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This is a professional development blog primarily for teachers in Spring Branch ISD.

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I chose to complete Einstein problem 5 on pg.117, which asks one to find the perimeter of the figure. As I said in my A-HA response, it would help if the children were taught to create two figures out of the original polygon, a rectangle and a triangle. The children would have to draw a vertical line from leg b (8 ft) to leg a (6 ft) to create the hypotenuse, and would then have to know that this length is equal to the missing length. Zaccaro does not teach the preceding strategy and he does not state that the hypotenuse is always opposite the right angle. Likewise, he does not state that the legs always make up the right angle. Furthermore, on pg. 108 he explains that the shorter sides of a right triangle are called the legs and that the shorter leg is usually referred to as side a. Unfortunately he continues explaining that the longer diagonal side is the hypotenuse, which is not the case in problem number 5, for it is vertical. Once the children have come this far, they would use the Pythagorean Theorem to find that the length of the hypotenuse is equal to10ft, thankfully a² + b² came out to a perfect square, 100ft². Finally, they would solve the problem by finding the sum (or the perimeter) of 8 + 6 + 12 + 10 + 12 = 48 ft.

ReplyDeleteSession 2 Question 3

ReplyDeleteI worked on problem #8 on page 104. Frankly I love pool problems and will definitely add this one to my repertoire. There are three hoses. A fills the pool in six hours, B fills in 2 hours and C fills in six hours. If all three are used how long will it take to fill the pool? Hose A fills 1/6 of the pool in one hour, hose B ½ in one hour and hose C 1/6 in one hour. If all three are used 1/6 + ½ +1/6= 5/6 of the pool is filled in one hour. For the remaining 1/6, we need 1/5 of an hour which is 12 minutes. So pool will be filled in 1 hour and 12 minutes.

I chose Einstein Question #6 on page 163. I tend to gravitate towards these types of problems that allow you to show the mathematical relationships among the unknowns, or variables. I began by showing everything in number sentences so I could see the relationships building and figure out how to solve the problem. Since there are "three times as many cars as trucks" I knew that C = 3T. "The number of motorcyles is twice the number of bikes", so M = 2B, and "there are four times as many trucks as bikes", leaving T = 4B. Since C = 3T, and T = 4B, then C = 3(4B) = 12B. Then, since everything is expressed in relation to B and if each B represents one part of the whole, there should be 19 equal parts in the whole (1 for bikes, 2 for motorcyles, 4 for trucks, and 12 for cars). Then, 285/19 = 15. Since cars is 12B then there must be 180 cars. Of course, my students would say they did guess and check!

ReplyDeleteIn response to SusanM on June 18th, don't they love guess and check? I think that is a natural consequence of multiple choice testing. Especially with problems like that, they don't even have to make a reasonable guess, just plug in each choice and see if it works.

ReplyDeleteI worked on the Einstein problem #3 on page. 125 because I made pancakes this morning. :-) It caught my attention. I worked with the information for the large pancakes first. I worked with the areas of a single large pancake first, and then multiplied to get the area of the 10 pancakes. I then got the area of a small pancake. All that was left to do was to divide the area of the 10 large pancakes by the area of a small pancake--getting a result of 160 small pancakes. Yummy!

ReplyDeleteI chose to solve Einstein Level question #5 on page 114- finding the perimeter of a figure. You can't find the perimeter until you find the missing length. First, you need to block off the figure into easier shapes to work with. That then gives you a rectangle and a triangle. Using the Pythagorean Theorem (a^2 + b^2 = c^2, you will find the missing length for the side of the triangle because that will help you with the rectangle too. a^2 (8^2) + b^2 (6^2)+ c^2. So 64 + 36 = 100. You then need to find the square root of 100 to determine the length of the missing side of the triangle since the theorem squares the lengths. The square root of 100 is 10. Since this missing side of the triangle is being shared by the rectangle part of the figure, that means you have found the missing side of the rectangle and you may now find your perimeter. DO NOT count the 10 twice. The 10 for the triangle's side should not be added in for the perimeter because it is on the inside of the figure, not the outside like we need for perimeter. So, add 12+10+12+6+8 to equal a perimeter of 48 units.

ReplyDeleteThe problem I picked was problem number 3 on page 150. It is exactly the same type of problem that Zaccaro used in his explination on page 138. He simplfied the process by using a chart to display the accumulated interest. I did my problem using the same method. The answer is $8861.34.

ReplyDeleteProblem 10, page 164: A bathtub is filled with 100 gallons (12,800 cups) of water. A cup of milk is then poured in and mixed thoroughly with the 100 gallons of water. A cup of the liquid is then taken out and placed in a bowl.

ReplyDelete1) What fraction of the liquid in the bowl is milk?

The fraction is the ratio of the amount of milk (1 cup) to the total amount of liquid (milk and water) (1,601 cups), or 1/1601.

2) What is the ratio of milk to water?

milk : water 1 : 1600

I did the Einstein level question on pg 163 question #1.

ReplyDeleteI really enjoy solving this problem. At first I did it making a ratio and proportion to solve. This is the lesson that Zaccaro teaches in his chapter. Then I solved it the way I thought a majority of my students would begin doing it. I thought about how my students would explain their thinking to me. I thought it would be very challenging and I plan to use several of these for my brain stumpers of the week with my GT students next year

I did Einstein level question #5 on page 114. I found the missing side by using the pythagorian theorem on the right triangle on the opposite side. The missing side a^+b^=c^ and c^=100 so it was 10. Then I added all the sides together to get a perimeter of 48 sq. feet. I think many kids would think this was easy, and enjoy exploring this theorem with other shapes that include a right triangle.

