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Problem Solving Strategies

Problem Solving Strategies

Yunianto
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Problem Solving Strategies

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  1. Problem Solving Strategies UmmySalmah Wahid Yunianto

  2. Problem Solving Strategies Organizing Data Logical Reasoning

  3. Organizing Data(Mengorganisasi Data) Everyday Life Problem-Solving Situation Shopping Trip Traveling Plan

  4. Organizing Data(Mengorganisasi Data) Mathematics Problem • 72,43,98,57,87,89,67,23,56,89,91,88,72,75,66 • 23,43,56,57,66,67,72,72,75,87,88,88,89,91,98 Suppose we are asked to find the median score for the following group of 15 test scores:

  5. Problem 1 Find the sum of the terms in the series 202- 192+ 182- 172+ 162- 152+ ... + 42- 32+ 22- 12

  6. Solution Using the problem-solving strategy of organizing data, we can view this series as: (202- 192) + (182- 172) + (162 - 152) + ... +(42- 32) + (22- 12) Factoring each pair of parentheses: (20 - 19)(20 + 19) + (18 - 17)(18 + 17) + (16 - 15)(16 + 15) + ... + (4 - 3)( 4 + 3) + (2 - 1)(2 + 1) = (1)(20 + 19) + (1)(18 + 17) + (1)(16 + 15) + ... + (1)(4 + 3) + (1)(2 + 1) = 20 + 19 + 18 + 17 + ... + 4 + 3 + 2 + 1 = 210

  7. Problem 2 Four married couples belong to a theater club. The wives’ names are Alice, Barbara, Christa, and Edith. The husbands' names are AI, Frank, Fred, and Ernest. Who is married to whom? Use the following clues to determine the couples. AI is Edith's brother. Edith and Fred were once engaged, but "broke up" when Edith met her present husband. Christa has a sister, but her husband is an only child. Alice is married to Ernest.

  8. Solution Alice Christa Barbara Edith Al X X YA X Frank X X X YA Fred X YA X X Ernest YA X X X

  9. Problem 3 A E G F H K B C D How many triangles are in figure below?

  10. Solution: We have 9 new triangles (FBH, AFC, BHC,AFK, KDC, AKC, FBC,HKG, EHC) We have 5 new triangles (ABG, BGD, AGE, BEC, ABE) We have exactly 1 triangle (ABC) A There are 17 triangles E G F H K B C D We have 2 new triangles (ABD, ABC)

  11. Problem 4 125,250 Given the sequence of integers 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, . . ., where each positive integer, n, occurs in a grouping of n consecutive terms, how many terms are needed so that the sum of the reciprocals is 500?

  12. Problem 5 Each of the 10 court jewelers gave the king's adviser, Mr. Pogner, a stack of gold coins. Each stack contained 10 coins. The real coins weighed exactly 1 oz each. However, one and only one stack contained "light" coins, each having had exactly 0.1 oz of gold shaved off the edge. Mr. Pogner wishes to identify the crooked jeweler and the stack of light coins with just one single weighing on a scale. How can he do this?

  13. Solution: Let us try to solve the problem by organizing the data in a different fashion. We must find a method to vary the deficiency in a way that permits us to identify the stack from which the counterfeit coins are taken. Label the stacks 1, 2, 3, 4,...,9, 10. Now, take one coin from stack 1, two coins from stack 2, three coins from stack 3, four coins from stack 4, and so on. We now have a total of 1 + 2 + 3 + 4 + ... + 8 + 9 + 10 = 55 coins. If they were all true, the total weight would be 55 ounces. If the deficiency is 0.5 oz, then there were 5 light coins, taken from stack 5. If the deficiency is 0.7 oz then there were 7 light coins, taken from stack 7 and so on. Thus, Mr. Pogner can readily identify the stack of light coins and, consequently, the jeweler who had shaved each coin.

  14. Problem 6 The basement of an apartment building has six tanks of oil. A measuring gauge for the oil stored in the basement is set up to measure five tanks at a time. Following are the results of these measurements: Without tank A there were 2,000 liters of oil. Without tank B there were 2,200 liters of oil. Without tank C there were 2,400 liters of oil. Without tank D there were 2,600 liters of oil. Without tank E there were 2,800 liters of oil. Without tank F there were 3,000 liters of oil. How much oil is there in each tank?

  15. Solution: B + C + D + E + F = 2,000 A + C + D + E + F = 2,200 A + B + D + E + F = 2,400 A + B + C + E + F = 2,600 A + B + C + D + F = 2,800 A + B + C + D + E = 3,000

  16. 5A+5B+5C+5D+5E+5F = 15,000 A+B+C+D+E+F = 3,000

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