This is a professional development blog. We'll be discussing Challenge Math by Ed Zaccaro. Our discussions will be focused on gifted children.

Tuesday, July 20, 2010

Session 4 - Question 3

Complete three of the "Math Contest" questions from different contests, located on pages 307-326, by using the author’s information, and share your experience. Site the page number and the question number in your answer.


  1. Math Contest 1, p. 307, #3: You are looking for the area of the shaded part. Given that the area of the sqare is 144 sq. in., you know that each side has a length of 12 inches. That means that your diameter for the circle is 12 in. That helps you find the area for the circle since a circle's area is pi x radius squared. Your radius is 1/2 the diameter which gives a radius of 6 in. in this scenario. So, 3.14 (which is pi) x (6 squared= 36)gives a product of 113.04 for the circle's area. You then need to find the area of the triangle b/c it takes up part of the circle's area which means that 113.04 is not your final answer. Aagain, knowing that the square's sides have a value of 12 in., that gives you the base of your triangle. Using the formula for a traingle's area a= (b x h)/2, base(12) x height (6)= 72 and then divide by 2. the triangle's area is 36. To find the area for the shaded part, you must subtract the triangle's area (36) from the circle's area (113.04) and you get a difference of 77.04 which is the area for the shaded part.
    Math Contest 2, p. 308, #4: You are finding the length of the missing side of the triangle. You have a 30 degree angle and an opposite side of 8.661 in. This will be used to find the missing side's length (side b). Use the formula: [side a (8.661)/b] = [tangent of 30 degrees (.5774)/1]. That will give you the equation of .5774b = 8.661. Isolate the variable so you have b = 8.661/.5774. Side b has the value of 15 in.
    Math Contest 3, p. 309, #3: How many microns are in 8 meters? Use the given info.: a micron equals 0.001 of a millimeter, and your background knowledge of 1000 mm eual a meter and 8000 mm equal 8 meters. Your formula requires division since you are going from a smaller unit to a larger one. You will take the 8000 mm and divide them by the micron (0.001). 8000/0.001 = 8 million microns. There are 8 million microns in 8 meters.

  2. Page 307 #5: I had to find the price of a computer on sale. I used multiplication and then subtraction to figure out the final price. This was a fairly straightforward problem.
    Page 309 #5. I set up the problem as a fraction, and did cross multiplication to figure out how much orange juice I needed to have.
    Page 313 #1. I used a table and made a function machine, to figure out the 500th term in the given sequence. Once I figured out the pattern between the given numbers, I could make an equation to come up with the final answer, 3,493.

  3. Problem #1, p.309,(pp.37-42, Algebra): Let Michelle's age = m. Then Bob's age = 4m, and Sarah's age = 2m. m + 4m + 2m = 84 (the sum of all three ages). 7m = 84. Divide both sides of the equation by 7. Therefore, m = 12. Michelle is 12 years old.
    Problem #4, p.317, (p.152, Ratios and Proportions): Write a proportion, 28/1.28 = 3164/x. The scale factor is 113 (3164 divided by 28). So, the cost of the gasoline for the trip is $144.64 (1.28 times 113).
    Problem #4, p.325 (p.249, Simultaneous Equations): Add the two equations to eliminate 'x'; 15y = 30; y = 2; x = 11; substitute for 'x' and 'y' to check.

  4. I did pg 307 number 4. I used the section on algebra to arrive at my answer of the bat costing $9.50. I also did pg 308 number 5. I used the section on volume and fractions to help arrive at my answer of the volume of the quarter being .043 inches cubed. Finally, I did pg 309 number 1. I also used the algebra section to help set up the equation and arrive at my answer that Michelle age is 12. I also had to use a little guess and check for that one by setting up a table…the non-gt way!

  5. nlopez in response to PKassir on 7/22/10- Page 307 #5: was a fairly straightforward problem. I was actually surprised that such a "simple" problem was included here, but hey, we need all levels and it gives us a real world situation that this could be applied to.

  6. nlopez in response to tatumt on 7/23/10- Problem #1, p.309,(pp.37-42, Algebra): Thank you for such a simple, straightforward way to solve this. Your solution made me think that I will use your example when I am covering variables this year.