ReplyDeleteI did Einstein problem number 1 on p.163 to try out the author's examples by replacing the numbers with numbers in the problem. I liked the way the author explained the ratio at the beginning of this chapter. I divided the 80 pounds by the total of 512 pounds (the total of the family) and determined that Mark would be entitled to 15.6%. Then I multiplied the 64 ounces of ice cream by that percentage and determined he would get 9.984 ounces. This worked relatively well.

ReplyDeleteI solved problem 2 on page 125, mainly because I miss teaching these kinds of questions.

ReplyDeleteI like the multi-step nature of the problem, since you have to work backwards from the outside rectangle to the inside triangle. I also liked that the problem spiraled back to include Pythagorean theorem from the diagonal of the smaller square.

I could see that some of my GT students could work these in groups and with a little guidance, but I think that a lot of them would struggle with the process without some practice.

In response to SusanM and SDawson--I agree they do love guess and check!! Anything that gives them an opportunity to be the detective and solve a mystery!!

ReplyDeleteIn response to SDawson--This pool problem reminds me of an exemplers that we do each year. The one you did has a higher degree of difficulty and is great to use when teaching fractions. I really liked your explanation, it was very easy to follow.

ReplyDeleteI worked on question #3 on page 125, the "pancake question". Using the simple formulas and working backwards, I found the same amount of batter that makes 10 pancakes of an 8 inch diameter, will make 160 pancakes of a 2 inch diameter. My first attempt was on a fraction question, but that didn't go so well. It was easy to plug in the formulas and work backwards to solve this one, much less paper, much less frustration!

ReplyDeleteI completed problem #4 on page 103. Both people in the problem hiked the same distance but one arrived earlier. I had to do several steps to solve the problem: subtract the two fractions by finding a common denominator, multiplying by 60 and then converting the improper fraction into the number of minutes. I finally came up with 6 minutes later. This problem was a little more challenging then I thought it would be.

ReplyDeleteI completed problem #3 on page 103. I spent a lot of time teaching fractions this year and love this kind of problem. First I had to add two fractions that did not have common denominators. I had to change the fractions by multiplying the numerator and the denominator so that the two denominators had the same number. I turned 1/6 into 7/42 and 1/7 into 6/42. We knew how much money and the fraction of money the remaining sibling received. I added 6/42 and 7/42 to determine the other two siblings share and then subtracted that from 42. The 3rd sibling had a share of 29/42 of the total. I knew her share was $23,490. I multiplied (29/42)x=23,490. Then I had to divide (29/42) by each side to isolate x. When I divided, I flipped the fraction to become 42/29 and then multiplied. I could not simplify since 29 is a prime number. I was able to determine that the total of money left was $34,020. Then I was able to multiply that number by 1/6 to determine that child A received $5,670. I love multi-step problems!!!

ReplyDeleteI solved problem #1 on p.133. It was a basic volume problem just with double digit numbers, which I think that even my non gt students would be able to solve. I think you had to infer that it wasn't round or oval b/c the measurements didn't have radius or diameter. That is one thing we do and our students do is just assume that it would be rectangular. Not that it was that hard of a problem, but made me think about what shape it was before I started.

ReplyDeleteI solved problem #6 on p. 104 about Fractions since it was a multi-step problem. As I read the problem, I drew a box and cut it in fractions coloring the ones that they told me that were used. Then I plugged the money left and found out the other three parts of the 3/4. ($20 each part and total of the 4/4 =80) After that I divided the total of the 3/4 by 5 (since it belonged to the 5/8 part= 80/5=16) to help me find out how much was each 1/8 fraction ($16). Finally I multiplied 16*3 =48 +80= $128 was the amont she strated with.

ReplyDeleteIn response to ReneeR and SDawson, I wouldn't mind the guess and check if they actually did it systematically and gave some organization to their work, such as in a table or something like that. Often when they use guess and check, their work is all over the place, they can't figure where they started and where they've been, and sometimes they forget the most important part...the checking. I do a lot of modeling with this strategy to make sure they understand how to use it effectively! :-)

ReplyDeleteI have chosen the problem 5 on page 51 because I love the algebra problems

ReplyDeleteLet p = pennies and n=nickels, dimes would equal 2n and quarter would equal 6 n

So the penny would equal 59-9n

So when you plug in for the how much you have it would be

.59-.09n + .05n + .20n + 1.50n = 7.23

So combining like terms give you

.59-.59 -.09n + .05n + .20n + 1.50n = 7.23-.59

1.66n = 6.64

Divide by 1.66 and n=4

You have 4 nickel, 8 dimes, 24 quarters, and 23 pennies.

I feel drawn to these inscribed circle and squares so I choose Problem2 on page 163.

ReplyDeleteYou have a circle inscribed in a square. You know the circle has the area of 78.5{ (PI r2 ) 3.14 x 52} 5 being the radius and 10 is the diameter and the side of the square so the area of the square is 100. Going to the inside square you know the diameter diagonal(Base ) is 10 and the radius ( height ) is 5 so the area of the triangle is .5 x 10 x 5 =25 and put 2 triangles together and you have area of the square is 50.