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Heuristics

Solving Word Problems. Using. Heuristics. What is Heuristics ?. Thinking skills are skills used in solving mathematical problems, such as classifying, comparing, sequencing, analysing parts and wholes, identifying patterns and relationships, induction, deduction, and spatial visualisation.

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Heuristics

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  1. Solving Word Problems Using Heuristics

  2. What is Heuristics ? Thinking skills are skills used in solving mathematical problems, such as classifying, comparing, sequencing, analysing parts and wholes, identifying patterns and relationships, induction, deduction, and spatial visualisation. Examples of heuristics are drawing a diagram, making a list, using equations, using guess-and-check, looking for patterns, making suppositions, acting it out, working backwards, working before-after, re-stating the problems, simplifying the problem and solving part of the problem.

  3. Making Suppositions – Mr Zanizam • Working Backwards – Mr Long • Working Before and After – Mr Joseph Lim • Re-stating the problem/Simplifying the problem – Mrs Elizabeth Simon

  4. Making Suppositions In a coop, there are 6 spiders and chickens. If there are 36 legs in all, find the number of chickens in the coop. Step 1 – Assume all animals are chickens. 6 chickens’ legs  2 × 6 = 12 Step 2 – Find the difference in the number of legs between each pair of animals. 8 – 2 = 6 Why do we need to find the difference? Since spiders have more legs than chickens, it is easier to “hide” their legs and assume all animals are chickens. Therefore, each spider is hiding 6 legs.

  5. Making Suppositions In a coop, there are 6 spiders and chickens. If there are 36 legs in all, find the number of chickens in the coop. Step 3 – Find the difference between the number of chickens’ legs and the total number of legs of the two animals. 36 – 12 = 24  (hidden spiders’ legs) Step 4 – Find the number of spiders. 24 ÷ 6 = 4 Step 5 – Find the number of chickens. 6 – 4 = 2 Checking: 4 × 8 = 32 (spiders’ legs) 2 × 2 = 4 (chickens’ legs) 32 + 4 = 36

  6. Making Suppositions There were 15 Math Olympiad questions. 5 marks were awarded for every correct answer and for each wrong answer, 1 mark was deducted. Sam scored 33 marks. How many questions did he answer correctly? Assuming all questions  5 x 15 answered correctly = 75 Difference in marks 75 – 33 = 42 Marks deducted for each  5 + 1 wrong answer = 6 No. of wrong answers  42 ÷ 6 = 7 No. of correct answers  15 – 7 = 8

  7. Patrick had a sum of money. On the first day, he spent of the money and donated $30 to charity. On the second day, he spent of the money he still had and donated $20 to charity. On the third day, he spent of the money he still had and donated $10 to charity. In the end, he had $10 left. How much did he have at first? sdrawkcab gnikrow

  8. Patrick had a sum of money. On the first day, he spent of the money and donated $30 to charity. On the second day, he spent of the money he still had and donated $20 to charity. 30 On the third day, he spent of the money he still had and donated $10 to charity. spent d 120 ÷ 3 = 40 90 + 30 = 120 60 ÷ 2 = 30 40 + 20 = 60 30 + 60 = 90 1st Day 40 2nd Day 20 3rd Day spent d 10 + 10 = 20 10 10 2 × 20 = 40 spent d left In the end, he had $10 left. How much did he have at first? 40 × 4 = 160 He had $160 at first. Checking: 160 ÷ 4 = 40 (spent on 1st day) 160 – 40 – 30 = 90 (left after 1st day) 90 ÷ 3 = 30 (spent on 2nd day) 90 – 30 – 20 = 40 (left after 1st day) 40 ÷ 2 = 20 (spent on 3rd day) 20 – 10 = 10 (left after 3rd day)

  9. Example 2 John had some books. He gave May half of the books plus 1 book. He gave Mike half of the remaining books plus 2 books. If he had 5 books left after this, how many books had John at first?

  10. John had some books. He gave May half of the books plus 1 book. 7 × 2 = 14 14 + 1 = 15 2 1 Mike May 2 + 5 = 7 He gave Mike half of the remaining books plus 2 books. If he had 5 books left after this, how many books had John at first? 5 15 × 2 = 30 He had 30 books at first. Checking: 30 ÷ 2 + 1 = 16 (given to May) 30 – 16 = 14 (left after giving to May) 14 ÷ 2 + 2 = 9 (given to Mike) 14 – 9 = 5 (left after giving to Mike)

  11. Working Before and After

  12. Before and AfterExample 1 Clement had thrice as many apples as pears at his stall at first. After he sold 129 apples and threw away 8 rotten pears, he had only half as many apples as pears left. How many pears had he left?

  13. Clement had thrice as many apples as pears at his stall at first. After Before 1 part 1 part 1 part Apples 129 sold apples 8 8 Apples 8 pears thrown away Pears left Pears 1 part After he sold 129 apples and threw away 8 rotten pears, he had only half as many apples as pears left. How many pears had he left?

