Question number 11. The size of a rectangular box is A by B by C with A being less than B and C being greater than B. If you increase A or B or C by a given positive number the volume of the box also increases. In which of the following cases is the increase of the volume of the box the greatest? So here to illustrate the situation I have drawn a box and I have chosen the dimensions so that the edge here A that's the smallest dimension next is B and C clearly from the picture is the largest dimension. Now if we increase each of these dimensions one at a time we notice that we are increasing the volume of the box in terms of multiples of areas of the faces. So let me make that more precise. The face here with dimensions B by C this this face over here has area B times C and the face over here that face has area we have to multiply A by B so that's area A B and the top face over here that face has area C by A or A C. Now B and C are the greatest numbers so the product BC is the greatest and we see that that corresponds to this front face having the greatest area out of all the other two possibilities. So if we increase the dimension A what we would be doing is imagine we are slicing this box and on top of it we're adding more faces with area BC. So it makes sense that the greatest change in volume will occur if we increase dimension A. So let me just make a note of that increasing increasing A causes the volume to increase in increments of B times C which is greater than either of the other faces A C or A B. Hence we should choose answer A as the answer here to the problem. Question number 12. In a soccer match the winner gets three points the loser gets zero points and in the case of a tie each team gets one point. Four teams A B C and D take part in a soccer tournament. Each team plays three games one against each of the other teams. At the end of the tournament team A has seven points and teams B and C have four points each. How many points does team D have? So here in a rubric like this what we would know is that the sum of the entries in row A is equal to seven and the only way that can happen is we have three plus three plus one. So two wins and one tie. Similarly for row B we have a total of four and that can only happen with a three and a zero and a one. So a win a tie and a loss and C has the same track record. D we don't know but in D we will have a number here A representing D's score against A, a number B and a number C for the results of the matches between D and the teams B and C. And so far we know that the maximum number of points here in the tournament is three points per game times six games so that would be 18 points. And we know that team A is responsible for seven points, team B for four, team C for four and then we don't know what team D did so we have plus A plus B plus C and this is at most equal to 18. So A plus B plus C is at most 18 minus 15 or 3. And so the partitioning of 3 is then given as follows. So this tells us that then 3 is equal to a 1, a 1 and a 1 or is equal to a 3 plus a 0 plus a 0. Now we know immediately that the last partitioning cannot happen. This cannot happen because then we would not have any ties in in the tournament and we do have some ties. So what about 1 plus 1 plus 1? If that was the case then we wouldn't have the correct number of points. The total number of points would be then equal to the 16 that we already have maximum plus 3 that would be 19 points. So this cannot happen either. So the next possibility is that so A plus B plus C is not equal to 3. It also cannot be equal to 2. It has to be equal to 1. And 2 is not possible so it must be that A plus B plus C is equal to 1. And so we have initially chosen the correct answer but we are now sure of it. So D earns 1 point. And our answer here B is correct. Question number 13. The radii of two concentric circles which are shown in the diagram on the right and which I have enlarged over here are in proportion 1 to 3. AC is the diameter of the big circle and BC is a chord of the big circle which is tangent to the smaller circle here. The length of AB is 12. So what is the radius of the big circle? We know immediately a few pieces of information that should be marked here on the circle. Let's draw in the line segment from A to B. That is equal to 12. And what we are looking at here is a triangle that is inscribed in a circle and its diameter is exactly the hypotenuse. So by an old result from geometry we know that this is in fact a right triangle here. That's a 90 degree angle. Now we can draw in the radius of the smaller circle. Let's call it lowercase r. So this would be the radius of the smaller circle and the radius is perpendicular to anything that's tangent to the circle which we know is the chord BC. So the angle right here is also a 90 degree angle and we are looking at similar triangles. So starting from the information given in the problem we know that capital R is to little r as given in the problem 3 is to 1. And now looking at our triangles capital R here in red is along the segment AC which is the diameter and that is in proportion to AB in exactly the following way. So that is as AB is to excuse me that is as AC. AC is to AB and AC is exactly equal to 2 times the radius of the big circle and AB is 12. So now we have that 3 is to 1 as 2r is to 12. We can cross multiply here and we see that 36 is equal to 2r or dividing by 2 the length of the radius of the big circle is exactly 