  7. Contest #9: problem 4: I selected this problem because I loved teaching this when I coached a MathCounts program. The best way to "count" the squares is with an organized list:
    one 6 by 6; four 5 by 5; nine 4 by 4; sixteen 3 by 3, twenty-five 2 by 2; and thirty-six 1 by 1.
    Quickly the students figure out that all they need to do is add the first six perfect squares to get 91.

    Contest #10: problem 5: I enjoy these types of logic problems. I wrote three equations initially with the names and then started substituting much the same way the author did in his hamster/dog/turtle/cat question. I came up with 15 years for Sara.

    Contest #11: problem 1: I'm not sure what approach the author would have recommended here. I guess I could have made a table where I substituted the different values of x until I got the same results. I found it much faster to use my knowledge of exponenets to write an equation with x on both sides and solve to get x = 8.

  8. I solved problem #5 on page 307,it was finding the new cost after 45% percentage off. This something I am not to terribly good at when I am at the store, but now I can do it!!!
    #1 on page 309, I had to find Michelle's age based on the facts about the other people in the problem. I struggles with this a couple of times in the Einstein level, but was able to solve this one fairly quickly. I do like these kind of problems, but I have to be in the right thinking mode to solve them.
    #4 on p. 317, You had to find the cost of gas for a whole trip. This one challenged me the most because there were several steps involved and several problems that had to be combined together to find the final answer.
    Overall I enjoyed solving and challenging myself with these math problems here and throughout the whole book.

  9. Math contest #2 question #3--Each die has 6 sides so we need to multiply 6 x6x6x6 = 1296. Since you only have one shot at rolling all ones the answer is 1/1296.
    Math contest #3 question #5--If you have 5 oz of OJ and 7oz of MW you need to figure out what part 5 is 7 so divide 5 by 7. This gets you .7142... Now you take .7142 and multiply by 134.75, your answer is 96.249 or 96.25
    Math contest # 9 question #3 For this problem I created a progression chart to show the 7X relationship between the nickels and quarters. When you reach 77 quarters and 11 nickels it equals 19.80

  10. In response to nlopez in response to PKassir on 7/22/10- Page 307 #5: I was also surprised at how easy that problem was--all of the problems I did could be easily included for my regular students.--Not as challenging as the other problems in the book.

  11. In response to kohlerj Contest #10: problem 5: I love these problems also. The students enjoy using the guess and check method for these types of problems. The dino pic next to it is prcious.

  12. I actually randomly selected these problems, and they turned out to be less than challenging then other problems I have solved previously.

    Contest 7 page 313 #2. Solved using algebra. 9x/12=36… x=48. I did not do a lot with variables last year, but I definitely want to introduce them to students who understand them next year. I feel that they can make math problems much simpler, even though it does create a bit of temporary abstract.

    Contest 7 page 313 #1 Patterns- the pattern is the previous numbered term*7. So the 500th term would be 499*7=3493.

    Contest 1 page 307 #2. I used a proportion. 5/x=6/88.5 and cross multiplied. The oak tree is 73.75 feet tall.

    Contest 1 page 307 #5. Percents- 45% of 80 is $36. $80-$36=$44.

    Again, I think students would enjoy working independently or in partners/small groups to solve these problems. The varying level of rigor allows children to collaborate and dig deeper into their knowledge of the concepts.

  13. In response to RatliffB on 7/28, I agree that those pesky percent problems can be tricky, especially when pricing sale items. In mental math, I find out what 10% is, then multiply by 4 and add 1/2 of the 10%. That sounds complicated... i.e.- 45% of 80 is 4*8=32+4=36. $36 is the savings. 80-36=44.... ok so it might be complicated, but the sales are usually just 10 or 20% :-)

  14. The first math contest problem I did was on p.308, contest #2 and problem #3. I chose this problem as it was similar to an Einstein problem I had done before and I applied the same strategy and then checked my answer with a calculator. I was building on a past challenge and success since I had to work through the previous problem and ask for ideas from others.
    The second math contest problem I did was on p.315, math contest #9 and problem # 3. This is a type of problem that my fourth graders encounter each year and I thought it was pertinent to try this type of problem again.
    I did it on my own, and then went back to the corresponding examples provided in the book to check to see if I was following those steps and arrived at the correct answer. Prior to this my students and I would have used "guess and check" which still works but can be time intensive.
    My third problem is on p. 317, math contest #11 and problem # 1.
    Again I went back to where this type of problem was presented and used those strategies as examples,plugging in the information from this problem to arrive at the answer. Having the graphics and examples
    were reassuring and easy to follow.

