Question number 11. What does the expression given here simplify to? And what we have is a square root with four sets of parentheses inside, so four terms, and looking closely at each of those terms, we see that in the first set of parentheses, 2015 plus 2015, well, we'll just write that as two times 2015 for now. Then in the second set of parentheses, I'm subtracting 2015 minus 2015, so that is equal to zero. 2015 times itself, I will write as 2015 to the second power, and then 2015 divided by itself, well, that is another name for one. So now let me rewrite what I have over here. I have that my expression is going to be equal to two times 2015 plus 2015 squared plus one, and all of that is inside of a square root, like so. Okay, so now let's stare at this for a little bit, and it kind of reminds me of a formula for the square of a binomial. So the sum of two terms squared comes out to be the square of the first term plus twice their product plus the square of the second term, and here it looks like one of the square terms is 2015, so let's make A 2015, and let's make the other square term one, like so, and so we have it exactly in the form that we need. Okay, so inside of the square root, I'll just rearrange my terms to make this absolutely clear, 2015 is going to be first, and that is squared, that is A squared, and I have twice the product of 2015, and I'll write in that one just to be absolutely clear, and then I have one squared, which is still one, all of that inside of the square root. Okay, and now let's rewrite the inside of the square root using our square of the binomial as quantity 2015 plus one, with A equal to 2015, B being equal to one, squared, inside of the radical, like so. So what I have is the square root of 2016, and that is B squared. Now, from algebra, we recall that the square root of a square is really another name for its absolute value, and since we're talking about a positive number, we can just remove that absolute value, so that is 2016. Okay, so that is our answer. This expression simplifies to, after all this work, 2016, and that is choice C. Question number 12. Ella wants to write a number in each circle in the picture in such a way that each number is the sum of its two neighbors. Which number must Ella write in the circle with the question mark? So here is that diagram. We have five numbers here that can be written in, and then we can decide on the number that replaces the question mark. So let me make a larger version here of this diagram, and let's start filling in numbers. We have some rules to follow. We know that each number is the sum of its two neighbors, and so right away with that in mind, we can add 3 to 5 and write 8 in here on the top spot. 8 is the sum of its two neighbors, and continuing on, we now know that 5 would be the sum of 8 and its other neighbor, which has to be minus 3. We need to add minus 3 to 8 to obtain 5, and similarly 3 would be the sum of its neighbors, 8, and we have no choice but to add minus 5 to 8 in order to obtain 3. Then moving on, minus 3 would be the sum of 5 and negative 8 on the left-hand side, and the remaining dot, we fill in the number that gives us negative 5 as a sum of 3 and then also negative 8. So negative 8 plus 3 is minus 5, and we have that here and we have that here in the bottom two circles. That allows us to fill in the question mark with a minus 16. Minus 8 plus minus 8 gives us negative 16, and that looks like answer B. However, looking carefully through our choices, we have answer E here claiming this situation is not possible, so before deciding on the correct answer, we must account for that eventuality. Let's double check our work. We have negative 8 here on the right-hand side, should be the sum of its two neighbors, so minus 16 plus minus 3 comes out to be minus 19, which is not negative 8. That is incorrect. Just to be absolutely sure, let's look here on the left-hand side, and we have negative 8 is equal to, according to our rules, negative 5 plus minus 16, and that is equal to negative 21, which is also not correct, so we don't have here true statements if we follow our rules. So Ella's prescription for filling in these empty spots in her diagram leads her to a value in the question mark that is inconsistent with her rules. Therefore, we conclude that minus 16 is not the answer, and this situation is not possible, and that would be answer E. Question number 13. Given five different positive integers, A, B, C, D, and E, we know that C divided by E is equal to B, the sum of A plus B is equal to D, and E minus D is equal to A. Which of the numbers A, B, C, D, or E is the largest? Before I begin analyzing our equations here, I'll underline the important piece of information. We know that we have different positive integers, so the fact that they're all distinct and the fact that they're all positive will become very useful in our problem. So let's begin by looking at the last equation here, E minus D is equal to A. E minus D is equal to A, and here I'll use that A is positive. I'll just need that it's positive. I don't really need to say that it's an integer at this point. This implies that E has to be in fact larger than D. We're taking away a positive integer, subtracting it from E, which is also a positive integer, and