18. And so that solves our question. The answer is B. Question number 14. How many triples ABC of integers with A greater than B, B greater than C, and C greater than 1 satisfy the inequality 1 over A plus 1 over B plus 1 over C is greater than 1. We can begin by looking at the minimal example and that is when ABC is equal to C can only be as small as 2 so we're going to have 4, 3, and 2. Okay then 1 over A plus 1 over B plus 1 over C is a half plus a third plus a fourth and that gives us exactly 13 over 12 which is bigger than 1. So this example here works. We have found such a triple. Now if C is at least 3 what happens? Then also 1 is going to be less than 3 here and then we have let's say that B is going to be 4 and A is going to be 5. So with C being at least 3 we have B is at least 4 and C is at least 5 and in that case 1 over A plus 1 over B plus 1 over C is going to be less than 1 third plus 1 fourth plus 1 fifth. That is equal to 47 over 60 which is less than 1. So this this possibility does not work. So C cannot be greater than or equal to 3. So we have that C is actually going to be equal to 2. Now if C is equal to 2 then we have to consider what happens if B is equal to 3. We did that already. So what happens if we are in the next case and B is at least 4. So if C is equal to 2 and like before B is at least 4 and C is at least 5 then what we compute is 1 over A plus 1 over B plus 1 over C is less than or equal to now we have a half plus a fourth plus a fifth and that gives us exactly 19 over 20 which is less than 1. So that's also bad. So C has to be equal to 2 and B has to be equal to 3. In that case we have only one variable remaining and we can finally say that thus 1 over A plus 1 over B plus 1 over C is going to be equal to 1 over A we don't know but it's going to then be a half and a third so that's 1 over A plus 5 6 and here A is greater than B which is equal to 3 and 1 over A plus 5 6 is going to be then greater than 1 if and only if provided that 1 over A if we subtract is greater than 1 sixth. So A could be equal to 4 or A could be equal to 5. A cannot be equal to 6 or greater. We have already discovered that A here can be equal to 4 but now we also know that A can be in fact equal to 5 so we have another choice here 5 and then B would be 3 and C would be 2 that's that's the second choice so we have in fact after all this found another candidate and there are two possibilities so the answer is C. Number 15. A, B and C are nonzero numbers and N is a positive integer. It is known that the numbers here given have the same sign which of the following is definitely true? The only observation we can make so far is that if we take two numbers of the same sign either both positive or both negative then their product must be positive. So we know that we know their product is positive. So let's start with that. So multiplying what we would have is the following factors so negative 2 appears on the left hand side so that's negative 2 to the power 2 and plus 3 and then we multiply that with a negative 3 to the power 2 and plus 2. Then we collect the exponents on A. On the left hand side or on the first factor we would have a 2n plus 2 and then from the other factor that's a 4n plus 1. The exponents on B are the sum of 2n minus 1 and 2n plus 5 and the exponent on C is 3n plus 2 added together to 3n minus 4. So now we have here we can factor out a negative 1 and that's negative 1 to the 2n plus 3. We have another negative 1 to the power 2n plus 2. Then what remains is a 2 to the power 2n plus 3 and a 3 to the power 2n plus 2. And on A we have the total exponent 6n plus 3. On B we have the total exponent 4n minus 4 and on C we have the total exponent 6n minus 2. And then let's take one more step over here. We can combine the powers on exponents of negative 1 and we see that it's negative 1 to the power 4n plus 5. Okay now let's look at the exponents. We know that the overall sign here is positive. It's a positive number and we see here that 6 times an integer minus 2 is even. So overall this is positive. Whatever C is, C raised to an even power is going to be positive. Now here we have another even number 4n minus 4 and so this factor here B raised to some power that's even is going to be positive. Now we run into some ambiguity here. 6n plus 3 that is an odd number. So we don't know what happens to A. If A is positive then this is a positive number. If A is negative then this is a negative number. These are just positive numbers multiplied together so this is this is positive. And then here we have a negative 1 to an odd exponent and so that is in fact equal to negative 1. Now if the total product has positive sign then the number here A raised to the 6n plus third power has to have the same sign or it has to be negative in order for the negatives to cancel. So we conclude that A to the power 6n plus 3 has to be negative so that the product with negative 1 will cancel the minus signs and the overall answer is positive. And so this can only happen if A is negative. If A is positive then raising it to any power will preserve the positive sign but if A is negative then raising it to an odd power will also have a negative sign at the end. So that is our answer here. A has to be a negative number. Question number 16. Six weeks is n factorial seconds. What is the value of n? Let's just begin by figuring out the number of seconds in six weeks. So we have six weeks and a week is seven days and a day is 24 hours and then an hour is 60 minutes and a minute is 60 seconds. So that's the number we need. And now that's going to be equal to n factorial. And we have to figure out the value of n. So let's get a prime factorization here, and then figure out the value of n after that. So 6 is 2 times 3, then 7 is prime. 