Estimation Station
January 16, 2015
Problem 1: 'Pages in a book'
- Click Here to see problem
- Initial Thoughts:
I think this problem is very reasonable, because in the description information is provided about a second (smaller) book, and the two books are placed next to each other for easy comparison. I also know that most pieces of paper are about the same thickness, so I was able to measure the relative size (thickness) of the larger book and compare it to the thickness of the little book.
- Helpful Information:
It was helpful to have the smaller book and the larger book together in the same picture, because then it is easier to compare the two to each other. It was also important to know how many pages are in the smaller book, because without that I wouldn't have anything to base my estimation on.
- My Estimation:
I think there are about 350 pages in the larger book. I am relatively confident in my answer because I used a ruler to measure the thickness of the smaller book (how it appeared in the image) and the thickness of the larger book, then calculate how many times the little book's thickness could fit into the larger book's thickness. I estimated that the little book could fit into the bigger book a little less than three times. 3 X 132 = 396, 396 - about 50 = about 350.
Problem 2: 'Length of a CD Case'
- Click Here to see problem
- Initial Thoughts:
This problem seems relatively simple, because it provides a photograph with a point of reference against which I can base my calculations. On the CD case is a coin, and the diameter of the coin is given, so it was easy to to estimate how many coins would fit along the edge of the case, and that would allow me to calculate its length.
- Helpful Information:
Having both the CD case and the coin in the same picture was very helpful because I could compare their relative sizes to estimate the length of the case. It was also very important for the diameter of the coin to be labeled, because the problem would be impossible without that piece of information.
- My Estimation:
I believe that the CD case is approximately 15 cm long. I believe it is 15 cm long because the coin is about 2.54 cm, and I estimated that about six coins could fit along the length of the CD case, and 2.54 X 6 = 15.24 cm. I am confident that this is a reasonable estimation, I could be more sure if I used a ruler to figure out exactly how many coins could fit along the edge of the CD case, but that wouldn't be an estimation.
Problem 3: 'Ground Coffee Beans'
- Click Here to see problem
- Initial Thoughts:
I think this estimation will be more difficult, because from the picture it is difficult to judge the approximate volume of the two bowls, and therefore difficult to compare them to each other.
- Helpful Information:
It is helpful to know how many coffee beans were ground to result in the first bowl. Also, seeing the bowls next to each other makes it possible to compare the two, but it is harder to guess how much bigger the second bowl is from the first. If the grounds were displayed in two identical bowls it might be easier to compare.
- My Estimation:
I estimated that it took about 35 beans to produce the amount of grounds in the second bowl. I guessed this because it looks like the amount of grounds in the second bowl are not quite double the amount in the first. So if 20 beans produced the first bowl, then 40 would be double, so I think about 35 would be a reasonable estimation.
Problem 4: 'Jelly Beans'
- Click Here to see problem
- Initial Thoughts:
I was initially a little bit confused by the problem, because I notice that there was one jar that was 62% full, and the other was 100% full. Also, I wasn't sure if the bounce ball was in the middle of both jars or only the smaller one. I assumed it was in both.
- Helpful Information:
Knowing the exact percent of how full the jars were was helpful, but just by looking at them it was hard to tell the size of the beans in comparison with the jar, so I don't know how many beans are in either jar. I am not actually sure what I am looking to estimate, I assume it is the number of beans in the smaller jar. It would have been helpful to have the number of beans in the larger jar given.
- My Estimation:
Assuming that I am looking for the number of beans in the less-full jar, I would assume there are about 38 beans. I am very unsure about this estimation, because I just estimated it by noticing how many beans are in a 'layer' of the jar, like how many are touching the bottom, then estimated how many layers there would be in the jar.
Problem 5: 'Toast'
- Click Here to see problem
- Initial Thoughts:
This problem sounded very difficult at first, because the only information I was given was a picture of the room, and the dimensions of toast. Also, I wasn't sure if the question wanted to know how many toasts could fit if you filled the room or just lined the floor with toasts. I am going to estimate lining the floors.
- Helpful Information:
Having a picture of the room with people in it gives me some indication of its size, but having the actual dimensions would have been more helpful. Also, maybe seeing a photograph with the actual toast in the room so I could compare the size (the toast is in a separate scale).
- My Estimation:
My estimation is that about 4200 pieces of toast could line the floor of our classroom. I am not very confident with this answer because I had to assume a lot of things, such as the length of the walls and the shape of the room (I assumed it was square). One side of the room is about 21 ft long (calculated with given scale) X 12 = 252 inches. The other side of the room is about the same, so I think the floor of the room would be about 63,504 in^2. If the length of the toast is 5, then against one wall you could fit about 50 pieces of toast. The other side of the toast is 3 inches, so along a wall you could fit 84 pieces of toast. So, 50 X 84 = 4200 pieces of toast.