the result is positive, so E has to be greater than D. So let's keep going. Let's look at the middle equation here, where the sum of two positive positive integers gives us a third integer distinct from the first two, and here both A is positive as before, and B is positive, so we know that the sum must in that case be greater than either of the terms in the sum, so D is greater than A and B is greater than B. So now what we have we can combine. We know that here D is greater than A and E is greater than D, so the implication here is by the transitive property of inequality, and we can conclude that E is in fact greater than A. Similarly, since D is greater than B and E is greater than D, let me write that down. D is greater than B and E is greater than D. We have by the transitive property that E is greater than B, so it looks like E is the largest so far, but we haven't finished analyzing all the equations. Let's look at the last one here. Finally, when we look at C divided by E, and the result is B, a positive integer, means that means that E is a factor of C. Okay, so it is a smaller positive integer than C, so C is greater than E, and C is in fact greater than B. Okay, so now use the transitive property again as before to conclude that the largest number here that we are looking at in our list of five is in fact C. It is greater than E, and we spent some time here before analyzing the last two equations to conclude that E is greater than A and also greater than B, but C is still larger, so the answer here to our problem is C, and that also happens to be choice C. Question number 14. The x-axis and the graphs of the functions f of x is equal to 2 minus x squared, and g of x is equal to x squared minus 1. Split the Cartesian plane into how many regions? So, we have a graphing question here, and we need to analyze a little bit what type of graphs can we expect. Here we have two degree two polynomials, so both f and g are quadratic are quadratic polynomials, so their graphs of both of them are parabolas. Okay, and a parabola has a vertex, and it either opens up or opens down. The basic graph we're looking at here is x squared, and we can make some more comments about either function. So, f of x here is the graph of the function x squared, then what we need to do to it is is reflect it about the x-axis, and then shift it two units up. Okay, so the minus sign here indicates that the graph is rotated about the x-axis, or reflected about the x-axis, and then we have a shift two units up. Then the graph of g of x is just x squared, and then we're going to shift this one unit down. Okay, finally, to be able to graph this, we need to know where the vertex lies, on which axis of symmetry, and we can say that both functions have a graph symmetric about the Okay so what we have here is a graph where I'll need the x-axis. Let me just sketch it like that so that's going to be my x-axis. And then I have a y-axis here that I want to draw. I have a graph in red that's going to be a reflection of x-squared shifted two units up so I'll sketch roughly something like this. And then I have a graph in blue that's x-squared shifted one unit down. And it might look something like this. And you can probably refer to the suggested solutions for a more detailed graph here. Now it looks like I have intersection points that are above the x-axis so we should verify that. We should solve for these intersection points and be sure that they occur here above the x-axis. That would be important for counting our regions. So that is when x is equal to which value? What we have to do is solve f of x is equal to g of x and when we do that we obtain 2 minus x squared is equal to x squared minus 1. That gives us the equation 3 is equal to 2x squared or 3 halves is equal to x squared. The square root of 3 halves is still a positive number and it is going to be greater than 1 so therefore the y value here is greater than 0. Okay we can simply make a replacement here x squared is equal to 3 halves so x squared 3 halves minus 1 is 1 half so the y value here has to be 1 half like that. Let me denote that here maybe in red. So that is when y is equal to 1 half. And then we're ready to count our regions. So we're going to have here region number 1 is below the blue graph above the red graph and above the x-axis. Region 2 below the x-axis. Region 3 over here. Then let's say this is 4, 5, 6 over here, 7, 8 and then I have region 9 here below the blue graph above the red above the x-axis and then finally below that is region 10. So these graphs and the x-axis together they split Cartesian plane into 10 different regions and it is very important to actually figure out that they do cross, they do intersect above the x-axis otherwise we would have a different number of regions. So the answer is 10 regions and that is choice D here in question number 14. Question number 15. The geometric mean of a set of n positive numbers is defined as the nth root of the product of those numbers. The geometric mean of a set of three numbers is 3 and the geometric mean of a set of three different numbers is 12. What is the geometric mean of the combined set of six numbers? Okay so let's follow the definition here of geometric mean and write down the two equations that are