In question number 11, we are asked to analyze the following situation. There is a table with height T and two gentlemen, one standing on the floor, the other one standing on the table, with some height difference between them. If we are told that first Adam stands on the table, Mike stands on the floor, then Adam is 80 centimeters taller than Mike. If they switch positions and Mike stands on the same table and Adam stands on the floor, then Mike is one meter taller than Adam. We are asked to determine the height of the table. So let us say that Adam's height will be denoted by A and that Mike's height will be denoted by M. Then we can write down two equations. If Adam stands on the table, then his total height above ground is the height of the table plus his own height, and that is 80 centimeters more than Mike's height. If they switch positions and Mike stands on the table, then he is 100 centimeters, or one meter taller than Adam. With this information, we end up with a system in three variables, and so let's see if we can try to do something about the solution. If we add the two equations together, we obtain 2T plus A plus M minus 80 minus 100 is equal to A plus M. So that's nice. We can subtract A and M from both sides and add 80 and 100 to both sides. So we obtain 2T is equal to A plus M minus A minus M plus 80 plus 100, or twice the height of the table is 180 centimeters, therefore the table's height must be equal to 90 centimeters. And so the answer to number 11 is C, 90 centimeters. Question number 12. Dennis and Mary are playing a game where the winner is determined by the toss of a coin. We are told that initially each player has at least 100 pieces of candy, and the game is played as follows. If the coin shows heads, the winner is Mary, and she receives two pieces of candy from Dennis. On the other hand, if Dennis is the winner, Mary has to give him three pieces of candy. We are told that after 30 games, the balance of candies won and lost is equal, so both players end up with the same number of pieces as at the start of the game. How many times did Dennis win? So assuming that the balance of candies won and lost is the same after several games, let's count up the hypothetical wins and losses for both Mary and Dennis. If after three games we have three consecutive wins for Mary, she receives six pieces of candy from Dennis. On the other hand, if Dennis wins the next two games, he will also receive six pieces of candy from Mary. So the games have to be played in the following proportion, where out of five games, Mary wins three, and Dennis wins the other two, like that. And we want 30 games to be played, so three-fifths is the same as 18 over 30, and two-fifths as 12 over 30. So out of 30 games, Dennis must win exactly 12 in order for he and Mary to keep the balance of candies won and lost. So the answer to number 12 is B, 12. Question number 13. Consider the following diagram. We have six circles arranged in a triangle inside of a rectangle, one of whose sides has length equal to six centimeters. Two of our circles are shaded in gray, and we have to determine the shortest distance between them. Now, that distance will lie along a straight line connecting the two diameters, so let's sketch that in, in yellow. And we also know that the circles are not only arranged in a triangle, but that triangle has equal sides, so if we were to connect the centers like that, that triangle in green is supposed to be an equilateral triangle. So let's copy it over here, larger, so we have something to refer to, and our job will be to determine the lengths of the legs in green, and then compare that to the leg in yellow. Okay, so in terms of diameters of our circles, we can see that this leg is equal to one diameter, or two radiuses, like that, let's say r is the radius of one of those circles, and the hypotenuse, the other green side, is equal to four times the radius. Okay, so let's compute the radius, and then we'll be able to use the Pythagorean formula to determine the length of this missing side here, that I will for now denote as b. We know that the one length that is given to us, six centimeters, represents three diameters, or six radiuses, and so one centimeter is our radius. So we have here the following relationship, b squared plus two squared is equal to four squared, b squared is then 12, or b is the square root of 12, which we can write as two times the square root of three. Okay, now the distance we are interested in is not b, it is b less two radiuses, because the yellow line is along the line connecting diameters, so we have here a radius and a radius, and that distance in between is the shortest distance that we are interested in. So we're almost done, that distance star is equal to b minus twice the radius, or two root three minus twice the radius, and we have in centimeters, two root three minus two centimeters, and so the answer is therefore c, two root three minus two. Question number 14. In Billy's room there are four clocks, however each one shows the incorrect time, either by two, three, four, or five minutes. One day Billy wanted to know the exact time by his clocks, and he saw that one shows six minutes to three, or the second one was three minutes to three, the third one two minutes past three, and the last one three minutes past three. So let's write down those times, and then all the other possibilities. So we have the first clock showing six minutes to three, so that's two fifty-four, three minutes to three, two fifty-seven, two minutes past three, three oh two, and three minutes past three, three oh three. Now, accounting for all the possibilities, we don't know which clock is how fast or how slow, so we take two minutes plus or minus in each direction, that gives