  15. On page 307: #4-A bat and a ball cost $10, with a $9 difference. What is cost of the bat? I solved by subtracting 9 from 10 which left me with a dollar. So the ball is $0.50 and the bat is that plus the 9 so, the bat is $9.50.

    Page 307: #5 A computer is $80 with 45% off, what is the new price? I multiplied 80 x .45 and got 36. I subtracted 36 from 80 to get the sale price of $44.

    Page 317: #1: Suppose six days before the day before yesterday is Sat. What day of the week is tomorrow? The day before the day before yesterday being Saturday would mean the day would be a Monday. Which means today is Sunday and tomorrow will be Monday.

  16. I completed:
    Contest #1, pg 307, # 2- Finding the tree’s shadow height required me to set up a proportion, solve the proportion through cross-multiplication, and then use algebra to solve x = 73.75 feet. Additionally, I used long division with decimals. There was a wealth of math going on in this problem, great for the kids!
    Contest #2, pg 308, #1- Finding out how long it will take Kristin to read the pages required me to understand units, to convert mixed numbers into decimals, to properly set up for long division, and to divide with decimals through hundredths. I could have also solved this problem by first converting into improper fractions.
    Contest #3, pg 309, #1- Finding Michelle’s age required me to understand algebra. I needed to use information to accurately assign 3 variables and to set up an accurate equation, to understand substitution, to know how to add fractions with non-common denominators (with variables as the numerators), and to reread the question and use substitution to find “m” instead of b!

  17. Math Contest 2 p.308,#2-Find the cost of the guitar before the tax was added required the use of algebra. I first set up the problem by trying to divide both sides by .05 but my answer did make sense. I then multiplied and subtracted from the total cost to get the price of the guitar.

    Math contest 6 p.312, #2- to find the area of the shaded part I had to use the formula for the area of a triangle and the area of circle. Since only the diameter was given I had to divide up the sqaure into trangles to figure out the area of the square and then find the area of the circle. After I found the two areas I subtracted to find the area of the shaded region which was 18.24 square inches.

    Math contest 19 p.325, #4- This required using elimination to find the value of x and y in two equations. I first found the value of y which was 2 and then I substituted the value of y in the second equation to find the value of x which was 11.

  18. On Math Contest 1 question #4, I used a chart and table to organize the information and eliminate possible scenarios that didn't fit the problem. The bat cost $9.50 while the ball is 50Cent which means that both are $10.

    Question #5- I first figured out what 10% of $80 and then created a table to calculate 45% of 80 would be $36. The new price for the computer would be $44.

    I also did question #2 by using cross multiplication in finding 6X= 442.5 then used division to find that the oak tree was 73.75 feet tall.

  19. 1. Math Contest 8 #2: The area of unshaded part of the square is 64 - area of 1/4 of the circle. If area of the square is 64 square inches, then one side or the radius of circle is 8 square inches. If radius is 8, then the area of 1/4 of the circle is pi*8 squared/ 4= 50.24 square inches.
    Unshaded area= 64 - 50.24= 13.76 square inches
    2. Math Contest 9 #1:
    Sixteen workers finished half of the road in 10 days. 4 more workers are added to the crew, how long will it take them to finish the road?
    Half of the road took 16 x 10= 160
    The other half will also be 160 but the worker number will be 20. So 160/20= 8 days.
    3. Math Contest 11 #5: The height and base of a triangle is given, perimeter is asked. Since height of the triangle is perpendicular to the base we can use Pythagorean theorem to find the legs of the triangle. The right triangles formed by the height and base are 3,4,5 triangles. 3 and 4 being the legs, 5 is the hypotenuse. Or
    3 squared + 4 squared = 9 +16 = 25
    square root of 25 is 5.
    Since each leg of the triangle is 5 and base is 6, perimeter is 5+5+6=16 inches.