described to us in the problem. So we have three numbers with a geometric mean equal to 3. What we need to do is give these numbers first names. So let A, B and C be the three numbers that have the geometric mean 3. Okay and now we can write this down in terms of equations. So 3 is equal to how do we compute the geometric mean? We have to take the product of these numbers so A, B, C multiplied together and we need the nth root and with three numbers that would be the cube root. So like so that would be the geometric mean of A, B and C. Okay now we'll let D, E and F have a geometric mean of 12 and that is written as 12 is equal to the cube root of the product of D, E and F like so. Again three numbers for the geometric mean requires cube root. Okay and the question here is asking us for the geometric mean of the combined set of six numbers and what we need to compute then is the following. I'm going to multiply these numbers all together so A, B, C that product times D times E times F that is in the radical and I need a sixth root now over here. So somehow I have to introduce a cube root into the equation and what I'm going to do here is recall that recall that if I have the square root here of X that's equal to X to the power 1 half I have the cube root of X and that's X to the 1 third and the square root of the cube root is something like this in radical notation and that is X to the 1 6th or here I have the 6th root of X and I'll use that property here of radicals and of exponents to break up my geometric mean of six numbers into something that's manageable something we have computed already. So first I'll write the 6th root here as the square root and of the cube root of my product like so and then I'll break up the product here of the cube roots into the product of two cube roots first of all of A, B, C and then D, E, F. Okay so that's some manipulation here with radicals. These numbers inside the square root I have already computed the geometric mean of A, B, and C that was 3 the geometric mean here of D, E, and F that was 12 so what I'm looking at here is the cube root excuse me I'm looking at the square root 36 so the square root of 3 times 12 36 and that comes out to be 6 so the answer here to problem number 15 is B that would be 6. Question number 16. In the figure to the right there are three concentric circles and two perpendicular diameters. If the three shaded figures have equal areas and the radius of the smallest circle is one what is the product of the three radii? So here's our figure let me move it and enlarge it we see the three concentric circles and we also notice that here we have the diameters they are perpendicular this is important for the following reason we can emphasize here in red the radius of the smallest circle that is known to be 1 so that length over here is 1 and we know that the shaded region here which I will emphasize in red is equal in area to the other two shaded regions and knowing that the diameters are perpendicular allows us to conclude that with respect to the inner smallest circle in the region I have shaded here in red is exactly a quarter of that circle because the diameters of that smaller circle are perpendicular they divide the circle into four equal regions and that allows us to compute the area here so the area of this region in red is equal to 1 fourth the area of the circle with radius 1 which is just pi times the radius squared so pi times 1 squared or that would be simply pi over 4 and that is the area of that shaded region and it is also the area of the other two shaded regions we know that the three shaded figures have equal areas we have figured out it is pi over 4 now we need the other radii so let me emphasize here the next radius we're going to compute in blue I will call this radius 2 r2 like that and that radius is the radius of a quarter circle here that I will emphasize shade in blue like so now that order circle consists of a shaded region with known area pi over 4 and the unshaded region here in white is exactly equal to the area of the neighboring here shaded region so together we have two regions with areas of pi over 4 therefore the area here is twice that or pi over 2 so let's write that down we have in blue what I myself shaded as area of pi over 2 and that's equal to 1 4th the area of the circle with radius that I called r2 so r2 squared like so and we can now solve for r2 divide both sides by pi multiply by 4 so we have 2 is equal to r2 squared and the square root of 2 would be equal to that second radius and we can repeat the same procedure here for the third and remaining radius which I'll emphasize here in orange that area there that I'll shade in orange like so consists of the shaded region in gray and the two regions of known areas pi over 4 each so the area of the region here in orange that would be 3 pi over 4 three of those and that's equal to 1 4th the area of a quarter circle here with radius 3 r3 and so that is that's this radius over here that would be r3 okay and then we can solve this equation for this radius multiply both sides by 4 divide both sides by pi so we end up looking 3 is equal to r3 squared and the square root of 3 would be the third radius okay and now we have the lengths of 3 radii and we need the product of them