Problem 21. Winnie's vinegar wine water marinade contains vinegar and wine in the ratio of 1 to 2, and wine and water in the ratio of 3 to 1. Which of the following statements is true? A. There is more vinegar than wine. B. There is more wine than vinegar and water together. C. There is more vinegar than wine and water together. D. There is more water than vinegar and wine together. E. There is less vinegar than either water or wine. To understand what's going on in this problem, let's write several equations to understand the relationships between the ingredients. First, let's say that V is equal to vinegar. W is equal to wine, and H is equal to water. So if we read this problem, we see that the amount of vinegar, which is V, is equal to one-half the amount of wine. And that the amount of water, H, is equal to one-third the amount of wine. So let's try checking these various statements. Statement A says that there is more wine than vinegar. But if we look at this equation right here, we see that actually vinegar is equal to half the amount of wine. So we know that statement A is false. Let's check the second statement. There is more wine than vinegar and water together. So let's add vinegar and water, which is equal to three-sixths wine plus two-sixths wine, which is equal to five-sixths. So let's write this equation like this, just so that it's very clear. And we see that actually it's true, that there is more wine than vinegar and water together, because vinegar and water only equal five-sixths of the amount of wine. So we found our true statement, which is B. Problem 22. Kangaroos hip and hop play jumping by hopping over a stone, then landing across so that the stone is the midpoint of the segment traveled during each jump. Picture 1 shows how Hop jumped three times, hopping over stones marked 1, 2, and 3. Hip has the configuration of stones marked 1, 2, and 3 to jump over in this order, but starts in a different place as shown on picture 2. Which of the points, A, B, C, D, or E, is his landing point? To solve this problem, we have to remember that each stone is exactly in the middle of a jump. So if we start here, we know that stone 1 is in the middle of a jump. And if we know that there are two squares on one side of it, there have to be exactly two squares on the other side. And so since these are jumps in a straight line, we know that the first jump will end right here. Now we know that this stone number 2 is in the middle of a jump. And stone 2 is in the middle of the second jump. So if we draw a line, because the jumps are straight from the start to the middle of the jump, we see that we've gone over four squares and gone up one square. So we have to continue that line. So let's count over four squares, 1, 2, 3, 4, and up one square. And so we know that the second jump will end at this point right here. And since we know that the third stone is in the middle of the third jump, let's draw a line that will connect them like this. And we see that that line has gone down three squares and over two squares. So now we have to continue with that jump because the stone is only the middle of it and count down three squares and over two squares, and we see that the third jump will end at point D. So the correct answer is D. Problem 23. There were 12 children at a birthday party. Each child was either 6, 7, 8, 9, or 10 years old, with at least one child of each age. Four of them were six years old. Of the group, the most common age was 8 years old. What was the average age of the 12 children? To begin solving this problem, let's write down the facts that we know. We know that there are 12 children at a party, and we know that their ages are 6, 7, 8, 9, and 10. We know that four of the children are 6, and we also know that at least one child is each of the ages that are mentioned. So we know that there has to be at least one child who is 7, at least one child who is 9, and at least one child that is 10. And we know that most of the children are 8, so that means that there have to be more than 4 children who are 8. But let's first add up how many children we have accounted for already. So that's 4 plus 1 plus 1 plus 1, which is equal to 7. And if we subtract that from 12, we get 5. And we know that that's more than 6, so we know that there have to be 5 children who are 8 years old. So to find the average age of the children at the party, we have to multiply 6 times 4, add 7, add 8 times 5, plus 9, plus 10, and divide that by the 12 children, and we get an answer of 7.5, which is D. Problem 24. Rectangle ABCD is cut into four smaller rectangles, as shown in the figure. The four smaller rectangles have the properties A, the perimeters of three of them are 11, 16, and 9. B, the perimeter of the fourth is neither the biggest nor the smallest of the four. What is the perimeter of the original rectangle ABCD? Let's start by labeling the sides of the smaller rectangles. Let's label this one P, this one Q, and this one R, and this one S. And now let's start by looking at the smallest of the rectangles, this one. We know that its perimeter is equal to 2 times P plus S, and that since it is the smallest, P plus S, and that since it is the smallest of the rectangles, we know that it's equal to 11, because that's what's given in the problem over here. And then let's look at the largest of the rectangles, and we can write an equation for its perimeter as well. We know that its perimeter is equal to 2 times Q plus R, and that's equal to 19, because we know that from right here. And we can rewrite these equations, we can add them together rather, and say that 2 times P plus S plus 2 times Q plus R is equal to 30. And now