Problem number 11. Positive numbers a, b, c, d, and e have the following property. a, b equals 2, b, c equals 3, c, d equals 4, and d, e equals 5. What is the value of e over a? Now, given the information in the problem, our strategy should be to find a way to write d in terms of c, and then write c in terms of b, and write b in terms of a. So that way we can indirectly write d in terms of a. And here's how we would go about doing that. So going in order, here's how we would list it out. We have b is equal to 2 over a. All we're doing is taking a, b equals 2 and dividing both sides by a. Now we know that b is equal to 2 over a. We also know by similar logic that c is equal to 3 over b. But since b is equal to 2a, c is equal to 3 over 2a. And since we have a fraction in the denominator, we can take the reciprocal and multiply it. So that would be 3a over 2. And again, this is basically just dividing 3 by 2 over a, which is the same as multiplying 3 by a over 2. So remember that that's how you divide by a fraction. It's the same as multiplying by its reciprocal. So we have c equals 3a over 2. And now d is equal to 4 over c. And c is equal to 3a over 2. So it's like we're taking 4 and multiplying it by 2 over 3a, the reciprocal of 3a over 2. And from that we get 8 over 3a. And now we have e is equal to 5 over d. And 5 over d is equal to 5 divided by 8 over 3a, where a is in the denominator. And that's the same as multiplying by 3a over 8. So that will be 5 times 3a, which is 15a, all over 8. So that's what e is equal to. And now that we've rewritten e in terms of a, going back, the original fraction was e over a. So that's just taking this fraction, which is equal to e, and dividing it by a. And what that does is it just cancels out the a in the numerator. So we're left with 15 over 8. So our final answer is a. Problem number 12. An indiscreet neighbor asked Mrs. Novak how old she was. Mrs. Novak replied, if I live to be 100, then my age is 2 thirds of the time I have left. How old is Mrs. Novak? From this we can actually write down an equation. Let's go ahead and call x the age of Mrs. Novak, and that's what we're trying to find. We're trying to find x. So we have x is equal to 2 thirds of 100 minus x, where 100 minus x is the number of years Mrs. Novak has to live if she lives to be 100, because she's already x years of age, so she needs 100 minus x more years. And x is equal to 2 thirds of this number, as stated in the problem. So now we're just going to solve the equation. We multiply both sides by 3 halves. So we have 100 minus x equals 3 halves of x. We add an x to both sides, and that gives us 100 is equal to 5 halves of x. And we isolate x by multiplying both sides by 2 fifths. 100 times 2 fifths is equal to 40, so we have x equals 40. And therefore our final answer is b. Problem number 13. The rectangle in the picture is divided into 6 squares. The length of a side of the smallest square is 1. What is the length of a side of the largest square? So let's go ahead and call the side length of the largest square x. Now what does that tell us about the side lengths of all the other squares? Well, looking at this particular side here, the right side of the biggest square, we can tell that the left side of the second biggest square is equal to x minus 1, because it shares it with the side of length 1 from this little square. So this big square has side length x. This guy has side length x minus 1. And by similar reasoning, this square will have side length x minus 2, since this whole side is x minus 1. And so this side of the third largest square has to be x minus 1 minus 1 more because of this little square. So we have x, x minus 1, and x minus 2 for the side lengths. And then by the exact same reasoning, we have the side length of this little square being equal to the side length of this little square, which are x minus 3. So let's go ahead and label those. So again, the side lengths are x, x minus 1, x minus 2, x minus 3, x minus 3, and 1 for this little square. So what does this tell us? Well, we can actually write down an equation from this. We see that x plus 1 is actually equal to x minus 3 times 2 because we have these two sides being equal in length to this big side along with the side of the little square with side length 1. So with this equation, we distribute out the 2 to the x minus 3. We have x plus 1 equals 2x minus 6. We subtract x from both sides, and we add 6 to both sides, and that gives us x equals 7. And so that's the side length of the biggest square, and our final answer is D. Problem number 14, a box of chocolate costs $10. There is a coupon inside each of the boxes of chocolate. With 3 coupons, you can get an additional box of chocolate free. What is the greatest number of boxes of chocolate you can get for $150? This problem at first seems very, very easy, but it's actually very sneaky. It tries to punish anybody who tries getting through it too quickly. So in particular, let's start by counting how many boxes we can actually buy. Each one costs $10. We have $150. 