24 would be 6 times 4. So 2 times 3 times 2 squared. That's 6 times 4. And then I have 60 twice. So 60 would be, for example, 15 times 4. So 2 to the second power times 3 times 5. That's 15 times 4. And everything is squared. So we have a 1 over here. Let's just multiply by 1, even though 1 is not prime. That's part of the factorial. And then times, let's see how many powers of 2 do I have. I have 1, 2, plus 2, and then plus 4 from the parentheses. I have 3 to the power 2 from outside the parentheses, plus 2 more from inside the parentheses. No 4s. Times 5, that appears twice. And then times 7. So that's my prime factorization. And then I will try to get a factor of each positive integer here. So I have a 1, I have a 2, I have a 3. And then what remains is the exponent on 2 has to be reduced by 1. So now I have an 8. So that's 7. The exponent on 3 is reduced by 1. So that's a 3. 5 squared times 7. So let's keep going. 1 times 2 times 3 times 4 times 5. And then the exponent on 2 has to be reduced by 2. 3 is so far untouched. 5 appears one factor less, and then 7. Now I have 1 times 2 times 3 times 4 times 5. Now I have a 6, so 2 times 3, and a 7. And so the exponent on 2 and 3 has to be reduced by 1, 2 to the fourth times 3 squared. 5 still remains. And so now we have here 7 factorial. And I would like to have an 8 inside the parentheses. So that would be 2 to the third power. That's an 8. And 9 would be 3 squared. And then I still have a 2 times 5, which is a 10. So I have now used up all my numbers. And I have 7 factorial times 8 times 9 times 10. And that's another name for 10 factorial. So after all this, we have figured out that exactly 10 factorial seconds is equal to 6 weeks. The answer is D. Question number 17. The vertices of a cube are numbered 1 to 8 in such a way that the result of adding the four numbers on the vertices of a face is always the same for all faces. Numbers 1, 4, and 6 are already set on some vertices, as shown here to the right. What is the value of x, where x is the value assigned to the corner here on the right face we can see? So first of all, we should calculate the value for the vertices on each particular face. We have exactly six faces. And they come in three pairs of opposite faces. And that is important to note, because on each pair of opposite faces, we have, in fact, all eight numbers. So each pair of opposite faces sums 2. 1 plus 2 plus 3 plus 4. We add up all the numbers we see. And that is 36. So then on each face, the vertices have to sum 2. We take 36 times the three pairs that we have, and then divide by 6. And this gives us 3 times 6, or 18. So on each face, the vertices add up to 18. And now we can calculate the missing value here below x. Let's call that value y. So we know that 1 plus 6 plus 4 plus y is 18. And this gives us that y is 7. And now we are missing a value over here in the front. Let's call that vertex z. So now we know that x plus z plus 4 plus y is equal to 18 also, but we know the value of y. So x plus z is 18 minus 4. y is 7 minus 7. So that comes out to 7. And in how many different ways can x plus z become 7? So let's say that x is equal to 1 and z is equal to 6. Well, that can't be, because both 1 and 6 already have happened. So this is not possible. Then we go to x is equal to 2 and z is equal to 5. That is a possibility. Then we go to x is equal to 3 and z is equal to 4. That can't happen, because 4 is already in use. And if we then increase x to 4, z has to be 3. We've done that already. If we increase x to 5, then we just keep running back through the possibilities. So in fact, the only possibility here is that x is equal to 2 and z is equal to 5. So we have found our answer. The value of x in the missing vertex here indicated in the diagram is 2, and the answer is A. Question number 20. The label on a package of cream cheese reads 24% total fat. The same label also reads 64% fat and dry matter. What is the percentage of water in this cream cheese? So the total contents of the package consists of dry matter and water. So let me just abbreviate and say D plus W. Now we know that 24% of the total is fat. So fat is equal to 0.24 of the total, so that's D plus W. And we also know that 64% of the dry matter is fat. So that's also 0.64 times the dry matter D, like that. What we have to find is the percentage of water and the total. So we need W divided by the total, which is D plus W. So let's find the relationship between D and W, the dry matter and the water, and the percentage of 1 in the total package. We do that by solving the equation here that tells us the amount of fat for D in terms of W. So multiplying 0.24 D plus 0.24 W is equal to 0.64 D. And then we can subtract 0.24 D from both sides. That leaves us 0.24 W is equal to 0.4 D. And we would like to solve for D. So we can multiply by 10 on both sides. We have 2.4 W is equal to 4 D. And then divide by 4 on both sides. So 0.6 W is equal to D. And now we can simplify the expression here on the right, which is exactly what we have to calculate. So W divided by D is 0.6 W plus another W. That gives us, after canceling W's, 1 divided by 1.6. That is 1 