what is that product and that would be the product is now 1 multiplied together with r2 which was the square root of 2 multiplied together with r3 which square root of 3 that gives us the square root of 6 and in our answers that is answer a so the product of the 3 radii here is the square root of 6 or 8 question number 17 an automobile dealer bought two cars he sold the first one for 40% more than he paid for it and the second one for 60% more than he paid for it the money he received for the two cars was 50% more than what he paid for both the ratio of the prices the dealer paid for the first and the second car is which of the following I'll begin by assigning variables to the prices dealer paid for his automobiles so let's say that the dealer paid X dollars for the first car and Y for the second car now with that what we have to do in the problem is find the ratio of the prices that the dealer paid for the first and the second car so what we have to do is compute the ratio of your X to Y or X over Y okay and with that task in mind we'll need to write down some equations what we know is that the first car was sold for 40% more the second car was sold for 60% more and the money received for the two cars was 50% more than what was paid for them so we're talking about profit here and that allows us to write down some equations so for profit what we're going to have here are three equations first for the first car we have a profit of 40 so maybe we can write this down as follows um the original price was multiplied by 1.4 and then the car was sold so we have a 0.4 or 40 percent uh profit of the original amount paid then for the second car it's 60 so 1.6 times y minus y is 0.6 y and then we have the total profit so it was 54 percent more than what he paid for both what he paid for both was x plus y that is multiplied by 1.54 then the price of both is subtracted and we have a profit of 54 percent okay so the individual profits and the total profit when we add them up they should equal so that's what we do next now equate here the profits and what we get is the following 0.54 times x plus y is equal to that's the total profit that should be the total of the individual profits so 0.4 x plus 0.6 y and we can solve this equation for the ratio of x to y so we'll subtract 0.4 from both sides and we'll obtain here 0.54 x minus 0.4 x on the left hand side here on the right hand side i'll subtract 0.54 y from both sides of 0.6 y minus 0.54 y and that gives me 0.14 x is equal to 0.06 y and i need the ratio of x to y so i'll divide both sides by y and also both sides by 0.14 so x over y is 0.06 divided by 0.14 and we can simplify this to 3 over 7 in the end and that is the ratio we were looking for and we answer the question by saying well the ratio of x to y the price paid for the first and the second car is 3 to 7 3 over 7 and another way of expressing that is with a with the colon so that the answer c over here 3 to 7 question number 18 the following is the multiplication table of the numbers 1 to 10 what is the sum of all 100 products in the complete table there are several different ways of summing up these 100 integers here that appear in the table and i'll just focus on one method and that is to use the distributive law here to set up a multiplication problem whose end product consists of the sum here of exactly these 100 numbers so we will do that as follows i'll take product here of the number 55 but written as the sum of the first consecutive 10 integers itself and this is 55 times 55 so 55 squared let's say that is the correct answer the following reason if i take the number 1 over here and distribute it to this entire second factor well then i have 1 plus 2 plus 3 plus 4 all the way up to 10 and that is exactly the first row or the first column in my table if i continue and take the number 2 and distribute it here to the entire second factor then likewise now i have 2 times all the terms over here so that's 2 and 4 and 6 and so forth all the way up to 20 and that's exactly the second row or the second column in my table and just to continue the pattern to the end if i take 3 and multiply here by everything in the second factor likewise i'm ending up looking at 3 plus 6 plus 9 and so forth that would be the next visible column here so that would be these numbers here like that or the next row which is which is not written in this table and then finally if i take 10 times everything then i obtain the last row here on the table so that distributive law applied to this product does give me all the terms in my sum we can add up the first 10 positive integers they add up to 55 so 55 squared is the answer here that however doesn't appear exactly in that form in our choices so we have to simply multiply 55 by 55 and that comes out to 3025 after doing that multiplication and that is answer d over here 3025 question number 19 a two-digit number with digits x and y can be written in this notation here we juxtapose x with y to create a two-digit number and then draw a line over both let a b and c be different digits in how many ways can we choose the digits a b and c such that this inequality here is preserved and that inequality is the two-digit number in our notation a b is less than b c and that is less than c a so these digits