tag:blogger.com,1999:blog-1509799006346376000.post428269346700701600..comments2011-05-18T03:39:47.749-05:00Comments on PGP in SBISD: Session 3 - Question 2atxteacherhttp://www.blogger.com/profile/15216583790234498239noreply@blogger.comBlogger16125tag:blogger.com,1999:blog-1509799006346376000.post-74915366241485859112011-05-18T03:39:47.749-05:002011-05-18T03:39:47.749-05:00I like the questions in your blogs...Keep updating...I like the questions in your blogs...Keep updating...Essay Writinghttp://www.essayprovider.com/noreply@blogger.comtag:blogger.com,1999:blog-1509799006346376000.post-36744682773008855862010-08-07T23:53:00.239-05:002010-08-07T23:53:00.239-05:00Chapter 24, Question 1...King Arthur wanted a circ...Chapter 24, Question 1...King Arthur wanted a circular table built that was large enough for 16 people. If each person had 4 feet of space, what should the diameter of the table be? (Round to the nearest foot.) Before I could solve, I had to find the formula for circumference as I could not pull that out of my memory bank.<br /><br />16*4=64<br />64=x*3.14<br />64/3.14=x<br />20=x<br /><br />Chapter 23, Question 1 (pg. 243) had to do with squaring. A student put a mystery number through a square root machine. When the answer came out, she put it through another square root machine. The number 2 came out of that machine. What is the mystery number.<br /><br />Honestly, I answered this wrong because I didn't read the "number 2 came out." My answer was 81...9...3. I worked it backwards but it was still wrong. So the correct way for me would be:<br /><br />2*2=4...4*4=16 Answer is 16.<br /><br />LBranon, I concurred in an earlier post about letting the students take on the challenge of learning this and not relying on me teaching it.<br /><br />Head Squirrel, I worked that same problem and learned that Stanley won't make it into college. I did not like that problem and was willing to work another one!MWE Libraryhttps://www.blogger.com/profile/08627462403594399565noreply@blogger.comtag:blogger.com,1999:blog-1509799006346376000.post-48759928836980996262010-08-03T08:19:01.299-05:002010-08-03T08:19:01.299-05:00Wow, I can't believe Viking's conquering t...Wow, I can't believe Viking's conquering the geometry question with the answer of pi. It will be interesting to see if any of our GT kids will get that one. I agree with KHarrell on the GT students love for the logic problems. It will be a challenge for me to teach it. But then again, the students will most likely take on the challenge themselves and bypass my instruction. Ownership!LBranonnoreply@blogger.comtag:blogger.com,1999:blog-1509799006346376000.post-59117518213486945052010-08-03T08:14:14.035-05:002010-08-03T08:14:14.035-05:00I chose Chapter 12 Einstein level, problem #1. Th...I chose Chapter 12 Einstein level, problem #1. This sounded interesting with figuring shadows. The facts tell the height of Brianna as 4 feet and the length of her shadow as 3. Then it tells her dad's shadow's length of 4 feet 6 inches. I just used the good old ratio technique to figure the inches in fractions to equalize the equation. 48/36=?/54. I multiplied 48 x 54 and then divided that answer by 36 and ended up with 72 and divide that by 12 inches to get 6 feet tall. <br />The second one I solved was in Chapter 22, number 3. I just solved for n. I added the fractions of 1/8 and 7/8 to get 1n +18=20. I took 18 from 20 to solve for n of 2. It will be interesting to see if the GT students will grab on to the abstract nature of algebra. I think they will, knowing their love of the challenge.LBranonnoreply@blogger.comtag:blogger.com,1999:blog-1509799006346376000.post-55968722937973841442010-07-29T16:38:47.691-05:002010-07-29T16:38:47.691-05:00Becky Lee