divided by 16 over 10. So 10 over 16, or 5 eighths, is the number we are looking for. Now, the answers are in percentages. So we recall that 1 eighth is a half of a quarter. So that's 0.125. And so 5 eighths would be 5 times 0.125. And that comes out to 0.625 after multiplying. And we have our percentage. This is the same as 62 and 1 half percent. So the answer would be B. Question number 19. The line L passes through vertex A of the rectangle A, B, C, and D. And we know that the distance from C to L is 2. The distance from D to L is 6. The distance from A to D is twice the distance from A to B. So in my larger diagram here, in a second, I will label these quantities. So let's go ahead and do that. This dimension from L to C is 2. This dimension from L to D is 6. And these are always given along the shortest distance, so along perpendicular lines. And we have two right angles here, consequently two similar right triangles. So let's label the dimensions here. I will call the edge AB here Z. So CD has length Z. I will call the small hypotenuse here of the small right triangle X. And the edge here is going to be Y. Now, I will project X down onto AD by drawing a line through the vertex C towards the segment AD, but parallel to the line L, like that. So then this allows me to say that that's distance X here also, and Y is then here as well. Now, by similar triangles, we have that the hypotenuse, X plus Y, of the large right triangle is to the hypotenuse of the small right triangle, which is X, as 6 is to 2. So that's 3. And we then see that dividing here by X, 1 plus Y over X is equal to 3, or consequently, Y is equal to 2X. So we can just say this is 2X, and we can just say this is 2X. Now, we also have from the problem the information that 2Z is equal to exactly X plus Y. And X plus Y is 3X. So we have a relationship between Z and X also. And the last relationship that I will exploit here is the area. We know that the area is equal to Z times 3X. So let's say first Z times X plus Y, but X plus Y is 3X. So the area is 3XZ. And we can write the area in another way. We have two triangles plus a parallelogram. So the two triangles I'm talking about are the ones here with dimensions Z by 2X. And the parallelogram is the one I created by copying the blue line. So the area of the two triangles combined would be 2X times Z. So let me write exactly 2XZ. And the area of the parallelogram is going to be its height, which is 2. And then I have to give the length of the blue line here a name. I will call it W. So the dimension W is exactly also the length here. And then we have just moved it over. So using the area relationship, we see that subtracting 2XZ from both sides, we see that XZ is equal to 2W. And W is the hypotenuse of one of those triangles. So W squared is equal to 2X quantity squared plus Z squared. So here in my original equation for area, I will square both sides. So X squared Z squared is going to be 4W squared. And so that's then equal to, by substituting, that is equal to 4 times 2X squared becomes 4X squared and Z squared. And I do have a relationship between Z and X, which is right over here. So using that relationship from up here, we see that Z is equal to 3 halves times X. So Z squared, now I know what that is. Let's replace Z squared by 9 4ths X everywhere. So on the left-hand side, I have X squared times 9 4ths X squared. And here I have a 16X squared plus 4 times 9 4ths X squared. And I can then cancel out a factor of X squared. So we have 9 4ths X squared is equal to 16 plus 9, which is 25. And then X squared is going to be equal to 100 divided by 9. And X squared is therefore 10. Excuse me, X is therefore 10 over 3. If X is 10 over 3, then Y is twice that. So Y is going to be 20 over 3. And so finally, at the very end, if we add these up, X plus Y is 30 over 3. And that gives us 10 as the length of the segment AD. So that was an involved problem. But finally, we have the answer. And that is A. Question number 20. The function f of X equal to AX plus B satisfies the equalities if we compose it with itself three times. And evaluate at 1, the value is 29. And if we compose it with itself three times and evaluate at 0, the value is 2. What is the value of A? So let's just compute these compositions, starting with f evaluated at 1. We see that that is going to be A plus B. So then, f evaluated at A plus B is going to be A times A plus B plus B, which is A squared plus AB plus B. And the last thing we have to do here is evaluate f at A squared plus AB plus B. And that would give us A times all of this plus B, which is A cubed plus A squared B plus AB plus B. Then again, repeating the same calculation, starting with f is equal to 0, we see that that is B. f evaluated at B is AB plus B. And f evaluated at AB plus B is going to be A times AB plus B plus B, which is A squared B plus AB plus B. And so what we have here is then that the value of f evaluated at f evaluated at f of 1, like this, is equal to A cubed plus the value of f evaluated at f evaluated at f, then evaluated at 0. After all, we see that this expression here is exactly our expression from above. And so this is then equal to A cubed plus 2. And the whole thing is equal to 29. So that tells us that A cubed is equal to 27. And so we conclude that A is, in fact, 3. So the answer here is C.

geometry

algebra

reasoning

volume maximization

soccer tournament

concentric circles

cream cheese water content

function composition