missing the target 25% of the time. How many times did Bob fire at the target? So first, let's name some variables. Let's say that t is the total number of times he fired, x is the total number of times he hit the 5, y is the number of times he hit the 8, and z is the number of times that he hit the 10. The first thing that the problem tells us is that he hit the 10 point mark as many times as he hit the 8 point mark. So therefore we know that y must be equal to z. Then the problem tells us that Bob managed to earn 99 points. So we know that 5x plus 8y plus 10z is equal to 99. Using the first equation, let's substitute z for y to get 5x plus 8y plus 10y equals 99. Of course, we can combine this further into 5x plus 18y equals 99. The next thing the problem tells us is that he missed the target 25% of the time. So x plus 2y is equal to 3 fourths t. The reason it's 2y is because we're accounting for the number of times he hit the 8 mark and the 10 mark, and we know they're the same value. So instead of doing y plus z, we can just do 2y. So next, let's solve for x in terms of y. So let's move all the irrelevant stuff to one side, and then divide by 5 to get x equals 99 minus 18y over 5. Now we know that 18y has to be sandwiched between 0 and 99 inclusive. This means, of course, that y only has a certain number of values. 0, 1, 2, 3, 4, or 5. Now how do we figure out which one it actually is? Well, we know that x has to be a whole number, since he couldn't have shot a fractional number of arrows. So by just guessing and checking, when y equals 3, we can see that x equals 9. So that's great, but the problem asks us how many times did he fire at the target, so we have to solve for t. So if you remember our equation for t was x plus 2y equals 3 fourths t. Of course, plug in for x and y, multiply by 4 thirds, and we get that t equals 20. So the question asks us, how many times did Bob fire at the target? The answer is 20 times. Letter D. Problem number 26 states, in the convex quadrilateral ABCD, length AB is equal to length AC. In addition, angle BAD is equal to 80 degrees, angle ABC is equal to 75 degrees, and angle ADC is equal to 65 degrees. See illustration. What is the measure of angle BDC? Okay, so first it's important to note that the 80 degrees does not measure angle DAC, but rather angle DAB. So first let's try to figure out what angle DCB is. So obviously we know that the sum of all the angles of a quadrilateral have to add up to 360 degrees. So we have A, B, C, and D, of course we don't know one of them, so we can solve. We plug in all of our values, combine like terms, subtract 220 from both sides, and we get that angle C is equal to 140 degrees. In this case angle C refers to the angle of the quadrilateral, so angle BCD. Now, the problem told us that AC and AB were congruent. This means that triangle ABC is isosceles. Therefore, the two angles ABC and ACB are the same. That means that that angle is 75 degrees, and since the whole angle was 140 degrees, angle ACD is 65 degrees. Of course, now we know two of the three angles in triangle ABC, so we can solve for the third one. Of course, the sum of all the angles is going to be equal to 180 degrees. So now we plug in 75 for both of them, and we get that angle A is equal to 30 degrees. So CAB is 30 degrees, and DAB was 80 degrees, so DAC is 50 degrees. The problem told us that these two lines were of the same length. And we can see that triangle ACD is isosceles, so therefore, this length is also the same. Through the transitive property, we know that AD and AB are congruent, making a large isosceles triangle through the quadrilateral. Since we know that the angle that is not repeated plus twice the angle that is repeated equals 180 degrees, we know the angle that is not repeated, that's 50 degrees plus 30 degrees, so 80 degrees. So then we subtract 80 from both sides and divide by 2 to get that angle ABD and angle ADB are both 50 degrees. Of course, previously angle ADC was 65 degrees, so now angle ADB is 50 degrees and BDC is 15 degrees. So the question asked us, what is the measure of angle BDC? The answer is 15 degrees, letter B. Problem number 27 states. The illustration shows a cube and its layout with face ABCD labeled. The cube is cut along the bold lines into two identical solids. Which of the following pictures represents the layout of the cube with the bold lines properly marked? So here we have our cube and its net. So first let's start out with the front face. Here we have a line that goes from the top right vertex C to the midpoint of the edge AD. So on our net it would look something like this. Next let's analyze the next face. The one to the left. This is going to be the face that shares the common edge AD. So that's going to be to the left. And we can see that it goes from the middle of DA all the way to the bottom left corner. So that would look like this. Now let's go the other way. Let's go to the right. To the face to the right of edge CB. And this one goes from the top to the middle. So it would look like this. And then finally let's go to the last one. The