us two fifty-six, two fifty-two, two fifty-nine, two fifty-five, three oh four, three o'clock, exactly, three oh five, three oh one, and now we see if there is a time that appears in each list, and we don't have such a time yet. If we can find an hour, or a displayed time that matches across the lists, then that must be the correct time. So let's see. Three minutes off in each direction, two fifty-seven, two fifty-one, three o'clock, exactly, two fifty-four, three oh five, two fifty-nine, three oh six, three o'clock again, and again, we don't have a displayed time that matches all the lists, so the next possibility is off by three minutes, two fifty-eight, two fifty, three oh one, two fifty-three, three oh six, two fifty-eight, three oh seven, two fifty-nine, and again, two fifty-nine appears in three lists, but not in all four, so we again don't have a number common to all possibilities, so last thing to try, and it is five minutes off in either direction, so two fifty-nine, two forty-nine, and now with that, we can stop, because we have a two fifty-nine in each list, therefore that must be the correct time, if each clock is off, plus or minus, and they all display the same time. The answer to question fourteen is therefore D, two fifty-nine. Question number fifteen. Consider the following diagram. A semicircle is inscribed in a triangle with sides of length five, twelve, and thirteen. What is the radius of the inscribed semicircle? So let us draw in that radius, we can refer to it, I will connect the center of the semicircle with the hypotenuse like so, and then we observe that since the circle is inscribed inside the triangle, the hypotenuse must be tangent to the circle, and the radius is perpendicular to any tangent line, so what we have created here is in fact a right triangle. So let us invert that, so it is easier to look at, we have a triangle like this, a right triangle in fact, and let me call this the radius of our circle, so that is R, the side on the hypotenuse, that will be twelve minus the radius, and then the leg here, I will just give that a name, A, like so. Since we have one right triangle, we would like to see if the other triangle was also a right triangle, so by applying the Pythagorean theorem, we can check that five squared plus twelve squared is twenty-five plus a hundred and forty-four, one hundred and sixty-nine is thirteen squared, so the larger triangle is also a right triangle. And now we have a relationship between the sides, and the green triangle, the smaller triangle, the radius R is to its hypotenuse twelve minus R, as five is to its hypotenuse in the larger triangle, so thirteen, and by cross-multiplying we can solve for R, so five times twelve minus R is thirteen R, thirteen R is sixty minus five R, so eighteen R is sixty, That gives us the radius as 60 over 18, which is 10 over 3. And so the answer is b, 10 over 3. Question number 16. How many four-digit numbers are there for which the 100th digit is 3 and the sum of the other digits is also 3? So what we are trying to construct is a number that has 1, 2, 3, 4 digits, with the number 3 appearing in position 2. And then the remaining three digits add up to 3. So let's enumerate, first of all, the partitions of 3 into 3 positive integers. We can write 3 as 3 and two 0s, as 1 plus 2 with a 0, and then finally as the sum of three 1s. And that should give us all the possible four-digit numbers with the stated property. So with the first partition, we have 3, 3, 0, 0. It's a four-digit number. Notice that we cannot start off with a 0 if we want to create a four-digit number, so that's the only possibility. With the second partition of 3 into 1 and a 2, we obtain the numbers 2, 3, 1, 0, 2, 3, 0, 1, 1, 3, 0, 2, 1, 3, 2, 0 are the only possibilities. Again, we cannot begin with a 0. And finally, for the last partition, we have no choice but to write 1, 3, 1, 1. And those are all the possible numbers with the stated properties. There are six of them, and so the answer is E, 6. Question number 17, consider the following 3 by 4 grid. Kanga is writing numbers in the cells of this grid so that the sum of every row is the same and the sum of every column is also the same. What number should she write in the shaded square? This is essentially a guess and check type of a calculation, so I'll make an educated first guess and then have to enumerate all the possibilities associated with each guess and simply check if one of them satisfies the conditions of the problem. So by summing the entries in the second and the third column, I see that 7 plus y is equal to x plus 4, and this tells me that 3 is equal to x minus y. So x, the way I have chosen it, is a number bigger than y and exactly by 3. So the guesses will be as follows. I will let x be 9, and then y has to be 6. Then I will guess x is equal to 8, y is equal to 5, and so forth. So hopefully, I will not have to repeat this calculation very many times before I arrive at the correct answer. So let's begin. With x equal to 9 and y equal to 6, we obtain for the sum of the entries in the first row a 15. And so that has to be the sum of the entries in the remaining two rows. And we know that the sum of the entries in each column has to be equal to 4 plus 3 plus 6, a 13. So with that, I know that I must use a 5 in this cell to make the column add up to 13. And that forces a 4 over here to obtain a 15 as the sum of the entries in that row. And we see that in order to have the last column sum to 13, I must use a number that will not make the entries in the third row sum to 15. For example, with 2, I have 6 plus 1 plus 2 makes a 15. But 2 plus 4 plus 2 is not a 13. So guess number 1 does not satisfy the conditions of the problem. And then we continue like so. It turns out