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

Published byClara Carson Modified over 9 years ago

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Presentation on theme: "Proudly Presents Heuristics Approach Solving Challenging Primary Mathematical Problems Adrian Ng Principal Trainer."— Presentation transcript:

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

2 G UESS AND C HECK ( 3 GUESSES + L OOK FOR PATTERN ) 2

3 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 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 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 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.lengthno.length 045x0=0105105x24= 2520 2520-0= 2520

7 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.lengthno.length 045x0=0105105x24= 2520 2520-0= 2520 145x1=45106106x24= 2544 2544-45= 2499 2520-2499= 21 (pattern) ( 0 ) 2520-0= 2520 (gap) Gap/pattern 2520/21=120 +120 0+120= 120 120x45 =5400 120+105 = 225 225x24 =5400

8 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.lengthno.length 045x0=0105105x24= 2520 2520-0= 2520 145x1=45106106x24= 2544 2544-45= 2499 2520-2499= 21 (pattern) ( 0 ) 2520-0= 2520 (gap) Gap/pattern 2520/21=120 +120 0+120= 120 120x45 =5400 120+105 = 225 225x24 =5400 120+225=345Ans: 345 balloons

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

10 S YSTEMATIC LISTING 10

11 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 1 st 2 nd 3 rd 4 th 5 th 6 th 1 2 11 1 1 3 11 1 1 4 11 1 1 5 11 1 1 611 1 1 Group Friends 4 4 44 4 4 Total (each) 11 1 1

12 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 1 st 2 nd 3 rd 4 th 5 th 6 th 111 1 1 2 11 1 1 3 11 1 1 4 11 1 1 5 11 1 1 611 1 1 Group Friends 4 4 44 4 4 Total (each) 2.00pm to 4.30pm  150min

13 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 1 st 2 nd 3 rd 4 th 5 th 6 th 111 1 1 2 11 1 1 3 11 1 1 4 11 1 1 5 11 1 1 611 1 1 Group Friends 4 4 44 4 4 Total (each) 2.00pm to 4.30pm  150min 150min/6 = 25 min per group

14 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 1 st 2 nd 3 rd 4 th 5 th 6 th 111 1 1 2 11 1 1 3 11 1 1 4 11 1 1 5 11 1 1 611 1 1 Group Friends 4 4 44 4 4 Total (each) 2.00pm to 4.30pm  150min 150min/6 = 25 min per group 25x4 = 100 min Ans: 100 min

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

16 16 Thank You

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