Problem number 17. Which of the following pictures cannot be made by using only the figures like the one shown here? We will call this our figure with four squares. To solve this problem, we will look at each of the figures given in answers in turn and figure out whether or not it can be made out of the figure with four squares shown in the problem. First, let's take a look at a close-up of figure A. Here it is. And here is one of the pieces that looks like the figure in the problem. We see right away that we can easily place it in here on the left side of the figure. And now we just need to add a figure like that that's upside down. And we see that we can easily make figure A using two of the pieces that we are given. So A will not be the answer because we can make it out of those pieces. Now let's take a look at figure B. Here's a close-up of figure B. If we take our original figure with the four squares, we easily see that we can set it here on the right side of the figure. And then we can take one that has been rotated and put it here. So we see that we indeed can make figure B out of these pieces. So figure B also will not be the answer. Now let's look at figure C. Here is figure C. If we start with our original piece, we can set it on the bottom. And then all we need is one that has been turned upside down and set it right here. So we can make figure C out of the pieces. And so C is also not one of the answers. Next we will try making figure D. This one is going to be a little harder to figure out. We can start by taking our original piece the way it's placed and maybe putting it here. Then we can try to put one of the pieces on the side. Trying to see that maybe if we take one that has been rotated upside down and put it here, we can just add a fourth piece that has been turned the other way. And we can make this square out of these pieces. So we can make figure D. And again, this will not be the answer because we're looking for the figure we cannot make out of the pieces. Now it's time to try to make figure E. Here is the close-up. We can try to put our figure unrotated here, but this is the only way it will fit, which leaves us with two unconnected areas of two squares each, so that's not going to work. Maybe if we rotate the figure a little, we can put it on this side here, but again, we are left with a shape that does not look like the piece. If we try to put the figure turned upside down, we can set it here, which does not allow us to add another one, or over here, which leaves us an L shape instead of the shape that we are looking for. If we had rotated our figure another way, it will not fit at all, so that's not an option. So figure E is the only figure we cannot make using the shapes that we were given, which means that the answer is E. Problem number 18. Which tile should go in the middle of the pattern in the picture? Hint, you might need to rotate the tiles for the answers. I will try to solve this problem by connecting the colored lines inside the pattern. So I will start with the blue line, connect it like this, the red line from here all the way across the tile, and then the yellow line. And just to make things easier, I am going to outline our middle tile. There we go. So the tile that we are looking for has the red line going from one side to the opposite side, and both the blue and the yellow lines cross the red line. So the answer is tile A. You can see the red line going from this side to this side, and both the blue and the yellow lines cross it. Problem number 19. Amy used six equal small squares to build the figure. What is the least number of equal small squares you should add to the picture in order to obtain a larger square? Hint, a square has four equal sides. Here is a close-up of our figure. A square needs to have four equal sides. Since we cannot move any of the small squares that make up the figure we are given, we need to figure out whether its height or its length are bigger. Looking at the height of the figure, we see that there are three rows. One here, second one here, and the third row is here. So the height is three. Now let's look at the length. This way, there will be one, two, three, four, four columns. So we see that the length is larger. Our final square will need to be made out of four small squares by four small squares. Looking at our figure again, we need to see how many small squares we need to draw in. We can make it into a rectangle first. And then we will need to add a row either on top or the bottom. I'll put it right here. And let's just make it look like a grid. Now we can count the number of squares that we added. One, two, three, four, five, six, seven, eight, nine, ten. We added ten squares, so we need to add ten small squares to make a larger square. And a way of checking our work is to see that we made a square that is four by four or has four columns and four rows, which means that it's made out of 16 small squares. We originally had six small squares in the figure, so we will subtract those from the final number of squares. 16 minus 6 is 10. Problem number 20. Five sparrows set on a wire as shown in the picture. Each sparrow chirped only once to each bird it saw on the side it faced. For example, the second sparrow chirped one time. In total, how many times did they chirp? Here's a close-up of the picture. We will look at each sparrow and see how many are next to it on the side which it's facing. This is the first sparrow. Here's the second one, third, fourth, and this is the fifth sparrow. So we're already told that the second sparrow chirped one time. We can see that this sparrow's head faces to the left and there is only one sparrow to the left of it, so it chirped once. The first sparrow is looking to the right and there are one, two, three, four sparrows next to it and where it's facing. The third sparrow is looking to the right again and it sees two sparrows, so it chirps twice. The fourth sparrow is looking to the left and it sees one, two, three sparrows. And the fifth sparrow is looking to the right, so it doesn't see any sparrows. So now we need to add these figures to find out how many times they chirped in total. We have 4 plus 1 plus 2 plus 3 plus 0. So we know that we'll do this by grouping them. 4 plus 1 is 5, plus 2 plus 3 is 5, and then the 0 is just 0. So 5 plus 5 is 10, and plus 0 is still 10. So our answer is D. In total, the sparrows chirped 10 times. Problem number 21. Which pattern can we make using all 5 cards given below? Hint, since we will be placing cards one on top of another, we will not necessarily see all the cards in the final pattern. First, let's take the green square. Here it is. Because we can see that all the patterns are laid out on top of the green square. In pattern A, we can take the two pink squares and put one in either corner of the green square. Here's the