we can rewrite this equation in this way. We can say P plus P plus S plus S plus Q plus Q plus R plus R is equal to 30. And we can rewrite this equation again, where we say P plus Q plus P plus Q plus S plus R plus S plus R is equal to 30. And we can simplify this again, and say 2 plus P plus Q plus 2 times S plus R is equal to 30. And if you realize what I've done here, I've just written out an equation for the perimeter of this big square. And so we know that that perimeter is equal to 30, and the correct answer is B. Problem 25. Kanga wants to arrange the 12 numbers from 1 to 12 in a circle in such a way that any neighboring numbers always differ by either 1 or 2. Which of the following pairs of numbers have to be neighbors? Now let's begin by just trying to arrange the numbers according to the rules that we've been given. Let's start with the highest number, which is the number 12. And let's put it over here. It doesn't matter where we put it on the circle, because the circle is symmetrical on every axis. So it doesn't matter if we put the 12 up or down or left or right. Let's just start here. And then let's move on to the next number, which would be 11. And since the rule is that neighboring numbers can only differ by 1 or 2, really the only place that it can go is right next to the 12. So then let's go to the next number, which is 10. Now the number 10, again, it can only be different by either 1 or 2 from its neighbors. And since 10 is 2 away from 12, it could fit right here. And then let's go down to the number 9. And since the number 9 is 2 away from 11, really this is the only place that it could go. And then let's go to the next number, which is number 8. And again, this is the only place that it can go because it's 2 away from the number 10. And then let's put the number 7 over here. And then following our logic, which we've established now, the number 6 has to go here. The number 5 has to go here. The number 4 has to go here. The number 3 goes here. The number 2 goes here. And the number 1 goes here. And really because we've arranged these numbers so that they are only different from each other by 1 or 2, this is really the only possible way that you can arrange these numbers. So if we look at the choices we've been given, D is the only choice that holds true, so D is the answer. Problem 26. Peter wants to cut a rectangle of size 6 by 7 into squares with integer sides. What is the smallest number of squares he can get? In this problem, we want to cut this rectangle into smaller shapes that are all squares. Let's start by drawing the largest square that we can in this rectangle. This square is going to be a 6 by 6 block square. But that leaves us with 1, 2, 3, 4, 5, 6 smaller squares that we have to divide up this rectangle into. So our first attempt yielded 7 squares, which I think we can do better, so let's keep trying. So now let's try to draw a slightly smaller square. Let's draw one that is 5 by 5 blocks and see what happens. So if we have 5 by 5 blocks, we can create larger squares over here, which is great. So we have 1, 2, 3 squares on this side, but then we have a whole bunch of small squares over here, and so we have one, two, three, four, five smaller squares. So altogether we have one, two, three, four, five, six, seven, eight, nine squares in our second attempt, which is even worse than our first one. But let's keep going and try a smaller square again. So now let's try a smaller square again, a square that's four by four, and then let's see what can happen. We can divide up this side into squares that are three by three, and we can divide up this side into squares that are two by two, which is great. This way we have created five squares, which to be honest is the correct answer, and you really cannot turn a rectangle into any squares less than five. The correct answer is B. Problem 27. Some cells of a square table of size four by four were colored red. The number of red cells in each row was indicated at the end of it, and the number of red cells in each column was indicated at the bottom of it. Then the red color was eliminated. Which of the following tables can be the result? This is a logic problem, so the best way to solve it is to go through the choices that are given, and to see if any of the choices break the rules that have been laid out in the problem. So let's look at choice A. According to this table, there should be four filled spaces in the top row, but if we look in the bottom row, we see that the first column says that there are zero colored squares in it. So we know that there's something wrong with choice A, and we can cross it off. Now let's look at choice B. An easy way to check if the logic is consistent is to add up both sides, because altogether each column and each row have to equal the same amount of red squares. So if we add up the rows over here, we have one plus two plus one plus three, which is equal to seven, and if we add up the columns, we have two plus two plus three plus one, which is equal to eight. So we know that there is some logical inconsistency here, there's something wrong, so we know that choice B is wrong. Now what we can do with choice C is we know that there can be no red squares in these two rows, but this column says that there are three red squares, but we only have two spaces for them, so we know that there's something wrong with choice C. Now let's look at choice D. Let's add up the the four rows, we have two plus one plus two plus two, which is equal to seven, and let's add up the columns, and we have two plus one plus two plus two, which is also seven. So this is looking really good. Let's just double check E and see if there's