150 divided by 10 is equal to 15. So we can buy 15 boxes, and now since it's 3 coupons gets us a free box, we will get 15 divided by 3, which is equal to 5 more free boxes. Now again, this is because we have 5 sets of 3 coupons being equal to 15. So we have 5 new free boxes that we received after paying for these 15. And now in this 5, we get 3 more coupons, which gives us another free box. So you would think that you have, okay, 15 plus 5 plus 1 more free box from the 3, but then we actually have 1 more extra piece of chocolate or extra box of chocolate because when we take out the 3 from this 5, we are left with 2 more chocolates on the side plus the extra one we get for free. So that actually adds up to another 3 chocolates, which are another 3 coupons, which give us 1 more chocolate. So we have the 15 that we buy, the 5 that we get from the coupons from the 15, and then we get 1 coupon from this, and then 1 more coupon from what remains after we get the first coupon. So we have 15 plus 5 plus 1 plus 1, which is equal to 22. So our final answer is E. Problem number 15. Mr. and Mrs. Dobson have several children. The average age of the Dobson family is 18 years. Without the 38-year-old father, the average age of the family decreases to only 14 years. How many children are there in the Dobson's family? So this is going to be an algebra-heavy problem. Now, it's actually not so bad once we name all of the variables. So let's start by letting X equal the total size of the family. So X is equal to the number of children plus 2 parents. And then we know the age of the father. We don't know the age of the mother, so we're going to call that N. And then let's go ahead and call C the total, the sum, of the ages of the children. So we can actually write down a couple of equations here. Now, 38 plus M plus C. This is the sum of the father's age, the mother's age, and all of the children's ages, and we are dividing it by the total number of family members. In other words, this represents the average age, which we are given is equal to 18. And if the father leaves, then we're subtracting 38 from the numerator, and we're subtracting 1 from X since there's one less family member we're considering. So it's M plus C over X minus 1, and we see that this is equal to 14. Now, what we can do here, which is quite nice, is we can rewrite both of these equations so that we have M plus C on one side of the equation and everything else on the other side. So, for example, here, for the top equation, we multiply both sides by X, and then we subtract 38 from both sides, and we get M plus C is equal to 18X minus 38. And here we just multiply both sides by X minus 1, and we get M plus C equals 14X minus 14. Now, the reason why we do this is we know M plus C is equal to M plus C, so what we've successfully done is related two different expressions with an X in it. So we know 18X minus 38 is equal to 14X minus 14. And using this equation, we can just solve for X. And that would look like this. We go ahead and add 38 to both sides of the equation, and we subtract 14X from both sides, and that leaves us with 4X is equal to 24. Divide both sides by 4, and we get X is equal to 6. And X is the total number of family members, so we need to subtract 2 to account for the parents, and that will give us the number of children, which is what the question is actually asking for. And X minus 2 is equal to 4, so our final answer is C. Problem number 16. Two identical equilateral triangles with perimeters of 18 centimeters are overlapping with their respective sides parallel. What is the perimeter of the shaded hexagon? First notice that the side length of any of the sides of the hexagon is one-third of the perimeter of whatever triangle, whatever smaller equilateral triangle it's attached to. So for instance, this side here is one-third of the perimeter of this triangle. This side here is one-third of the perimeter of this smaller triangle. So overall, the perimeter of this hexagon should be one-third of the combined perimeter of all of these triangles. The sum of the perimeters of the two original triangles. So we know that the perimeter of one of these triangles would be 18, and that means that the sum of both of the triangle's perimeters would be 36. So the perimeter of the hexagon should be one-third of that, since each side is just one-third of whatever triangle it's attached to. So we have one-third of 36, which is equal to 12, which means our final answer is B. Problem number 17. What is the maximum number of digits that a number can have if every pair of consecutive digits is a perfect square? So let's go ahead and first list all of the perfect squares which have two digits. And here they are, 16, 25, 36, 49, 64, and 81. Now, the goal here should be to form as long of a number as possible where every pair of digits that are right next to each other is a perfect square. So we really like using the perfect squares which have numbers in common. For instance, 16 and 64, or 16 and 81, since you can write down 8, 1, 6, and that's 81 and 16. Or 16 and 64, since you could do 1, 6, 4. And notice, by the way, that 25 doesn't really lend itself to this sort of thing since it doesn't have any numbers in common, any digits in common with the other numbers. So probably our number is not going to have a 25 in it anywhere. And 36 also only has one number, one of its digits in common with another one, with 64, as does 81. But really, the nice number here is 16 because it matches up with both 64 and with 81. And 64 is nice because it matches up with 49 and with 16. So we would really like to use 16 and 64, and the numbers that naturally go with it are 81 and 49. So this is sort of a hand-wavy approach. It really is a little bit of trial and error. You sort of have to convince yourself. But this number, 8, 1, 6, 4, 9, 81,649, this is the most digits that you can get. You won't be able to attach another perfect square anywhere. And you've already used the numbers that are the best for linking numbers together, like 16, 64, and so on. So this is going to be the biggest number you can have just by sort of trial and error. And thus we conclude that our final answer is A. Problem number 18. A square of area 125 square centimeters was divided into 5 parts of equal area, 4 squares and 1 L-shaped figure, as shown in the picture. Find the length of the shortest side of the