one on the back of the cube. That one goes from the middle to the very bottom. So that one would look like this. So that's our answer. That's how the net would look. So the question asked us. Which of the following pictures represents the layout of the cube with the bold line properly marked? The answer is this one. Letter A. Problem number 28 states, each letter in the expression shown to the right represents a digit other than zero. Different letters represent different digits, and the same letter always represents the same digit. What is the smallest whole number that can be the value of this expression? So first, before we even talk about how to minimize this, let's simplify this expression algebraically. So let's cancel out terms that appear both in the numerator and the denominator. So for example, this a and this g. Those are the only ones that repeat, so we can just straight up remove them. Now in order to minimize this, what we want is we want the lowest numerator and the highest denominator. So obviously, in the numerator, o repeats twice, so it would be best if that was the lowest number possible. But what numbers can we use? Well, the problem said that we can use any digit that's not zero, so 1, 2, 3, 4, 5, 6, 7, 8, or 9. So since o is used twice, we should probably say that it is equal to one. Since we're multiplying by one, that doesn't change the value, so we can just remove those. Now since we want the numerator as small as possible, let's use small digits. So we'll say k is 2, a is 3, and n is 4. Now, we probably shouldn't put a 5 for r, because ideally, we want the prime factorization of the numerator and the denominator to be similar, so they can cancel out and result in a smaller number. This means we shouldn't pick digits with rare primes like 5 or 7. So 5, not a good choice. Let's instead pick 6. Now in the denominator, let's try to cancel out as many of these numbers as possible. In the numerator, we have 144. Luckily, 8 and 9 multiplied together is 72, which is half of 144. So when canceled out, we get a 2. This is the lowest number we can possibly get. So the question asks us, what is the smallest whole number that can be the value of this expression? The answer is 2, letter B. Problem number 29 states, the shape in figure 1 is made of two rectangles. Two of the side lengths are given. The shape was cut along the dotted lines and rearranged into the triangle in figure 2. The length of side x is equal to... Okay, so you have the piece of paper that's cut, and then it's rearranged into the triangle on the right. So first, we have to notice that the triangle on the left was rotated 270 degrees. So on the right side, on the triangle side, on the left, that edge is 11, and on the bottom, we have x. That's the length that we want to figure out. Of course, one of the components of x is already given to us. We know that this length is 13. But for the other one, we have to use some algebra. So we know that this length right here is 24, because we have 11 plus 13. And this triangle on the bottom right is rotated 180 degrees around. So that bottom face is the top face, making that length 24. So x is the sum of 13 and 24, or 37. So really, this question is less complicated than it looks. So the question asked us, the length of side x is equal to... The answer is 37. Letter B. Problem number 30 states, each of three boys, Adam, John, and Carl, made a statement as follows. Adam, the distance between John and me is greater than twice the distance between Carl and me. John, the distance between Carl and me is greater than twice the distance between Adam and me. Carl, the distance between John and me is greater than twice the distance between Adam and me. At least two of these statements are true. Which of the boys is telling a lie? So here we have Adam, John, and Carl, and in order to solve this problem, let's assume they're all telling the truth for as long as possible before we start getting to things that are impossible. So first, Adam says the distance between John and him is greater than twice the distance between Carl and him. So that could look something like this. If the distance between Adam and Carl is x, then the distance between Adam and John is greater than 2x. Okay, so this is possible. Next, let's take a look at what Carl says. Carl then says the distance between John and him is greater than twice the distance between Adam and him. Okay, this also makes sense. All what this tells us is that the distance between Carl and John is also greater than 2x. So, so far, we're going pretty good. Next, let's take a look at what John is saying. John is saying that the distance between Carl and him is greater than twice the distance between Adam and him. That doesn't make any sense. Because if we rearrange it, this is what he's saying is true, which would make Carl and Adam liars. But the problem says that only one of them is lying. So the one that's lying must be John. So the question asked us, which of the boys is telling a lie? The answer is John, letter B.

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