are distinct that is important a b and c are different digits for our inequality to be preserved what we need to choose are digits a b and c that are in the same order as the inequality so here in other words we need a to be less than b and to be less than c well that should make sense because the smallest number a b should have the smallest tense digit and the largest number c a here should have the largest tense digit in the middle we have b c and b would be smaller than c greater than a so we need our tense digits to fall in order like that where the elements that we're choosing here they belong to the set of the first nine positive integers and of course we don't use zero for either of our three integers because we cannot have zero as the tense digit of a two-digit number okay so again we go back to the fact that our digits here are distinct they're different and we note that any three numbers and they have to be distinct they can be ordered as a is less than b is less than c as long as i have three different numbers three different digits like that i can order them in such a way without any problem so for example if a is just one b is two and c is three what we have is a b would be 12 that would be less than b c which comes out to be 23 and that has to be less than c a which then comes out to be 31 and we have here a is less than b is less than c as required okay so now we have to count the number of ways we can choose a b and c and that is the number of ways we choose three elements from a set of nine distinct elements without repetition so we need a combination here for this calculation and we say that there there are nine choose three ways to select our three different integers or three different digits from our set of nine so nine choose three that is by definition nine factorial divided by three factorial divided by six factorial so that's nine times eight times seven times six factorial the six factorial cancel and then we have one times two times three here so that's nine times eight seven divided by three divided by two that gives us three times four times seven which is 12 times seven or another name for that is 84 that is choice a in our answers 84 that is the number of ways we can select our three digits a b and c satisfy the conditions of the problem question number 20 for a standard die the sum of the numbers on opposite faces is seven the figure shows two identical standard dice what number may be on the not visible face on the right here marked by the question mark i'll make an enlarged version of our diagram and then we will label the faces on the right die over here with as many values as we can so let's look at the face that's opposite the one marked with six dots to add up to seven that face has to contain exactly one dot and opposite the face with four we need to have three dots so where the question mark is we have a choice of two or a five because those are the numbers we have not yet used now the key information in the problem is that the figure we're looking at shows two identical standard dot so on the left die somewhere on it there is a face with six dots it has to be in fact here opposite of one and if we could rotate that up and place it in the same way the right die is positioned we can probably figure out where the face with two or five dots is in relation to the question mark so let's rotate our die here in such a way that the face with six dots sits on top so let me make a copy here of our die and i'll erase on the left side here the the dots because we will be in fact rotating a little bit so after our rotation here what we will have is a situation like this where on the front face now we have two dots like so on the left face we'll have three and after a rotation they'll look like this and on the top face we'll have six dots and they will in fact look like this because what i need is one more rotation here to obtain the same arrangement as the die on the right now how do i know this well we have opposite opposite the face with three dots here we know that the obstructed face hidden from view on the left die that i'll mark here with uh with the line so that face over there that one's opposite three it must have exactly four dots on it and we need a rotation uh around which axis well we need a rotation around the center here like that and looking top down it has to be clockwise so like so in order to position that face with four dots out front where the four dots are already sitting on the on the front on the right die and when we do that uh the six dots on the top will fall in line with the position on the right the face with four dots will come out front the face with three dots will go in the back and here where the question mark is what we will see is whatever is the face opposite the the front face that we see here with two dots well that's the face with five dots isn't it so five is going to be right here on the on the back face opposite the face with two dots and after our rotation it ends up in this position where the question mark was so that is how we figure out that not two or five not just two but exactly five dots belong on that face with the question mark so the answer here is a only five may be on that face

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