One of the Einstein levels I worked thr...Becky Lee<br /><br />One of the Einstein levels I worked through was Ch. 22: I can use algebra to solve problems. I always had trouble with algebra as a student and decided to face my "fear" and go for it. I approached it slowly and enjoyed the pages leading up to the Einstein level problems on p. 231. #2 on that page is about finding out the page that was being read in a book - perfect question for a librarian! :) I messed up the first time (yes, algebra is still a struggle for me), but after leaving it for awhile and coming back to it, I worked it out again and got the correct answer - checked it in the back! Yay! I have to say, though, that making equations is still not my favorite thing to do. <br /><br />Another question I worked through was in Ch. 25, I can answer that. Love working with these logic problems and classifying things based on the information given. Sadly, I discovered in question 1 on p. 265 that poor Stanley won't get into college if he drinks coffee! I also enjoyed #3 about the radius where most people get in car accidents (30 mi from their home) b/c that's where they drive the most. These are such fun problems to answer!Head Squirrelhttps://www.blogger.com/profile/05945475398892377884noreply@blogger.comtag:blogger.com,1999:blog-1509799006346376000.post-33416027915376166352010-07-29T12:30:55.300-05:002010-07-29T12:30:55.300-05:00Since geometry was about the only math I ever real...Since geometry was about the only math I ever really enjoyed, I chose the chapter 24 fantastic formulas (pages 254-255) for my Einstein problems. I got carried away and did all five. Question 1 about King Arthur’s knights needed two steps. Find the amount of linear feet needed, and then plug that into the circumference formula to find the needed number for the diameter. Question two had a bit of a trick. You have the diameter of the lake and are looking for the circumference. But the circumference is not the answer. You must divide it in half to get the walking distance. Question three is a straight plug-it-into-the formula. The diameter of the circle is 250K miles. The orbit is pi. Multiply the two numbers together. The answer to question four is another one of those questions which seems complicated but it is not. You will get the same answer because that is the constant: pi. Question five is easily solved if you realize that there are nine square feet in a square yard, so the price should be $3 a square foot.<br />I also did question 1 in chapter 23, page 243, on squaring machines. I thought this out two steps backwards. The answer is 16. The square root of 16 is 4, and the square root of 4 is 2.Vikinghttps://www.blogger.com/profile/15430684776690050496noreply@blogger.comtag:blogger.com,1999:blog-1509799006346376000.post-82881450632111532132010-07-28T09:02:08.647-05:002010-07-28T09:02:08.647-05:00I solved several of the Einstein problems -- almos...I solved several of the Einstein problems -- almost one for each chapter, but then, I really like math! I'll talk about 2 that really had me thinking. For chapter 22 (I can use algebra to solve problems),p. 231, question 1. I almost messed up with creating the equation. I was going to fast (something our gifted kiddos tend to do), and didn't include the fact that the ages all added up to 17. Hmm, had to go back and redo it. <br /><br />Another problem I solved was in chapter 25 (I can answer that question). I tend to love logic problems, but the ones about having 2 containers and having to fill one to a certain level without marking it, etc. just baffle me! I solved question 2 on p.265. I had to draw this one out on paper in order to get it, but I finally did. I don't think we need to explain our answer, and so I won't put it all here -- it took 11 steps, using the author's method of explanation.Eleanornoreply@blogger.comtag:blogger.com,1999:blog-1509799006346376000.post-46081985946217392192010-07-27T15:10:29.116-05:002010-07-27T15:10:29.116-05:00In response to melscales, I also had fun solving t...In response to melscales, I also had fun solving the ladder rung problem, but the only way I could do it was by drawing a picture and counting up and down the rungs as the rat climbed as you did. My question is, would there be some kind of algebraic formula that could be written to solve this problem? I sure couldn’t come up with one. Any ideas??MillieHnoreply@blogger.comtag:blogger.com,1999:blog-1509799006346376000.post-52554375233418045662010-07-27T14:32:47.236-05:002010-07-27T14:32:47.236-05:00in response to NanetteG - I hadn't really thou...in response to NanetteG - I hadn't really thought about introducing the kids to Einstein. You've got my brain going on ways to mesh the math lessons in the book to a research project on Einstein or other inventors. hhhmmm...