Problem 9. Sophie arranges balls on a stairway in a pattern as shown in the picture. How will the balls be arranged on the step of the question mark? We need to figure out the pattern and how many balls there will be on the step of the question mark. First, let's number the steps. There's the first step, second, third, fourth, fifth, sixth, seventh, eighth, and ninth. We also see this way that on each step, the number of balls is equal to the number that we gave that step. There's one ball on the first step, two on the second, three on the third, four on the fourth, five on the fifth. So, we would see from this that there would be nine balls on step number nine. Now, if we look carefully at the pattern, we see that the balls are arranged so that each group starts with a black ball and then is followed by a ball that's white, and then again there's a black ball, and again white, black, and white. So, we'll be looking for something that has nine balls. One, two, three, four, five, six, seven, eight, nine. And the first one is black, the third one is black, the fifth one is black, and so the seventh ball is black, and the ninth, the final ball, is black. And this looks like it would be answer D. Let's check if there are nine balls here. One, two, three, four, five, six, seven, eight, nine. Yes, there are. So, D has the pattern that we should have on the step with the question mark. Problem 10. Lisa's hens lay white eggs and brown eggs. Lisa puts six eggs in the box shown below. Two brown eggs cannot touch each other. At most, how many brown eggs can Lisa put in the box? To solve this problem, the easiest thing is to draw several pictures of the box and figure out where we can put the brown eggs. Let's start with one picture. Let's see what happens if we put a brown egg in the upper left-hand corner. If this egg is brown, then the one next to it cannot be brown. Let's mark that purple. And the one under it cannot be brown. So we can put a brown egg on the bottom of the second column, and then again the one next to it cannot be brown, so we can have a brown egg in the upper right-hand corner as well. That gives us three eggs. Let's try another picture. This time, let's start with the brown egg in the lower left-hand corner. Again, mark the ones that now cannot be brown purple. This one can be brown, and again the one next to it will not be brown, so the one in the lower right-hand corner can be brown. Let's try another picture. If we color one of the eggs in the middle column brown, let's see this upper one, and the one next to it, or the one under it, cannot be brown, so the ones in the bottom corners can be. The same happens if we color the bottom one in the middle brown. There are three that cannot be brown, so three can be. Let's try some other options, see if we get something better. If we color one on one side, the one above it can't be brown, and then if we were to skip the ones in the middle together, then we can only have another brown egg. That would only give us two brown eggs, so it would not give us more than three. I don't think there's any way that we can get more than three brown eggs. So the answer will be C. Three is the maximum number of brown eggs that she can put in the box. Problem 11. One year has 12 months. Kenga is one year and three months old now. In how many months will Kenga be two years old? And it doesn't matter in which month Kenga was born, because each year has 12 months. Let's try solving this by drawing two calendars. It doesn't matter what the month names are, but I just made it from January to December to make it look familiar. She's already one year and three months old, so we can color in all 12 months in one year, and three more months in another year. That leaves us with the number of months until she is two years old, which will be 1, 2, 3, 4, 5, 6, 7, 8, 9. Nine months. Another simple way of doing this is to just do the arithmetic. Since Kenga is one year and three months old now, it will be less than a year until she is two years old. So we subtract 3 from 12. 12 minus 3 is 9. Either way, our answer is E. She will be two years old in nine months. Problem number 12. Granny went out to the yard and called her hens and her cat. All 20 legs ran to her. How many hens does Granny have? Their answer choices are 11, 9, 8, 6, or 4. Hint! First think about how many legs a cat has, and then think about how many legs each hen has. Here is Granny's cat. He has four legs. Now, let's look at one of the hens. A hen has two legs. So we know that all the animals that ran to Granny had 20 legs altogether. We will subtract from this the number of the legs that the cat has because we know there was only one cat. 20 minus 4 is 16. Now, these 16 legs belong to the hens, and each hen has two legs. If you remember, 2 times 8 is 16. Here's another way of saying 8 plus 8 is 16. So if there are eight hens and each one has two