  14. Before 1u 1u 8 8 1u 1u 8 8 1u 1u 8 8 Apples 1u 1u Pears 8 thrown away After Apples 129 sold 1u 1u 1u 8 thrown away Pears left 8 × 3 = 24 5 units  129 – 24 = 105 1 unit  105 ÷ 5 = 21 2 units  21 × 2 = 42 He had 42 pears left. Question: How many pears had he left?

  15. Before and AfterExample 2 James had twice as many pens as erasers at first. After he gave away 170 pens and 10 erasers, he had thrice as many erasers as pens left. How many pens had he at first?

  16. James had twice as many pens as erasers at first. Before 1 part 1 part After Pens Pens 170 given away Erasers 10 lost away 1 part Erasers left After he gave away 170 pens and 10 erasers, he had thrice as many erasers as pens left. How many pens had he at first?

  17. Before 1u 1u 8 1u 10 1u 1u 1u 10 Pens Erasers 10 lost away After 1u Pens 170 given away 1u 1u 1u 10 lost away Erasers left 10 × 2 = 20 5 units  170 – 20 = 150 1 unit  150 ÷ 5 = 30 6 units  30 × 6 = 180 180 + 10 + 10 = 200 He had 200 pens at first. Question: How many pens had he at first?

  18. RE-STATING AND SIMPLIFYING THE PROBLEM

  19. Aric has some 10-cent coins. Cathy has some 20-cent coins. Both have the same number of coins. Cathy has $1.20 more than Aric. How many coins does each of them have? PROBLEM 1

  20. Aric has some 10-cent coins. Cathy has some 20-cent coins. Both have the same number of coins. Cathy has $1.20 more than Aric. How many coins does each of them have? (A)10¢ No. of coins (C)20¢ Value of coins (A) 10¢ (C) 20¢ $1.20

  21. Aric has some 10-cent coins. Cathy has some 20-cent coins. Both have the same number of coins. Cathy has $1.20 more than Aric. How many coins does each of them have? What is the difference between them? 20¢ – 10¢ = 10¢ 10¢ 20¢

  22. Aric has some 10-cent coins. Cathy has some 20-cent coins. Both have the same number of coins. Cathy has $1.20 more than Aric. How many coins does each of them have? 10¢ Value of coins $1.20 20¢ 20¢ – 10¢ = 10¢ 120¢ ÷ 10¢ = 12 (sets of 10¢ and 20¢) Each of them has 12 coins.

  23. Devi had a number of 10-cent coins and Bala had a number of 20-cent coins. Devi had 2 more coins than Bala but Bala had $0.40 more than Devi.Find the amount of money that Devi had. PROBLEM 2

  24. Devi had a number of 10-cent coins and Bala had a number of 20-cent coins. Devi had 2 more coins than Bala but Bala had $0.40 more than Devi.Find the amount of money that Devi had. 2 (D)10¢ No. of coins (B)20¢ Value of coins (D)10¢ (B)20¢ $0.40

  25. Devi had a number of 10-cent coins and Bala had a number of 20-cent coins. Devi had 2 more coins than Bala but Bala had $0.40 more than Devi.Find the amount of money that Devi had. 2 10¢ No. of coins 20¢ The 2 10-cent coins will give a value of 2 × $0.10 = $0.20

  26. $0.20 Devi had a number of 10-cent coins and Bala had a number of 20-cent coins. Devi had 2 more coins than Bala but Bala had $0.40 more than Devi.Find the amount of money that Devi had. The value of the extra 2 coins can be found in the model below. (D)10¢ Value of coins $0.20 (B) 20¢ $0.20 $0.40 To make the no. of coins the same, we remove the 2 coins from 10¢ model. (D)10¢ Value of coins (B)20¢ $0.20 $0.40

  27. $0.20 Devi had a number of 10-cent coins and Bala had a number of 20-cent coins. Devi had 2 more coins than Bala but Bala had $0.40 more than Devi.Find the amount of money that Devi had. (D) 10¢ Value of coins (B) 20¢ $0.20 $0.40 $0.20 + $0.40 = $0.60 20¢ – 10¢ = 10¢ 60¢ ÷ 10¢ = 6 (sets of 10¢ and 20¢)

  28. Devi had a number of 10-cent coins and Bala had a number of 20-cent coins. Devi had 2 more coins than Bala but Bala had $0.40 more than Devi.Find the amount of money that Devi had. 2 × $0.10 = $0.20 $0.20 + $0.40 = $0.60 20¢ – 10¢ = 10¢ 60¢ ÷ 10¢ = 6 (sets of 10¢ and 20¢) 6 + 2 = 8 (No. of coins Devi has) 8 × $0.10 = $0.80 (Devi)

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