anything that we can find that's wrong with it. So E says that there are no squares in this row, and that there are no squares in this column, and it says that there are three squares in this row, as well as three squares in this row, and so we know that these squares have to be here, but this column says that there's only one square, so we know that there's something wrong with choice E. So now we know for sure that choice D was the only correct choice, and D is the answer. Problem 28. A square-shaped piece of paper has an area of 64 square centimeters. The square is folded twice, as shown in the picture. What is the sum of the areas of the shaded rectangles? To find the solution to this problem, let's try to understand what's going on geometrically. Let's begin by defining some of the distances. So let's say that the distance between points A and C is equal to 2h, and that means that since we're folding everything, every piece here in half, that this distance is going to equal to h, and that this distance is going to equal h over 2. That means that the distance from here to point C is going to equal to 3h over 2, and if we divide that distance in half, which is going to be this line over here, we get 3h over 4, which leaves us with this distance over here, which is equal to h over 4. So we've defined the proportions of these distances. So now let's see how they relate to the problem which we have here, which is a problem of finding the area of these two rectangles over here. We know that the area of a square is equal to one of its sides squared, and since we know that the total area here is equal to 64 centimeters, you know that 64 is equal to s squared, and so s is equal to 8. And since we know by the proportions that we've defined over here that the point M is equal to s over 2, we know that the line DM is equal to the side divided by 2, which is equal to 4. And here we know that the line D, sorry I've hidden this little N over here, but this is line DN, is equal to the side divided by 4. We know that this is equal to 2. So now we know the sides of one of the rectangles that we're looking for, and we know that the area of a rectangle is equal to the product of the two sides, so that's equal to 4 times 2, which is equal to 8 in centimeters squared. And since these are two identical rectangles, we know that the total of these two rectangles is equal to 16 centimeters squared, and that answer then is answer D. Problem 29. Abbott's house number has three digits. Removing the first digit of Abbott's house number, you obtain the house number of Ben. Removing the first digit of Ben's house number, you get the house number of Chiara. Adding the house number of Abbott, Ben, and Chiara gives 912. What is the second digit of Abbott's house number? In this problem, the digit 0 cannot be the first digit of a house number. So let's begin by saying that Abbott's house number is equal to ABC, Ben's is equal to BC, and Chiara's is equal to C. And then since we know that no house number begins with a 0, we can say that A, B, and C have to be between integers 1 and 9, and they have to be whole integers, obviously. And we can say that the sum of the house numbers is equal to 100A, because A is in the hundreds column, plus 10B plus C, that's Abbott's number, plus 10B plus C, that's Ben's number, plus C, which is Chiara's, is equal to 912. Therefore, 100A plus 20B, oops, 20B, sorry about that, plus 3C is equal to 912. And therefore, 20 times 5A plus B is equal to 3 times 304 minus C. And 304 minus C has to be a multiple of 20, which is on this side of the equation. So we know that C is equal to 4. Since we know C is equal to 4, we can say that 20 5A plus B is equal to 900, and 5A plus B is equal to 45. And B, therefore, is equal to 5 times 9 minus A. So B has to be a multiple of 5. And since the only multiple of 5 that's between 1 and 9, as we've defined earlier, is 5, so B has to equal to 5. So finally, A is equal to 912 minus 20 times 5 minus 3 times 4. Divide that by 100, and you get 8, which makes Abed's number then 854. Therefore, the second digit of Abed's house number is going to be C, right here. Problem 30. I give Anne and Bill two consecutive positive integers. For instance, 7 to Anne and 6 to Bill. They know their numbers are consecutive. They know their own number, but they do not know the number I gave to the other one. Then I heard the following discussion. Anne said to Bill, I don't know your number. Bill said to Anne, I don't know your number. Then Anne said to Bill, now I know your number. It is a divisor of 20. What is Anne's number? To begin, let's make a list of the first integers that we can encounter. So let's start with 1, 2, 3, 4, 5, 6, 7, and the list continues as you know. Now let's begin with a bit of logic that you can use to solve this problem. We know that anybody who would get the number 1 as their number would know that the other person would have a number 2. Anne doesn't have a number 1, and Bill gains this piece of information from Anne's first statement when she says, I don't know your number. And since Bill doesn't have 1, he would have known Anne's number, and he also admits to the fact that he doesn't know her number. So we know that neither of them have the number 1. If Bill had 2, he would know Anne's number would be 3, and since number 1 has already been X'd out, that's excluded. So he does not have 2, and Anne knows it. So Anne must have 3 to know that Bill's number at this exact moment after these first two statements. His number is 4, since 2 has already been eliminated, and 4 is a divisor of 20. So B is the correct answer.

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