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Proudly Presents Heuristics Approach Solving Challenging Primary Mathematical Problems

Adrian Ng Principal Trainer. Proudly Presents Heuristics Approach Solving Challenging Primary Mathematical Problems. Guess and Check ( 3 guesses + Look for pattern ).

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Proudly Presents Heuristics Approach Solving Challenging Primary Mathematical Problems

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  1. Adrian Ng Principal Trainer Proudly Presents Heuristics Approach Solving Challenging Primary Mathematical Problems

  2. Guess and Check(3 guesses + Look for pattern)

  3. String of 2 big balloons is 90cm. String of 5 small balloons is 1.2m. If both strings are of the same length, there would be 105 more small balloons than the big balloons. How many balloons are there altogether? 2009 PSLE question Big (90/2=45cm)

  4. String of 2 big balloons is 90cm. String of 5 small balloons is 1.2m. If both strings are of the same length, there would be 105 more small balloons than the big balloons. How many balloons are there altogether? 2009 PSLE question Big (90/2=45cm) Small (120/5=24cm)

  5. String of 2 big balloons is 90cm. String of 5 small balloons is 1.2m. If both strings are of the same length, there would be 105 more small balloons than the big balloons. How many balloons are there altogether? 2009 PSLE question Big (90/2=45cm) Small (120/5=24cm) Diff ( 0 )

  6. String of 2 big balloons is 90cm. String of 5 small balloons is 1.2m. If both strings are of the same length, there would be 105 more small balloons than the big balloons. How many balloons are there altogether? 2009 PSLE question Big (90/2=45cm) Small (120/5=24cm) Diff ( 0 ) no. length no. length 2520-0= 2520 0 45x0=0 105 105x24=2520

  7. String of 2 big balloons is 90cm. String of 5 small balloons is 1.2m. If both strings are of the same length, there would be 105 more small balloons than the big balloons. How many balloons are there altogether? 2009 PSLE question Big (90/2=45cm) Small (120/5=24cm) Diff ( 0 ) no. length no. length 2520-0= 2520 0 45x0=0 105 105x24=2520 2520-2499= 21 (pattern) +120 2544-45= 2499 1 45x1=45 106 106x24=2544 Gap/pattern 2520/21=120 2520-0= 2520 (gap) 0+120= 120 120x45=5400 120+105= 225 225x24=5400 ( 0 )

  8. String of 2 big balloons is 90cm. String of 5 small balloons is 1.2m. If both strings are of the same length, there would be 105 more small balloons than the big balloons. How many balloons are there altogether? 2009 PSLE question Big (90/2=45cm) Small (120/5=24cm) Diff ( 0 ) no. length no. length 2520-0= 2520 0 45x0=0 105 105x24=2520 2520-2499= 21 (pattern) +120 2544-45= 2499 1 45x1=45 106 106x24=2544 Gap/pattern 2520/21=120 2520-0= 2520 (gap) 0+120= 120 120x45=5400 120+105= 225 225x24=5400 ( 0 ) 120+225=345 Ans: 345 balloons

  9. Note: This question was provided by students who have sat for the 2009 examination. It may vary from the actual examination question.

  10. Systematic listing

  11. Six friends decided to rent computers from 2.00pm to 4.30pm. Four of them were playing while the other two would watch. If the cycle continues, and each of them played for equal number of minutes, how many minutes will each person get to play? 2009 PSLE question Friends Group 1 1 1 1 1 1 1 2 1 1 3 1 1 1 1 1 1 1 4 1 1 1 1 5 1 6 1 1 1 1 4 4 4 4 4 4 Total (each)

  12. Six friends decided to rent computers from 2.00pm to 4.30pm. Four of them were playing while the other two would watch. If the cycle continues, and each of them played for equal number of minutes, how many minutes will each person get to play? 2009 PSLE question Friends Group 1 1 1 1 1 1 1 2 1 1 3 1 1 1 1 1 1 1 4 1 1 1 1 5 1 6 1 1 1 1 4 4 4 4 4 4 Total (each) 2.00pm to 4.30pm  150min

  13. Six friends decided to rent computers from 2.00pm to 4.30pm. Four of them were playing while the other two would watch. If the cycle continues, and each of them played for equal number of minutes, how many minutes will each person get to play? 2009 PSLE question Friends Group 1 1 1 1 1 1 1 2 1 1 3 1 1 1 1 1 1 1 4 1 1 1 1 5 1 6 1 1 1 1 4 4 4 4 4 4 Total (each) 2.00pm to 4.30pm  150min 150min/6 = 25 min per group

  14. Six friends decided to rent computers from 2.00pm to 4.30pm. Four of them were playing while the other two would watch. If the cycle continues, and each of them played for equal number of minutes, how many minutes will each person get to play? 2009 PSLE question Friends Group 1 1 1 1 1 1 1 2 1 1 3 1 1 1 1 1 1 1 4 1 1 1 1 5 1 6 1 1 1 1 4 4 4 4 4 4 Total (each) 2.00pm to 4.30pm  150min 150min/6 = 25 min per group 25x4 = 100 min Ans: 100 min

  15. Note: This question was provided by students who have sat for the 2009 examination. It may vary from the actual examination question.

  16. Thank You

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