<br /><br />To KHarrell - I know the feeling! I kind of dreaded doing these last couple of problems, but once they are done, you feel such a sense of accomplishment. At least I did!melscalesnoreply@blogger.comtag:blogger.com,1999:blog-1509799006346376000.post-39516874477796656082010-07-27T14:29:30.140-05:002010-07-27T14:29:30.140-05:00I chose to work #5 on page 210. (I chose this cha...I chose to work #5 on page 210. (I chose this chapter because algebra was such a struggle for me!)<br />I first decided that since the number of cows was n, then the number of cow legs had to be 50n. To find the number of duck legs, you must subtract n from 50, then multiply times 2 because ducks have 2 legs. So the answer is :<br />Cow legs 50n<br />Duck legs 2(n-50)<br />If that makes sense to me after all these years of not getting algebra, then the kids will get it and miss out on the whole “algebra problem” that I have. (or had!)<br /><br />The second problem is chose is #5 on page 266. I enjoy these kinds of problems because you have use logic in addition to your math skills to find the answer.<br />I first drew a little picture of a ladder. I made marks for how high or low the rat climbed during each step of the problem. Then I counted to see how many rungs were below the starting point, which was the middle rung. There were 13 rungs below, so there had to be 13 rungs above. 13 + 13 + the middle rung = 27 rungs. Yay! the right answer!melscalesnoreply@blogger.comtag:blogger.com,1999:blog-1509799006346376000.post-44067608144877827942010-07-27T13:39:10.424-05:002010-07-27T13:39:10.424-05:00KHarrell - I dread this part as I know I will be w...KHarrell - I dread this part as I know I will be working for a while!<br />Chapter 25 Logic (maybe not so much math!)<br />I tried to answer question 2 - but after much trial - I gave up...I even read the answer in the back...and I wasn't even close. I was able to do the one in level 3, but this one was just too much!<br />So...then I tried question 3 - much easier - a simple one car accidents occur more often within 30 miles of one's home, does that me that it is safer to drive more than 30 miles from your home. This one made sense...it's not safer...you drive more often within 30 miles, so acccidents will happen more often. This one reminded me of the book Stories to Solve and More stories to solve by George Shannon...the PGP students love to figure out those logic problems.<br /><br />Problem 2 that I completed is problem # 1 in Chapter 10- If 9 ounces of food will feed a fish for 45 days, howe many days will 91/2 ounces last. I divided 45 by 9 and came up with 5..so that 5 days equal one ounce so the total would be 471/2 days for 91/2 ounces. My answer was correct, but discovered in a different way than the answer. As I've learned with my GT students...there is more than one way to get an answer.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-1509799006346376000.post-15478113324487145862010-07-26T13:32:41.470-05:002010-07-26T13:32:41.470-05:00Session 3-Question 2
I love Einstein! He has take...Session 3-Question 2<br /><br />I love Einstein! He has taken on a whole new meaning this assignment; however, it makes me question: do our young PGP students know anything about Albert Einstein? Is he only a cute illustration that gives them great tips? <br />I feel the children need to be introduced to this great mind in the form of a short biography or fact-finding mission, so they will be on a more “personal” basis with him as they work with him in this tome.<br />Einstein Problems<br />Chapter3 Problem 1<br />Per Einstein’s rhyme about the number of days in each month, December having 31 days, each day having 24 hours,<br />24 x 31= 744 hours in December x’s $00.15 an hour for electricity to heat Luke’s room =’s 744 x .15= $111.60 to heat his room the month of December.