that with guess number 2, we have the correct arrangement for the columns and the rows always to add up to the same number. And so what we have is a grid that looks like this. We have a 2, a 4, and then x is equal to 8, and then a 2. And then we obtain for the second row a 4, 3, 3, and a 6. And then 6y has to be equal to 5, 1. And then the number in the shaded square comes out to be a 4. Now we can check that all the rows sum up to the same number, 16, and all the columns sum up to the same number, 12. And so that is our answer here. The cell shaded in gray must contain a number equal to 4. So we choose b as the answer to problem 17. Question number 18. Three athletes, Kanga and Roo, took a break Three athletes, Kanga and Roo, took part in a marathon race. Before the race, four spectators discussed the athletes' chances. Now, after the race, it turns out that all of those statements were true. So we'll go back to the statements in a second. But using those statements, we have to determine the order in which Kanga and Roo, who were indeed the three top athletes in the race, finished. So after evaluating each statement, we'll have to compare it against the order of winners here. And evaluating the truth value of each statement will either accept or reject each possibility. So the statements are as follows. Statement number 1 says either Kan or Ga will win. So let's write that down like so. Kan or Ga comes in first. Statement number 2 reads, if Ga comes in second, then Roo will win. So that's a conditional. If Ga comes in second, so let's say Ga second, then we'll use that implication arrow. Roo comes in first, or in other words, wins. The third statement reads, if Ga comes in third, then Kan will not win. So another conditional, Ga comes in third implies that Kan is not first, does not win. And finally, the last statement reads either Ga or Roo will come in second. So Ga or Roo are second, like so. And now let's look at each statement and compare it against the given choices. Kan or Ga have to be first. So that's OK for A, B, not for C. Roo is first. And then the other choices are OK. Now, statement number 2 is a conditional statement, and it will be false only in one case. In that case, the premise is true, but the conclusion is false. So let's look for choices where Ga does come in second. So that's in choice A, for example. And then Roo has to be first, but he is not in this arrangement. So we have to get rid of possibility A. And the only remaining option where Ga is second has already been discounted. So let's go to statement 3, where we have another conditional. Let's look for the arrangement where Ga is third. So that is in B only. And in that case, Kan is first, but such a possibility would make this conditional statement false. So we cannot have B either. And finally, for the fourth statement, we have to make sure that Ga or Roo come in second. In D, that is indeed the case. Roo is second. But in E, Kan is second, so we cannot have that possibility. And that leaves us with choice D, which is the correct arrangement of finishing athletes. Ga is first, Roo is second, and Kan is third. Question number 19. The diagram shows a shape formed from two squares, a triangle with area 8 centimeters squared, and a shaded parallelogram. I copied the same picture over here so that I have more room to work with. Our job is to determine the area of the shaded parallelogram. So first, let's label the sides. The smaller square has a diameter of 1.5 centimeters. It has sides of length 4. The larger square has sides of length 5. And the parallelogram shares two of those sides. So we have a picture like this. Now we can draw in a diagonal of the parallelogram and observe that we have created two congruent triangles. However, we don't know the measure of these angles. And so we cannot immediately compute the area of the two congruent triangles. What we do know is that the one triangle with area equal to 8 centimeters squared, this triangle right here, is, in fact, congruent to one of the two triangles that make up our parallelogram. So I will suggestively fill that in using the same color and now justify why they are congruent. So notice that this angle right here is the same as this large angle over here. And over the white region, we have a right angle. Therefore, the red angle over here is the same as the angle inside of the triangle with 8 centimeters squared of area. So now by the congruency side, angle, side, we have this triangle being congruent to that triangle being congruent to the other half of the parallelogram. And as we know, the area of one of these, we conclude that the area of the parallelogram is equal to 2 times 8 centimeters squared, or 16 centimeters squared. And the answer is B. Question number 20. Anne wrote the following equation. 2012 equal to m to the power m times quantity m to the power k minus k for some positive integer values of m and k. What is the value of k? And such a question about factorization, it's always helpful to know the prime factorization of 2012, which is 2 to the power 2 times 503. And 503 is a prime number. So comparing that to the form in Anne's equation, we have the factor m to the power m matching 2 to the power 2. So therefore, we will use 2 is equal to m in our calculation. And now we must solve the following equation. 503 is equal to 2 to the power k minus k for k in some set of integers. So we recall that 2 to the power 10 is 1,024. 2 to the power 9 is 512, which is the closest power of 2 in the neighborhood of 512. And so we should probably work with k is equal to 9. And checking that, 2 to the power k minus k is 512 minus 9, which is indeed 503. Our answer to number 20 is k is equal to 9, or d.

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