- Click Here to see problem
- Initial Thoughts:
I think this problem is very reasonable, because in the description information is provided about a second (smaller) book, and the two books are placed next to each other for easy comparison. I also know that most pieces of paper are about the same thickness, so I was able to measure the relative size (thickness) of the larger book and compare it to the thickness of the little book.
- Helpful Information:
It was helpful to have the smaller book and the larger book together in the same picture, because then it is easier to compare the two to each other. It was also important to know how many pages are in the smaller book, because without that I wouldn't have anything to base my estimation on.
- My Estimation:
I think there are about 350 pages in the larger book. I am relatively confident in my answer because I used a ruler to measure the thickness of the smaller book (how it appeared in the image) and the thickness of the larger book, then calculate how many times the little book's thickness could fit into the larger book's thickness. I estimated that the little book could fit into the bigger book a little less than three times. 3 X 132 = 396, 396 - about 50 = about 350.
Problem 2: 'Length of a CD Case'
- Click Here to see problem
- Initial Thoughts:
This problem seems relatively simple, because it provides a photograph with a point of reference against which I can base my calculations. On the CD case is a coin, and the diameter of the coin is given, so it was easy to to estimate how many coins would fit along the edge of the case, and that would allow me to calculate its length.
- Helpful Information:
Having both the CD case and the coin in the same picture was very helpful because I could compare their relative sizes to estimate the length of the case. It was also very important for the diameter of the coin to be labeled, because the problem would be impossible without that piece of information.
- My Estimation:
I believe that the CD case is approximately 15 cm long. I believe it is 15 cm long because the coin is about 2.54 cm, and I estimated that about six coins could fit along the length of the CD case, and 2.54 X 6 = 15.24 cm. I am confident that this is a reasonable estimation, I could be more sure if I used a ruler to figure out exactly how many coins could fit along the edge of the CD case, but that wouldn't be an estimation.
Problem 3: 'Ground Coffee Beans'
- Click Here to see problem
- Initial Thoughts:
I think this estimation will be more difficult, because from the picture it is difficult to judge the approximate volume of the two bowls, and therefore difficult to compare them to each other.
- Helpful Information:
It is helpful to know how many coffee beans were ground to result in the first bowl. Also, seeing the bowls next to each other makes it possible to compare the two, but it is harder to guess how much bigger the second bowl is from the first. If the grounds were displayed in two identical bowls it might be easier to compare.
- My Estimation:
I estimated that it took about 35 beans to produce the amount of grounds in the second bowl. I guessed this because it looks like the amount of grounds in the second bowl are not quite double the amount in the first. So if 20 beans produced the first bowl, then 40 would be double, so I think about 35 would be a reasonable estimation.
Problem 4: 'Jelly Beans'
- Click Here to see problem
- Initial Thoughts:
I was initially a little bit confused by the problem, because I notice that there was one jar that was 62% full, and the other was 100% full. Also, I wasn't sure if the bounce ball was in the middle of both jars or only the smaller one. I assumed it was in both.
- Helpful Information:
Knowing the exact percent of how full the jars were was helpful, but just by looking at them it was hard to tell the size of the beans in comparison with the jar, so I don't know how many beans are in either jar. I am not actually sure what I am looking to estimate, I assume it is the number of beans in the smaller jar. It would have been helpful to have the number of beans in the larger jar given.
- My Estimation:
Assuming that I am looking for the number of beans in the less-full jar, I would assume there are about 38 beans. I am very unsure about this estimation, because I just estimated it by noticing how many beans are in a 'layer' of the jar, like how many are touching the bottom, then estimated how many layers there would be in the jar.
Problem 5: 'Toast'
- Click Here to see problem
- Initial Thoughts:
This problem sounded very difficult at first, because the only information I was given was a picture of the room, and the dimensions of toast. Also, I wasn't sure if the question wanted to know how many toasts could fit if you filled the room or just lined the floor with toasts. I am going to estimate lining the floors.
- Helpful Information:
Having a picture of the room with people in it gives me some indication of its size, but having the actual dimensions would have been more helpful. Also, maybe seeing a photograph with the actual toast in the room so I could compare the size (the toast is in a separate scale).
- My Estimation:
My estimation is that about 4200 pieces of toast could line the floor of our classroom. I am not very confident with this answer because I had to assume a lot of things, such as the length of the walls and the shape of the room (I assumed it was square). One side of the room is about 21 ft long (calculated with given scale) X 12 = 252 inches. The other side of the room is about the same, so I think the floor of the room would be about 63,504 in^2. If the length of the toast is 5, then against one wall you could fit about 50 pieces of toast. The other side of the toast is 3 inches, so along a wall you could fit 84 pieces of toast. So, 50 X 84 = 4200 pieces of toast.