<br />Chapter12 Problem 3<br />Per Einstein’s advice concerning the number of months in each of the seasons (3), Megan could figure out how many students in her school could have spring birthdays.<br />Number of students, 288 ÷ 3= 96 students who could possibly have spring birthdays.NanetteGnoreply@blogger.comtag:blogger.com,1999:blog-1509799006346376000.post-54690350406607362212010-07-22T11:38:16.149-05:002010-07-22T11:38:16.149-05:00I played with the Algebra problems since Algebra i...I played with the Algebra problems since Algebra is not my forte. New Math made higher mathematics very difficult since those of us who were subjected to it never got the correct foundation. I like the way the author explains that you have to turn algebra into sentances. <br /><br />p. 220 1/2n-8 =20 1/2n -8 = 20<br /> +8 + +8<br /> 1/2n = 28<br /> n=56<br /><br />p. 305 Carly: n+9<br /> Stacie: n+5<br /> Ben =n<br />Equation: 3n+14= 17<br /> -14 -14<br /> 3n = 3<br /> n = 1Of Life, Education, E-bay, Travel & Bookshttps://www.blogger.com/profile/17333382268292453140noreply@blogger.comtag:blogger.com,1999:blog-1509799006346376000.post-7752573198557089872010-07-22T09:19:31.506-05:002010-07-22T09:19:31.506-05:00Chapter 20: I Know a Math Language, p. 209, Einste...Chapter 20: I Know a Math Language, p. 209, Einstein Level, #1<br />Rick rode in his toy car for n minutes. To find out how many hours he rode, I knew to multiply n x 60 because there are 60 minutes in an hour. So, the answer is--> Hours: n x 60. <br /><br />Chapter 23: Radical and Squaring Machines, p. 244, Einstein Level, #4<br />To find the Mystery Number, I worked backwards by first finding the square root of 81 using either a calculator or previous knowledge that 9 squared equals 81. The second step was to take 9 and continue working backwards to find its square root by using a calculator or having previous knowledge that 3 squared, or 3 times 3, equals 9. The Mystery Number is therefore 3. Students will love the challenge of reversing what they know and applying it to this problem and possibly using a calculator app on an iTouch or iPad to find the square roots needed for the solution.FMoorenoreply@blogger.comtag:blogger.com,1999:blog-1509799006346376000.post-81590161822172540022010-07-22T07:53:24.974-05:002010-07-22T07:53:24.974-05:00Millie, you go girl! Your solution is a perfect ex...Millie, you go girl! Your solution is a perfect example of how students can arrive at the correct answer using another avenue…yours being the much shorter one!FMoorenoreply@blogger.comtag:blogger.com,1999:blog-1509799006346376000.post-23011688322695188702010-07-21T21:58:00.157-05:002010-07-21T21:58:00.157-05:00I solved problem #2 on page 220. First, per Einst...I solved problem #2 on page 220. First, per Einstein’s instructions, I combined all the things that were alike (1/4n+1/4n+1/4n = 3/4n and 8+5=13) which gave me 3/4n + 13 = 16. Then I tried to get the n alone on one side of the equation by doing the same thing to each side to be fair like Einstein stressed. So, I subtracted 13 from each side of the equation to get 3/4n = 3. Then I multiplied each side by 4 to get 3n = 12 and divided each side by 3 to get n = 4. <br /><br />The second problem I solved was #4 on page 266. I labeled the three tires A, B and C. Then I figured Phil could ride with a combination of tires A + B for 4 miles, then switch to tires B + C for 4 more miles and finally tires A + C for 4 miles. Therefore, he’s travelled a total of 12 miles and each tire has been used for 8 miles (one third of the 24 total miles that a pair of tires goes in 12 miles). <br /><br />I had a question – please let me know if I’m missing something. There is an example problem solved on page 258 that shows there are 5 steps needed to fill a 5 quart container with 3 quarts of water using only unmarked 5 and 2 quart containers. Wouldn’t it be possible to solve this problem in just one step by filling the 5 quart container full and emptying it into the 2 quart container, thereby leaving 2 quarts in the 2 quart container and the desired 3 quarts in the 5 quart container??MillieHnoreply@blogger.com