  20. In response to CKohl on July 29th, I agree that contest problems are less challenging than the Einstein problems. I think Zaccaro wanted to build a deeper understanding with Einstein problems and a speedy review with contest problems.

  21. Math Contest 1 p. 307 #3. You need to find the area of the shaded part-circle-excluding the triangle. Given the area of the square is 144 sq, then length of each side is 12 sq in. This means that the radius of the circle is 6 sq. After that, I found the area of the circle by multiplying 3.14(pi) by radius(6) square is a total of 113.04 sq inch. Next, we need to find the area of the triangle to exclude it from the area of the circle. The area of a triangle is (base x height) divided by 2. The base of the triangle is (12) x the height (6) =72/2=36 sq for the area of the triangle.
    Now we subtract the area of the circle including the triangle 113.04 minus the area of the triangle 36 = 77.04 which is the area of just the shaded part.
    This could be a somewhat challenging problem for the gt student because it encompasses three geometric figures and you have to remember the formulas to find their areas as well as the formulas to find their missing sides, and keep in track that you must subtract what is not being asked for.
    Math Contest 2 p. 308#2. You are asked to find the cost of the guitar before the tax was added. This was an algebra problem using variables.
    I wrote the equation x + .05x = $152.25
    1.05 x = 152.25
    152.25/1.05 = $145
    The guitar cost $145 before taxes.
    This was a fairly easy problem that students could immediately use it during their daily lives.
    Math Contes 13 p. 319 #1. Great problem for 4th grade probabilities.
    You are asked to find the probability that a couple will have all girls if they have 5 children. Each baby has a 1 in 2 chance of being a girl. So you multiply ½ x ½ x ½ x ½ x ½ =1/32 is the probability

  22. Math Contest 1, page 307, question 4, initially seemed to me to be algebraic and perhaps with different numbers I might have needed to use algebra. However, for this one, i basically reasoned it through after reading it. That bat has to be more than $9, otherwise the ball cost nothing since it has to be $9 less than the bat. So, the bat is $9.50 and the ball $0.50 for a total of $10.

    Math Contest 2, page 308, question 1 solved very easily by converting the fractions to decimals. Once I set it up as a division problem, I realized that 1.25 x 100 would be 125, which is 1.25 more than 123.75, making the answer 99.

    Math Contest 3, page 309, question 2 is the type of problem I like to pose to my students once they master basic area of rectangles in order to make a fairly dimple, straightforward concept a little more challenging and thought provoking. For this problem, I found the area of the whole outer triangle using 1/2bh, or 1/2 x 14 x 8 = 56 sq in, then subtracted the area of the missing rectangular piece, 3 x 4 = 12 sq. in, for an area of 44 sq. in for the shaded portion.

    I don't know if I was just more focused or what while completing these, but they did seem much easier than the Einstein level problems.

  23. I have chosen Math contest 5 question 2
    Because the bike is talking about 15 inch diameter I want to convert the miles to inches so 5280 x 12 =63360 inches in a mile.
    The circumference of the tire is then 3.14 x 15= 47.1 inches
    Then 63360 ÷47.1 = 1345

  24. I also like contest 4 number 4 on p 310
    If the circle diameter is 5 inches the radius is 2.5. Making the area of the circle 3.14 x 2.5 x2.5 = 19.625
    This makes the rectangle 5 inches wide and 10 inches long and the area would be 50 square inches.
    The triangle is then 5 x 5 x ½ = 12.5
    The shaded area is rectangle - circle and minus the triangle
    50 - 19.625 - 12.5 = 17.875 square inches
    Last I worked contest 20 number 5 on page 326 In a state with the sales tax of 6% a book cost 19.61 including taxs What is the price of the book before tax was added?
    This reminded me of the Singapore math so I drew a picture

    The entire book cost 106 %
    So n + .06n =1.06n
    And 1.06n = 19.61
    N = 18.50 with tax being 1.11

  25. I agree with
    “susanm said...
    I don't know if I was just more focused or what while completing these, but they did seem much easier than the Einstein level problems. “
    Here at the end of the summer the contest seems easier than the challenge.