legs, that would make the 16 legs that we need. So the correct answer is C, 8. Problem number 13. In Baby Roo's house, each room is connected to any neighboring room by a door. You can see the picture. Baby Roo wants to get from room A to room B. What is the least number of doors that he needs to go through? To solve this problem, we will draw paths and find one that goes through the least number of doors. Let's start drawing paths from room A to room B. Going through longer rooms will probably take us through fewer doors, so first let's try going to the longer rectangular room right next to A. That goes through one door. Go up here, two doors, three doors, and four doors gets us to B. We can try going through the large square room. Start the same way. That's one door, two, three doors, and then we have two more. We have to go to four and five. If we go across from A, we go through one, two, three, four, five doors. If we were to go straight up, we would go through one, two, three, four, five doors again. So far we see that we can go through four doors and get from room A to B. It also seems that going through any other rooms or combination of doors will not make the path go through fewer doors. It could only make it go through more doors. So, the answer is B, to go through four doors. Problem number 14. There are 12 rooms in a building and each room has two windows and one light. Last evening, 18 windows were lit. In how many rooms was the light off? If a light is on in a given room, then it will show through both the windows. Here is a picture of a building with 12 rooms where every room has two windows. Now, let's turn on the lights in 18 windows. One, two. Remember, the light just shows through both windows in a given room. Three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen, fourteen, fifteen, sixteen, seventeen, eighteen. Which leaves us one, two, three, four, five, six windows where the lights are off. But, since we're asking about the number of rooms where the lights were off, we will have one, two, three rooms with the lights off. So, the answer is B, three. Problem number 15. Mary is walking along the road and she reads only the letters located on her right side. Moving from point 1 to point 2, what is the word she will read? Remember that as Mary moves along the path, she will turn. So, we need to bear in mind which side is her right, based on which way she is facing. I will start by drawing a picture of Mary, as seen from above, starting at point 1. Here's her head. And as she is looking ahead on the path, I will mark her right arm with pink, and I will mark her left arm with blue. So, as she comes to the first letter, she is still facing the same direction, so her right arm will be right here, and her left arm will be on this side. So, we see that K is one of the letters on her right side. As she proceeds and comes to the turn, her right arm is right here, her left arm is here. So, we see that A is on her left side, so we will not read that one. She keeps going. At this turn, her right arm remains on the outside, her left arm is here. So, as she gets to the letter N, it is on her right side. And here is the left arm. She keeps going. The next turn, her right arm is still on the outside, her left arm will be on the inside. So, when she gets to the letter G, here is her right arm, left arm, so G is not one of the letters she reads. At the next turn, her right arm is still on the outside, her left arm is right here. When we get to A, here is the right arm, A is on the right, her left arm is here. Keep going. When she gets to R, we see that it is on the other side than A, so it is on her left side, so she will not read that letter. She comes to the letter O, here is her right arm, so we see that this O is on her left side, she will not read it. And we are going to make the final turn right here. Here is the right arm and the left arm. So, when she gets to the letter O right here, it is on her right side, here is the right arm and the left arm is here. So, the letters that we have picked out, let me line them up again in green, were K, N, A, and this O. K, N, A, O. And that is answer A. Problem number 16. The sum of John's and Paul's ages is equal to 12. What will the sum of their ages be in 4 years? Hint, remember that each of the boys will be 4 years older in 4 years. Here is a picture of John and here is a picture of Paul. The problem tells us that right now, John's age and Paul's age add up to 12. In 4 years, John's age will be John's age now plus 4, and Paul's age will be Paul's age plus 4. You see that we are adding the 4 years twice. Once for John's age and once for Paul's age. So, since right now their age adds up to 12, we need to add 4 two times to that which makes it 12 plus 8 which is 20. So, 4 years from now, the sum of their ages will be 20 which is answer E.

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