other. And then we can take both of the blue triangles and put one here, and the other right on top of it. So we can make the pattern shown in answer A. So that works for us. Let's make sure that this is the answer, and that we cannot make any of the other patterns. If we try to make pattern B, we can leave one of the blue triangles right in the corner there, and the pink squares where they are, and then we can rotate the other triangle and place it in the upper corner. Here's the triangle that has been rotated, and place it right here. However, we are out of blue triangles, so we used both of them. One, two, and we only had two to start with. So there's no way that we can add these two triangles to the other side, so we cannot make pattern B. In pattern C, we already see that we do not have long strips like these on the side, so we cannot make it. We can't cut these cards, so C doesn't work. In D, we would need a pink triangle, because if we move the pink squares around, there, take this one and rotate it, and we still need to add the pink triangle right here, but we do not have a pink triangle. So we cannot make pattern D. And likewise, in pattern E, we would need a pink triangle right here. That's another one we cannot make. So we see that the only pattern that we can make using all these cards is pattern A. Problem number 22. In the picture, you see four ladybugs. Each one sits on a flower. The flower that each sits on has as many leaves as the difference of the dots on its wings, and as many petals as the sum of the dots on its wings. Which of the following flowers has no ladybug on it? To solve this problem, look at each ladybug and subtract and add the number of dots on its wings to figure out which flower it's sitting on. The flower that's left over will be the answer. As for the expression for the first ladybug, it has 1, 2, 3, 4, 5 dots on one of its wings and 2 on the other. So first we will subtract and we get 3. This is the number of leaves. So this one so far has 3 leaves. Now let's add these dots. 5 plus 2 is 7. So we're looking for a flower with 3 leaves and 7 petals. This one has 3 leaves and let's check if it has 7 petals. Petals are the pieces of the flower on top. 1, 2, 3, 4, 5, 6, 7. This first ladybug will be sitting on flower D. So that's taken care of. Let's look at the second ladybug. It has 1, 2, 3, 4 dots on one of its wings and 1, 2, 3 on the other. 4 minus 3 is 1. So we'll be looking for a flower with 1 leaf. And then the number of petals should be 4 plus 3, which is 7. The flower with the 1 leaf has 1, 2, 3, 4, 5, 6, 7 petals. So this ladybug will sit on flower B. Next, let's look at the third ladybug. Here we go. It has 1, 2, 3, 4, 5 dots on the top wing and 3 on the other. 5 minus 3 is 2. We have 2 flowers with 2 leaves. So let's mark those for now. 5 plus 3 is 8. So let's check which one has 8 petals. This one has 1, 2, 3, 4, 5, 6, 7, 8. This should be the right one. Just double-check. Flower E has 1, 2, 3, 4, 5, 6, 7 petals. Now we are left with just one ladybug. Let's do the math for this one. On the top there are 1, 2, 3, 4, 5, 6 dots. And there's 1 on the bottom. 6 minus 1 is 5. So we are looking for a flower with 5 leaves. It should be this one. 1, 2, 3, 4, 5. And then 6 plus 1 is 7. So there should be 7 petals. 1, 2, 3, 4, 5, 6, 7. So this last ladybug is sitting on the first flower. So we are left with flower E as the one on which there are no ladybugs. So the answer is E. Problem number 23. On each of 6 faces of a cube, there is one of the following 6 symbols. Clubs, diamonds, hearts, spades, a square, and circle. On each face, there is a different symbol. In the picture, we can see the cube shown in two different positions. Which symbol is opposite the square? Hint. On a cube, any of the 6 faces is opposite one face and next to the other 4 faces. We can tell from the picture which symbols are next to the circle. There's a circle in this picture and in this picture. The first picture shows us that the circle is next to the diamond and to the clubs. It is also next to the heart and the spade. That means that the only symbol that it is not next to is the square. Which means that the circle is opposite the square. And if the circle is opposite the square, then the square is opposite the circle. So the answer is A. The square is opposite the circle. Problem number 24. The numbers 1, 5, 9, 8, 10, 12, and 15 are divided into groups of one or more numbers. The sum of the numbers in each group is the same. What is the largest possible number of groups? To solve this problem, we will first find the sum of all the numbers. Then, based on the number of groups each answer lists, we will find the sum of the numbers in that group need to have. Let's start by adding all the numbers together. 1 plus 5 plus 8 plus 9 plus 10 plus 12 plus 15 is equal to... Well, let's go ahead and rearrange our numbers to make it easier to add them. 1 plus 9 is going to be 10. Plus 5 plus 15, that makes 20. Plus 8 plus 12, which is another 20. And we are left only with the number 10. So that gives us 10 plus 20 plus 20 plus 10, which is 60. So yes, we can make one group, and now we know that the sum would be 60. Let's see if we can make two groups. If we had two groups, the sum in each group would be 60 divided by 2, which is 30. Let's see if we can make groups with a sum equal to 30. We already saw when we did our addition that we can find numbers that add up to 10 and to 20. So we can do 5 plus 15, that's 20, plus 10. That would be one of our groups. And the second group would be 1 plus 8 plus 9 plus 12, which also is going to be 30. Remember, 8 plus 12 is 20, 1 plus 9 is 10. So yes, we can make two groups. Let's see if we can make three groups. The sum would need to be 60 divided by 3 is 20. Let's see how we can make 20. Remember, 8 plus 12 is 20, 5 plus 15 is 20, and then we have 1 plus 9 plus 10. That is also 20. So we can make three groups. If we try to make four groups, each sum would need to be 60 divided by 4 equals 15. So one of our groups could be just the number 15 by itself. Then let's see if we can use the number 12 and any of the other numbers to make a sum of 15. We can add the 1 to it, which would make our sum be 13. And then we need a 2, which we do not have. If we try 12 and the next smallest number, which is 5, we get 17. And any of the other numbers will also make the sum even bigger. So there's no way to make groups where the sum is equal to 15, so we cannot make four groups. Now let's try making five groups. 60 divided by 5 is 12. We do have a number 12, but we also have a number 15. And there's no way to make a group with a sum of 12 using this number 15, because it is already too big. So we cannot make five groups. So our answer is C. Three groups is the largest number that we can make.

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