L-shaped figure. So what we're looking for is the length of this short side here or this short side here. And for our strategy, it doesn't matter which one we pick. So let's go ahead and just pick the top one. We'll start by labeling this top side to the best of our ability. We have 5 parts all of equal area. So it's 125 divided by 5, and that's equal to 25. So each of these squares has an area of 25 square centimeters. Now, that means that the side length is 5 for each of these squares since the area is 5 times 5, which equals 25. And we don't know what this side length is yet, so we're just going to call it x. So what we can label so far are the side lengths 5, 5, and x. Now, what do we know about the big side length of this square? We know that it should be the square root of 125 since the area of the square is 125. So we can actually go ahead and make note of that. We're going to write down the square root of 125, and we're going to rewrite it. We're going to pull out a 25 from it, and the square root of 25 is 5. So this is just equal to 5 times the square root of 5. So that should be the length of the total side of this large square. We also know that the side length is equal to 10, 5 plus 5, plus x. And what that means is that x is equal to 5 square root 5, the total length, minus the length of these two sides, which is 10. So x equals 5 square root 5 minus 10. And we don't have that answer anywhere in that form, but we can pull out a 5 from both of these expressions, and that gives us 5 times the quantity square root of 5 minus 2. And so our final answer is going to be E. Now in order to find the answer to this problem, the key is to assume that we pull out the most inconvenient set of balls as possible. And by inconvenient, I mean we pull out as many balls as we can that we're still violating the problem, the condition given in the problem, which is 7 balls sharing a color. So we can start by pulling out 6 balls. They could be red and blue, or they could all be red and green. But in this case, let's assume that there's a mix of them. The logic is pretty much the same no matter how you divide it up. Now in this case, we have 6 red halves, and we could have either 6 blue halves, 6 green halves, or 3 blue and 3 green. Now we can't have any more red balls, because then we would have at least 7 since we already have 6. One more red ball and the problem is satisfied. And remember, we want to go as long as possible without satisfying the problem. So now we dip into the blue-green balls. And now we see if we would have all 6 of the balls being blue, then pulling out the blue-green balls wouldn't really work here because we would have a 7th blue ball. And the same thing holds if we would have all green. So that's why we have 3 blue balls and 3 green balls with the other half being red. Now we go to the blue-green ones. And now we have 6 halves of each color. So 6 red halves, 6 green halves, and 6 blue halves. We've reached a situation where regardless of what kind of ball we pull out next, there's going to be at least 2 colors that find their 7th half. So this is as inconvenient as we can get for ourselves in terms of trying to hold off on satisfying the problem. That means that we only need 1 more ball here and we're going to be guaranteed to have 7 balls that share color. And there are 9 balls currently, so 9 plus 1 is equal to 10. And thus our final answer is D. Problem number 20. 2 squares with each side equal to 1 overlap in the way shown in the picture. What is the area of the shaded region? First of all, we're quickly noting here that this pink length is square root of 2 minus 1. And the reason for this is because the length of the diagonal is square root of 2, since it's a unit square. And the length of the diagonal is always square root of 2 times the side length, which is 1. And we have this side length of the square being 1. So it's the whole diagonal, square root of 2, minus the side length, which is 1. And that leaves us with square root of 2 minus 1. So one way to think about this shaded region is the area of the square minus this white triangle here. So if we can find out the area of this white triangle here and just subtract it from half of the square, then we're home free. So we know that half of the square is just equal to 1 half in area because it's a unit square. So the area is 1 times 1, which is equal to 1. And half of that is 1 half. So now, how do we figure out the area of this triangle? Well, we know that we're forming a 90 degree angle here. And we are also forming, then, a 45 degree angle in this corner here, since the diagonal perfectly bisects this angle. We have 90 degrees and 45 degrees, which means we also have 45 degrees here. And this means we have an isosceles triangle, where the base and the height must both have the same length. Because, again, it's an isosceles triangle, and these are the two equal sides. So it's square root of 2 minus 1 squared. And that times 1 half will be the area of this triangle. So we take the area of half of the square, which is 1 half, and we subtract 1 half base times height, which is 1 half square root of 2 minus 1 squared. And if we actually do all of that math, we end up... if we expand this out, it should be 2 minus 2 square root of 2 plus 1. And that's going to give us... and that's going to give us 3. We're going to get 2 plus 1 times 1 half, which is 3 halves. And we're doing 1 half minus 3 halves, which is equal to negative 1. 2 times 1 half, which is just square root of 2. So when you expand this out and actually do all of the arithmetic, you'll end up getting square root of 2 minus 1. And that's exactly equal to answer choice A,

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